Why study [functions](/page/Function) of a complex variable? The short answer is that complex differentiability is an absurdly powerful condition. In real analysis, knowing that a function is differentiable tells you almost nothing about its global behaviour: there exist real functions that are differentiable everywhere but increasing nowhere, or infinitely differentiable with Taylor [series](/page/Series) that converge to the wrong function. Complex analysis is a different world. A function that is complex-differentiable on an open set is automatically infinitely differentiable, equal to its own Taylor series, determined globally by its values on any tiny arc, and constrained by its [boundary](/page/Boundary) values through explicit integral formulas. The entire subject flows from one fact: the derivative of a complex function, if it exists, must be the same regardless of the direction of approach in the plane — and this single constraint, the Cauchy–Riemann equations, triggers a cascade of rigidity. This rigidity is not merely a curiosity. It is the reason that complex analysis is indispensable across mathematics and physics. Contour integration and the residue theorem provide the most efficient methods for evaluating definite integrals that arise in probability, number theory, and quantum mechanics. The Laplace and [Fourier transforms](/page/Fourier%20Transform) — whose inversion formulas are contour integrals — are the standard tools for solving linear differential equations in engineering and mathematical physics. Conformal mappings, which preserve angles because holomorphic functions locally act as rotation-dilations, solve boundary value problems for [Laplace's equation](/page/Laplace's%20Equation) in electrostatics, fluid dynamics, and heat conduction. The argument principle and Rouché's theorem give precise control over the location of zeros of polynomials and analytic functions, with applications to stability theory and control engineering. And the theory of uniform [limits](/page/Limit) of holomorphic functions underpins the rigorous foundations of special functions, from the Riemann zeta function to modular forms. The course develops in six parts. **§1 Complex [Differentiation](/page/Derivative)** establishes the foundations: the Cauchy–Riemann equations, conformality, [power series](/page/Power%20Series), and the connection to harmonic functions and potential theory. **§2 Contour Integration and Cauchy's Theorem** builds the integral theory — from the definition of contour integrals through the ML inequality and the Fundamental Theorem, to Cauchy's theorem in its full homotopy-invariant form, the Cauchy Integral Formula, and its consequences (Liouville, the Fundamental Theorem of Algebra, Taylor and Laurent series). **§3 Residue Calculus** develops the computational engine: residues, the residue theorem, techniques for real integrals (semicircular, indented, and keyhole contours), the argument principle, Rouché's theorem, and the open mapping theorem. **§4 Transform Theory** applies the machinery to the Fourier and Laplace transforms, the Bromwich inversion formula, and the solution of differential equations. **§5 Uniform Limits of Holomorphic Functions** establishes the remarkable closure property — locally uniform limits of holomorphic functions are holomorphic — and its applications to power series, normally convergent series, and limit interchange. **§6 Example Sheets** collects problems that synthesise the entire course. We assume familiarity with real analysis in $\mathbb{R}^2$ (limits, [continuity](/page/Continuity), real differentiability, partial derivatives, Riemann integration), linear algebra (matrices, eigenvalues), and basic point-set topology (open and [closed sets](/page/Closed%20Set), compactness, connectedness). No prior knowledge of complex analysis or algebraic topology is required. # Complex Differentiation At first glance, extending differentiation from $\mathbb{R}$ to $\mathbb{C}$ might seem routine — replace real limits with complex ones and proceed. Yet the requirement that the difference quotient $(f(z) - f(w))/(z - w)$ converges from *every* direction in the plane, not just from left and right, turns out to be extraordinarily restrictive. This single condition forces holomorphic functions to be infinitely differentiable, representable by convergent power series, and intimately connected to harmonic functions solving Laplace's equation. Complex differentiability reveals a hidden rigidity absent in real analysis, and this rigidity is the engine behind the powerful computational and theoretical tools of the subject. [motivation] ### Why Is Complex Differentiability So Much Stronger Than Real? In real analysis on $\mathbb{R}^2$, a function $F: \mathbb{R}^2 \to \mathbb{R}^2$ is differentiable at a point if it admits a linear approximation — its Jacobian can be any $2 \times 2$ matrix, giving four independent degrees of freedom. In complex analysis, the requirement that $f'(w) = \lim_{z \to w} (f(z) - f(w))/(z - w)$ exists as a single complex number means the linear approximation must be *multiplication by a complex number*. Writing $f'(w) = \alpha + i\beta$, the corresponding Jacobian is $\begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \end{pmatrix}$ — a rotation-dilation matrix with only *two* degrees of freedom. This halving of the degrees of freedom is the Cauchy–Riemann equations in disguise: the four partial derivatives $u_x, u_y, v_x, v_y$ are forced to satisfy two relations $u_x = v_y$ and $u_y = -v_x$. The geometric meaning is immediate. A general real-[linear map](/page/Linear%20Map) can shear, reflect, and stretch by different amounts in different directions. Multiplication by a nonzero complex number can only rotate and uniformly scale. This is why holomorphic maps with $f'(w) \neq 0$ are *conformal* — they preserve angles between curves — and why the theory of holomorphic functions is so much more rigid than real smooth function theory. ### The Cauchy–Riemann Equations as a PDE System The Cauchy–Riemann equations $u_x = v_y$, $u_y = -v_x$ are a first-order system of PDEs linking the real and imaginary parts of $f$. Differentiating once more and using symmetry of mixed partials gives $u_{xx} + u_{yy} = 0$ — Laplace's equation. So the real and imaginary parts of any holomorphic function are automatically *harmonic*. This connects complex analysis directly to potential theory: electrostatic potentials, steady-state heat [distributions](/page/Distribution), and ideal fluid flows are all governed by Laplace's equation, and [conformal maps](/page/Conformal%20Maps) provide the tool to transport solutions between domains. ### Holomorphic vs. Complex Differentiable A function can be complex differentiable at an isolated point without any of the powerful consequences above holding. The full theory — power series representations, Cauchy's integral formula, the identity theorem — requires differentiability in a *neighbourhood*. This motivates the distinction between complex differentiability at a point and holomorphicity (differentiability on an open set), which is the condition that unlocks the theory's full power. [/motivation] ## Foundations ### Domains [definition: Domain] A **domain** is a non-empty, open, path-connected subset $U \subseteq \mathbb{C}$. [/definition] Openness ensures that limits can be taken from all directions — the difference quotient $\frac{f(z) - f(w)}{z - w}$ requires $z$ to approach $w$ through a two-dimensional neighbourhood, not along a curve. Path-connectedness prevents the trivial pathology of a function with zero derivative being non-constant (on separate components, the function could take different constant values). We do *not* require simple connectedness at this stage; that stronger [topological](/page/Topology) condition becomes crucial for the existence of antiderivatives and logarithms. ### Complex Differentiability and Holomorphicity [definition: Complex Differentiability at a Point] Let $U \subseteq \mathbb{C}$ be open and $f: U \to \mathbb{C}$. The function $f$ is **complex differentiable at $w \in U$** if the limit \begin{align*} f'(w) = \lim_{z \to w} \frac{f(z) - f(w)}{z - w} \end{align*} exists as a complex number. The limit is taken over all $z \to w$ in $\mathbb{C}$ (not just along the real axis). [/definition] The crucial feature is that $z$ approaches $w$ from *every direction* in the plane. Along the real axis ($z = w + h$, $h \in \mathbb{R}$), the quotient reduces to a real derivative; along the imaginary axis ($z = w + ik$, $k \in \mathbb{R}$), it probes a different directional derivative. Requiring a single limit from all directions simultaneously is what produces the Cauchy–Riemann constraint. [definition: Holomorphic Function] A function $f: U \to \mathbb{C}$ is **holomorphic at $w \in U$** if it is complex differentiable on some open neighbourhood $B(w, \varepsilon) \subseteq U$. It is **holomorphic on $U$** if it is holomorphic at every point of $U$. [/definition] The distinction matters: complex differentiability at a single point is a local condition with limited consequences, while holomorphicity (differentiability on an open set) is what unlocks power series representations, Cauchy's integral formula, and the full machinery of the theory. A function can be complex differentiable at a point without being holomorphic there — for instance, $f(z) = |z|^2$ is complex differentiable only at $z = 0$ and holomorphic nowhere. [definition: Entire Function] A function $f: \mathbb{C} \to \mathbb{C}$ is **entire** if it is holomorphic on all of $\mathbb{C}$. [/definition] The polynomials, $e^z$, $\sin z$, $\cos z$ are all entire. Rational functions are holomorphic on their domains (the complement of their poles). The function $1/z$ is holomorphic on $\mathbb{C}^* = \mathbb{C} \setminus \{0\}$ but not entire. ### The Cauchy–Riemann Equations [definition: Cauchy–Riemann Equations] Let $f = u + iv$ be defined on an open set $U \subseteq \mathbb{C}$, with $u, v: U \to \mathbb{R}$. At a point $w = c + id \in U$, the **Cauchy–Riemann equations** are: \begin{align*} u_x(c, d) &= v_y(c, d), \\ u_y(c, d) &= -v_x(c, d). \end{align*} [/definition] In matrix form, the Cauchy–Riemann equations say the Jacobian $J_f = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}$ has the special structure $\begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \end{pmatrix}$, which is precisely the matrix representing multiplication by the complex number $\alpha + i\beta$. Matrices of this form commute with the rotation matrix $J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (representing multiplication by $i$), and the Cauchy–Riemann equations can be written compactly as $J_f J = J J_f$ — the Jacobian commutes with rotation by $90°$. ### Conformality [definition: Conformal Map] Let $f: U \to \mathbb{C}$ be holomorphic at $w \in U$. If $f'(w) \neq 0$, we say $f$ is **conformal at $w$**. A holomorphic map that is conformal at every point of $U$ is called a **conformal map** on $U$. [/definition] The condition $f'(w) \neq 0$ ensures the linear approximation $z \mapsto f(w) + f'(w)(z - w)$ is a non-degenerate rotation-dilation, which preserves angles. When $f'(w) = 0$, the map locally looks like $z \mapsto (z - w)^k$ for some $k \geq 2$, which multiplies angles by $k$. ## The Cauchy–Riemann Characterisation The central result of this section provides a concrete, computable criterion for complex differentiability using real partial derivatives. [quotetheorem:333] The forward direction extracts the Cauchy–Riemann equations by restricting the complex limit to the real and imaginary axes — two different paths that must give the same answer. Equating real and imaginary parts of the two limits yields $u_x = v_y$ and $u_y = -v_x$. For real differentiability, the existence of $f'(w)$ as a complex number means the best linear approximation to $f$ at $w$ is multiplication by $f'(w)$, which is manifestly a (real-)linear map $\mathbb{R}^2 \to \mathbb{R}^2$. The reverse direction is the more substantial claim: if $u$ and $v$ are real-differentiable and satisfy Cauchy–Riemann, the complex difference quotient has a limit. The proof assembles the real-linear remainder terms $o(|h|)$ from the two component functions, uses Cauchy–Riemann to factor the linear part as multiplication by $(u_x + iv_x)$, and divides by $h$. [citeproof:333] [example: Verifying Holomorphicity of $e^z$] Let $f(z) = e^z = e^{x+iy} = e^x \cos y + i e^x \sin y$, so $u = e^x \cos y$ and $v = e^x \sin y$. The partial derivatives are: \begin{align*} u_x = e^x \cos y, \quad u_y = -e^x \sin y, \quad v_x = e^x \sin y, \quad v_y = e^x \cos y. \end{align*} Then $u_x = e^x \cos y = v_y$ and $u_y = -e^x \sin y = -v_x$, so the Cauchy–Riemann equations hold everywhere. Since $u$ and $v$ are smooth (hence real-differentiable), the [Cauchy–Riemann characterisation](/theorems/333) gives $f'(z) = u_x + iv_x = e^x \cos y + ie^x \sin y = e^z$. So $e^z$ is entire with $(e^z)' = e^z$. [/example] [example: $f(z) = \bar{z}$ Is Nowhere Holomorphic] Let $f(z) = \bar{z} = x - iy$, so $u = x$ and $v = -y$. Then $u_x = 1$ and $v_y = -1$, so $u_x \neq v_y$ everywhere. The first Cauchy–Riemann equation fails at every point, so $f$ is nowhere complex differentiable. Geometrically, $z \mapsto \bar{z}$ is a reflection (orientation-reversing), not a rotation-dilation (orientation-preserving), so it cannot be the linear part of a holomorphic map. [/example] [example: Insufficiency of Cauchy–Riemann Without Real Differentiability] Define $f: \mathbb{C} \to \mathbb{C}$ by $f(0) = 0$ and \begin{align*} f(x + iy) = \frac{x^3 - y^3 + i(x^3 + y^3)}{x^2 + y^2} \quad \text{for } (x,y) \neq (0,0). \end{align*} At the origin, the partial derivatives computed along the axes give $u_x(0,0) = 1$, $v_y(0,0) = 1$, $u_y(0,0) = -1$, $v_x(0,0) = 1$, so the Cauchy–Riemann equations $u_x = v_y$ and $u_y = -v_x$ are satisfied. However, $f$ is not real-differentiable at the origin: along the line $y = x$, $f(t + it) = it$ while the linear approximation from the partial derivatives gives $(1+i)t + (-1+i)(it) = (1+i)t + (i + 1)t \cdot i$, which does not match. The limit $\lim_{z \to 0} f(z)/z$ does not exist (it takes different values along different rays), so $f$ is not complex differentiable at $0$. This shows that the Cauchy–Riemann equations alone, without real differentiability of $u$ and $v$, do not imply complex differentiability. [/example] ## Conformal Mappings The geometric meaning of the derivative is captured by the following theorem, which explains why holomorphic maps are central to applications in physics and engineering. [quotetheorem:334] The proof is a one-line application of the chain rule: $(f \circ \gamma_i)'(0) = f'(w) \cdot \gamma_i'(0)$, so multiplying a tangent vector by $f'(w)$ rotates it by $\arg f'(w)$ and scales it by $|f'(w)|$. Since both tangent vectors are rotated by the *same* angle $\arg f'(w)$, the angle between them is preserved. [citeproof:334] The condition $f'(w) \neq 0$ is essential. If $f'(w) = 0$, the local behavior of $f$ near $w$ is governed by the first nonzero term in the Taylor expansion: $f(z) \approx f(w) + c_k(z - w)^k$ for some $k \geq 2$, $c_k \neq 0$. The map $z \mapsto z^k$ multiplies angles by $k$, so a conformal map "folds" near a zero of the derivative. [example: The Joukowski Map] The map $f(z) = z + 1/z$ is holomorphic on $\mathbb{C}^*$ with $f'(z) = 1 - 1/z^2$, which vanishes at $z = \pm 1$. Away from these points, $f$ is conformal. The Joukowski map sends circles centred near the origin to curves resembling airfoil profiles — this is the classical application of conformal mapping in aerodynamics. Near $z = 1$, where $f'(1) = 0$ and $f''(1) = 2/1^3 = 2 \neq 0$, the map locally behaves like $z \mapsto z^2$, doubling angles. This creates the sharp trailing edge of the airfoil. [/example] [example: The Exponential Map is Locally Conformal] Since $(e^z)' = e^z \neq 0$ for all $z \in \mathbb{C}$, the exponential map is conformal everywhere. It maps vertical lines $x = c$ to circles $|w| = e^c$ and horizontal lines $y = d$ to rays $\arg w = d$. These families meet at right angles in both domain and range — a visible confirmation of angle preservation. The map is not globally injective (it has period $2\pi i$), but conformality is a local property and holds at every point. [/example] ## Power Series and Analyticity The following theorem establishes that convergent power series are holomorphic and can be differentiated term by term — the foundation for the equivalence of holomorphic and analytic functions. [quotetheorem:335] The key step is showing that the derived series $\sum n c_n (z-a)^{n-1}$ has the same [radius of convergence](/theorems/273) as the original — this follows from the Hadamard formula and the fact that $n^{1/n} \to 1$. The term-by-term differentiation is then verified by estimating the error in the difference quotient, which acquires an extra factor of $|z - w|$ from a double application of the geometric factorisation $(z-a)^n - (w-a)^n = (z-w) \sum_{k=0}^{n-1} (z-a)^k(w-a)^{n-1-k}$. [citeproof:335] This result, combined with Cauchy's integral formula (which shows that every holomorphic function has a local power series representation), establishes the fundamental identity: **holomorphic $\Leftrightarrow$ analytic** in complex analysis. This equivalence fails completely in real analysis — the function $e^{-1/x^2}$ is infinitely differentiable on $\mathbb{R}$ with Taylor series identically zero at $x = 0$, yet the function is not identically zero. [example: The Geometric Series and Its Derivatives] The geometric series $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$ converges on $B(0; 1)$. By [term-by-term differentiation](/theorems/335): \begin{align*} \frac{1}{(1-z)^2} = \sum_{n=1}^\infty n z^{n-1}, \qquad \frac{2}{(1-z)^3} = \sum_{n=2}^\infty n(n-1) z^{n-2}. \end{align*} Evaluating the first at $z = 1/2$: $\sum_{n=1}^\infty n/2^{n-1} = 1/(1 - 1/2)^2 = 4$. Each differentiated series has the same radius of convergence $R = 1$. [/example] ## Harmonic Functions and Potential Theory [quotetheorem:336] The proof is a direct computation: differentiate the Cauchy–Riemann equation $u_x = v_y$ with respect to $x$ and $u_y = -v_x$ with respect to $y$, then add. The cross-partials $v_{xy}$ and $v_{yx}$ cancel by equality of mixed partials (which holds because holomorphic functions are infinitely smooth), giving $u_{xx} + u_{yy} = 0$. [citeproof:336] This result has immediate physical applications. Steady-state heat flow, electrostatic potentials, and velocity potentials of irrotational incompressible fluid flow all satisfy Laplace's equation. Finding a holomorphic function whose real part matches given boundary conditions solves the corresponding physical problem. [example: Harmonic Functions from $f(z) = z^2$] For $f(z) = z^2 = (x+iy)^2 = (x^2 - y^2) + 2ixy$, we have $u = x^2 - y^2$ and $v = 2xy$. Check: $u_{xx} = 2$, $u_{yy} = -2$, so $u_{xx} + u_{yy} = 0$. Similarly $v_{xx} = 0$, $v_{yy} = 0$. Both are harmonic. The level curves $u = c$ (rectangular hyperbolae $x^2 - y^2 = c$) and $v = d$ (hyperbolae $xy = d/2$) form orthogonal families — this orthogonality is a general consequence of the Cauchy–Riemann equations and conformality. [/example] [example: Solving Laplace's Equation via Conformal Mapping] To solve $\nabla^2 \phi = 0$ on the first quadrant $Q = \{x > 0, y > 0\}$ with boundary conditions $\phi(x, 0) = 0$ for $x > 0$ and $\phi(0, y) = 1$ for $y > 0$: the map $f(z) = \log z$ sends $Q$ to the horizontal strip $\{0 < \operatorname{Im} w < \pi/2\}$. In the strip, the harmonic function $\Phi(w) = \frac{2}{\pi} \operatorname{Im} w$ satisfies the transferred boundary conditions. Pulling back: $\phi(x, y) = \Phi(f(z)) = \frac{2}{\pi} \operatorname{Im}(\log z) = \frac{2}{\pi} \arg(z) = \frac{2}{\pi} \arctan(y/x)$. This is the standard technique: conformally map to a simpler domain, solve there, and pull back. [/example] ## The Complex Logarithm The final result of this section connects complex analysis to topology. On $\mathbb{C}^*$, the equation $e^w = z$ has infinitely many solutions $w = \ln|z| + i(\arg z + 2\pi k)$, so $\log z$ is inherently multi-valued. The question is: on which domains can we choose a single-valued holomorphic branch? [quotetheorem:337] The proof constructs $\lambda(z) = \lambda_0 + \int_\gamma \frac{d\zeta}{\zeta}$ along a path from a basepoint $z_0$ to $z$. Simple connectedness ensures path-independence (by Cauchy's theorem for simply connected domains, since $1/\zeta$ is holomorphic on $D \subseteq \mathbb{C}^*$). Holomorphicity of $\lambda$ follows from differentiating under the integral, and the identity $e^{\lambda(z)} = z$ is verified by showing $z e^{-\lambda(z)}$ has zero derivative and equals $1$ at the basepoint. [citeproof:337] The theorem fails on $\mathbb{C}^*$ itself: going once around the origin changes $\int d\zeta/\zeta$ by $2\pi i$, so no single-valued branch exists on any domain containing a loop around $0$. This is the prototypical example of a topological obstruction in analysis — the fundamental group $\pi_1(\mathbb{C}^*) \cong \mathbb{Z}$ prevents global well-definedness. [example: The Principal Branch of the Logarithm] On the slit plane $D = \mathbb{C} \setminus (-\infty, 0]$, which is simply connected, the principal branch $\operatorname{Log} z = \ln|z| + i\operatorname{Arg} z$ (with $\operatorname{Arg} z \in (-\pi, \pi)$) is holomorphic with $(\operatorname{Log} z)' = 1/z$. The branch cut along the negative real axis removes the topological obstruction. On the slit plane $\mathbb{C} \setminus [0, \infty)$, a different branch (with argument in $(0, 2\pi)$) is holomorphic. No single branch works on all of $\mathbb{C}^*$. [/example] [example: Holomorphic Powers via the Logarithm] For $\alpha \in \mathbb{C}$ and $z \in D = \mathbb{C} \setminus (-\infty, 0]$, define $z^\alpha = e^{\alpha \operatorname{Log} z}$. This is holomorphic on $D$ (as a composition of holomorphic functions) with derivative: \begin{align*} \frac{d}{dz} z^\alpha = e^{\alpha \operatorname{Log} z} \cdot \frac{\alpha}{z} = \alpha z^{\alpha - 1}. \end{align*} The familiar power rule extends to complex exponents — but only on a domain where a branch of the logarithm exists. For $\alpha \notin \mathbb{Z}$, the function $z^\alpha$ is genuinely multi-valued on $\mathbb{C}^*$, and the branch cut is essential. [/example] ## Worked Example [problem] Let $f(z) = \frac{z}{z^2 + 1}$. Determine where $f$ is holomorphic, compute $f'(z)$, verify the Cauchy–Riemann equations at $z = 1$, and determine where $f$ is conformal. [/problem] [solution] **Step 1: Domain of holomorphicity.** The function $f(z) = z/(z^2 + 1)$ is a rational function, holomorphic wherever the denominator is nonzero. Since $z^2 + 1 = 0 \iff z = \pm i$, $f$ is holomorphic on $\mathbb{C} \setminus \{i, -i\}$. **Step 2: Derivative by the quotient rule.** \begin{align*} f'(z) = \frac{(z^2 + 1) \cdot 1 - z \cdot 2z}{(z^2 + 1)^2} = \frac{1 - z^2}{(z^2 + 1)^2}. \end{align*} **Step 3: Verification of Cauchy–Riemann at $z = 1$.** At $z = 1$: $f(1) = 1/2$ and $f'(1) = (1-1)/(1+1)^2 = 0$. To verify directly, write $z = x + iy$ near $(1, 0)$: \begin{align*} f(x + iy) = \frac{x + iy}{(x+iy)^2 + 1} = \frac{x + iy}{x^2 - y^2 + 1 + 2ixy}. \end{align*} Rationalising by multiplying by the conjugate of the denominator $D = (x^2 - y^2 + 1) + 2ixy$: \begin{align*} u + iv = \frac{(x+iy)(\overline{D})}{|D|^2} = \frac{(x+iy)((x^2 - y^2 + 1) - 2ixy)}{(x^2 - y^2 + 1)^2 + 4x^2y^2}. \end{align*} At $(1, 0)$: $D = 2$, so $u = x(x^2 - y^2 + 1)/(|D|^2)$ and computing partial derivatives gives $u_x(1,0) = 0$, $v_y(1,0) = 0$, $u_y(1,0) = 0$, $v_x(1,0) = 0$. The Cauchy–Riemann equations $u_x = v_y = 0$ and $u_y = -v_x = 0$ hold, consistent with $f'(1) = 0 = u_x + iv_x$. **Step 4: Conformality.** By the [angle-preservation theorem](/theorems/334), $f$ is conformal wherever $f'(z) \neq 0$. From Step 2, $f'(z) = 0 \iff z^2 = 1 \iff z = \pm 1$. Therefore $f$ is conformal on $\mathbb{C} \setminus \{i, -i, 1, -1\}$. At $z = \pm 1$, the derivative vanishes and angles are doubled (since these are simple zeros of $f'$, corresponding to $f$ locally behaving like a quadratic). [/solution] ## References 1. Sheratt, N., *Cambridge Part IB — Complex Analysis*, Lecture Notes. # Contour Integration and Cauchy's Theorem Having established the foundations of complex differentiation, we now turn to integration — and here the subject takes a dramatic turn away from real analysis. In the real setting, the [Fundamental Theorem of Calculus](/theorems/632) says $\int_a^b f'(x) \, dx = f(b) - f(a)$: the integral depends only on the endpoints. In the complex plane, this path-independence can *fail* — the integral $\int_\gamma f(z) \, dz$ along a closed path need not be zero, and the value can depend on which path is taken between two points. The central question of this section is: *when* does a contour integral depend only on the endpoints, and *when* does it vanish on closed curves? The answer involves a remarkable interplay between analysis and topology. The analytic ingredient is Cauchy's theorem: holomorphic functions integrate to zero over triangles. The topological ingredient is *homotopy*: two paths give the same integral if and only if one can be continuously deformed into the other within the domain. The synthesis is that on simply connected domains (those with no "holes"), every closed contour integral of a holomorphic function vanishes — and on domains with holes, the integral depends only on how the contour winds around the missing points. This section develops the full chain: the definition of contour integrals, the ML inequality, the Fundamental Theorem of Contour Integration, Cauchy's theorem (from triangles through star-shaped domains to the general homotopy version), the Cauchy Integral Formula, and its consequences including Liouville's theorem, the Fundamental Theorem of Algebra, Taylor and Laurent series, and Morera's theorem. [motivation] ### Why Does Integration in $\mathbb{C}$ Involve Topology? In real analysis on $\mathbb{R}$, there is essentially only one path from $a$ to $b$: the interval $[a, b]$. In $\mathbb{C} \cong \mathbb{R}^2$, there are infinitely many paths between two points, and they can wind around obstacles in topologically distinct ways. The simplest example is $f(z) = 1/z$ on $\mathbb{C}^* = \mathbb{C} \setminus \{0\}$: integrating around a circle centred at the origin gives $\int_{|z|=1} dz/z = 2\pi i \neq 0$, but integrating around a circle that does *not* enclose the origin gives $0$ (since $1/z$ has an antiderivative on any simply connected subdomain of $\mathbb{C}^*$). The integral "detects" whether the path goes around the puncture. This phenomenon has no real-variable analogue and is fundamentally topological: what matters is the *homotopy class* of the path (which paths it can be deformed into without leaving the domain), not the specific geometric shape. Making this precise requires the notion of homotopy from algebraic topology, and the Homotopy Invariance Theorem is the result that legitimises every "deform the contour" argument in the subject. ### The Logical Chain: Triangles → Stars → Homotopy → Simply Connected The development is not a single theorem but a chain of progressively stronger results: 1. **Cauchy for triangles** (the analytic core): uses only holomorphicity and the bisection argument, requires no topology. 2. **Cauchy for star-shaped domains**: builds an antiderivative using the triangle result; the star-shape ensures every point is connected to a centre by a line segment. 3. **Homotopy invariance**: the topological result that contour integrals are unchanged under continuous deformation of the path (the proof tiles the homotopy rectangle with small cells and applies the star-shaped result to each). 4. **Cauchy for simply connected domains**: an immediate corollary — in a simply connected domain, every closed loop is null-homotopic, so the integral vanishes. This chain makes clear *where* the topology enters (step 3) and *what it buys* (step 4 subsumes all earlier versions as special cases). The common textbook shortcut of proving only the convex case obscures this structure and leaves students unable to justify contour deformations in multiply connected domains. [/motivation] ## Paths, Contours, and Integrals ### Paths and Contours [definition: Path] A **path** in $\mathbb{C}$ is a continuous function $\gamma: [a, b] \to \mathbb{C}$, where $a < b$. The point $\gamma(a)$ is the **start point** and $\gamma(b)$ is the **end point**. The **image** (or **trace**) of $\gamma$ is the [set](/page/Set) $\gamma^* = \gamma([a, b]) \subseteq \mathbb{C}$. [/definition] [definition: Closed Path] A path $\gamma: [a, b] \to \mathbb{C}$ is **closed** if $\gamma(a) = \gamma(b)$. [/definition] [definition: Simple Path] A path $\gamma: [a, b] \to \mathbb{C}$ is **simple** if $\gamma(t_1) = \gamma(t_2)$ implies $t_1 = t_2$ or $\{t_1, t_2\} = \{a, b\}$. A simple closed path does not cross itself. [/definition] [definition: Piecewise $C^1$ Path] A path $\gamma: [a, b] \to \mathbb{C}$ is **piecewise $C^1$** if there exists a partition $a = t_0 < t_1 < \cdots < t_n = b$ such that $\gamma$ is continuously differentiable on each open interval $(t_{k-1}, t_k)$, and the one-sided derivatives exist and are finite at each $t_k$. [/definition] [definition: Contour] A **contour** is a piecewise $C^1$ simple closed path. By convention, contours are oriented counter-clockwise unless stated otherwise. [/definition] [definition: Reverse Path and Path Concatenation] Given a path $\gamma: [a, b] \to \mathbb{C}$, the **reverse path** is $\overline{\gamma}(t) = \gamma(a + b - t)$, traversing the same image in the opposite direction. If $\gamma_1: [a, b] \to \mathbb{C}$ and $\gamma_2: [b, c] \to \mathbb{C}$ satisfy $\gamma_1(b) = \gamma_2(b)$, the **concatenation** $\gamma_1 * \gamma_2: [a, c] \to \mathbb{C}$ follows $\gamma_1$ then $\gamma_2$. [/definition] Reversal negates the integral ($\int_{\overline{\gamma}} f = -\int_\gamma f$), and concatenation adds them ($\int_{\gamma_1 * \gamma_2} f = \int_{\gamma_1} f + \int_{\gamma_2} f$). These algebraic properties are what make contour deformation arguments work: replacing one path by another amounts to adding and subtracting integrals over auxiliary paths. ### The Contour Integral [definition: Contour Integral] Let $U \subseteq \mathbb{C}$ be open, $f: U \to \mathbb{C}$ continuous, and $\gamma: [a, b] \to U$ piecewise $C^1$. The **integral of $f$ along $\gamma$** is: \begin{align*} \int_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \, \gamma'(t) \, dt, \end{align*} where the right-hand side is a [Riemann integral](/page/Riemann%20Integral) of the complex-valued function $t \mapsto f(\gamma(t)) \gamma'(t)$, computed via real and imaginary parts. [/definition] The integral depends on the *oriented* path $\gamma$, not just its image: reversing the orientation negates the integral, and reparametrising (changing the speed but not the direction) leaves it unchanged. This parametrisation-invariance is what makes the integral a geometric quantity rather than a computational artifact. [definition: Length of a Path] The **length** of a piecewise $C^1$ path $\gamma: [a,b] \to \mathbb{C}$ is $\ell(\gamma) = \int_a^b |\gamma'(t)| \, dt$. [/definition] ### Antiderivatives and Star-Shaped Domains [definition: Antiderivative] Let $U \subseteq \mathbb{C}$ be open and $f: U \to \mathbb{C}$ continuous. An **antiderivative** of $f$ on $U$ is a holomorphic function $F: U \to \mathbb{C}$ with $F'(z) = f(z)$ for all $z \in U$. [/definition] [definition: Star-Shaped Domain] A domain $U \subseteq \mathbb{C}$ is **star-shaped** with **star centre** $a_0 \in U$ if for every $w \in U$, the line segment $[a_0, w] = \{(1-t)a_0 + tw : t \in [0,1]\}$ lies in $U$. Every convex domain is star-shaped (with any point as centre). [/definition] ## The ML Inequality The fundamental estimate for contour integrals: [quotetheorem:338] The proof applies the standard modulus estimate $|\int g| \leq \int |g|$ to the pullback $f(\gamma(t))\gamma'(t)$, then bounds $|f|$ by its supremum on the path. [citeproof:338] [example: Bounding an Integral Over a Semicircle] Let $\gamma_R$ be the upper semicircle $\gamma_R(\theta) = Re^{i\theta}$, $\theta \in [0, \pi]$. For $f(z) = 1/(z^2 + 1)$ with $R > 1$: \begin{align*} |f(Re^{i\theta})| = \frac{1}{|R^2 e^{2i\theta} + 1|} \leq \frac{1}{R^2 - 1}, \end{align*} and $\ell(\gamma_R) = \pi R$. The [ML inequality](/theorems/338) gives $|\int_{\gamma_R} f \, dz| \leq \frac{\pi R}{R^2 - 1} \to 0$ as $R \to \infty$. This "vanishing semicircle" estimate is the engine behind the evaluation of real improper integrals by contour methods. [/example] Note that Cauchy's estimate $|f^{(n)}(a)| \leq n!M/R^n$ (which appears below as a consequence of the Cauchy Integral Formula) is itself an application of the ML inequality to the derivative formula $f^{(n)}(a) = \frac{n!}{2\pi i}\int_{\partial B} f(w)/(w-a)^{n+1} \, dw$: bound the integrand by $M/R^{n+1}$, multiply by the length $2\pi R$, and divide by $2\pi$. This single estimate powers both Liouville's theorem and the growth characterisation of polynomials. ## The Fundamental Theorem and Antiderivatives [quotetheorem:339] The proof is a direct application of the chain rule and the real Fundamental Theorem of Calculus to $t \mapsto F(\gamma(t))$. [citeproof:339] The key corollary: *if an antiderivative exists, the integral depends only on the endpoints* — and in particular vanishes on closed paths. The converse is also true: [quotetheorem:340] The forward direction is immediate from the Fundamental Theorem. The reverse direction constructs the antiderivative by integrating from a fixed basepoint: $F(z) = \int_\sigma f \, dw$, where $\sigma$ is any path from $a$ to $z$. The hypothesis that all closed-path integrals vanish guarantees path-independence (since two paths from $a$ to $z$ form a closed loop), and differentiating $F$ recovers $f$. [citeproof:340] [example: $1/z$ Has No Antiderivative on $\mathbb{C}^*$] The function $f(z) = 1/z$ is holomorphic on $\mathbb{C}^* = \mathbb{C} \setminus \{0\}$, but $\int_{|z|=1} dz/z = 2\pi i \neq 0$. By the [Antiderivative Existence Characterisation](/theorems/340), $1/z$ has no antiderivative on all of $\mathbb{C}^*$. However, on any simply connected subdomain of $\mathbb{C}^*$ (such as the slit plane $\mathbb{C} \setminus (-\infty, 0]$), the principal logarithm $\operatorname{Log} z$ serves as an antiderivative. [/example] [example: Vanishing Integrals for $z^n$] For $n \in \mathbb{Z} \setminus \{-1\}$, the function $z^n$ has the antiderivative $F(z) = z^{n+1}/(n+1)$ on $\mathbb{C}^*$ (or on $\mathbb{C}$ if $n \geq 0$). For any closed path $\gamma$ in the domain: \begin{align*} \int_\gamma z^n \, dz = F(\gamma(b)) - F(\gamma(a)) = 0. \end{align*} The exceptional case $n = -1$ is the only integer power whose integral over a closed loop can be nonzero — a fact that will be explained by the residue theorem. [/example] ## Cauchy's Theorem: From Triangles to Star-Shaped Domains ### The Triangle Theorem (Goursat) The analytic core of Cauchy's theorem uses no topology — only holomorphicity and a bisection argument: [quotetheorem:341] The proof is one of the most elegant in analysis. Bisect the triangle repeatedly, choosing at each step the sub-triangle carrying at least $1/4$ of the total integral (the four sub-triangle integrals sum to the original because interior edges cancel). After $n$ steps, the selected triangle has shrunk to a point $z_0$ where holomorphicity provides a linear approximation $f(z) \approx f(z_0) + f'(z_0)(z - z_0)$. The linear and constant parts integrate to zero (they have antiderivatives), and the error term is controlled by the [ML inequality](/theorems/338): the integral is at most $\varepsilon \cdot \operatorname{diam}(T) \cdot \ell(\partial T) / 4^n$ on the $n$-th sub-triangle, while the lower bound is $|I|/4^n$. Cancelling $4^n$ and sending $\varepsilon \to 0$ forces $I = 0$. [citeproof:341] Note what the proof uses: only the existence of $f'(z_0)$ at one point (the limit of the bisection), not differentiability in a neighbourhood. This is why the result is sometimes called Goursat's theorem — Cauchy's original proof assumed continuous differentiability. ### The Star-Shaped Case [quotetheorem:342] The proof builds an antiderivative explicitly: define $F(z) = \int_{[a_0, z]} f \, dw$ by integrating along the straight segment from the star centre $a_0$ to $z$ (which lies in $U$ by star-shapedness). The triangle theorem shows that $F(z+h) - F(z) = \int_{[z, z+h]} f \, dw$ (the three segments $[a_0, z]$, $[z, z+h]$, $[z+h, a_0]$ form a triangle), and differentiating gives $F' = f$. The [Fundamental Theorem of Contour Integration](/theorems/339) then gives vanishing of all closed-path integrals. [citeproof:342] This result covers all convex domains (since convex implies star-shaped), but it does *not* cover domains with holes — the slit plane $\mathbb{C} \setminus (-\infty, 0]$ is simply connected but not star-shaped, and the punctured plane $\mathbb{C}^*$ is neither. To handle these domains, we need topology. ## Homotopy and Contour Deformation ### The Topological Framework The star-shaped Cauchy theorem raises a natural question: if two paths share the same endpoints and can be "continuously deformed" into each other within the domain $U$, must they give the same integral? The answer is yes, and making this precise requires the notion of *homotopy* from algebraic topology. [definition: Homotopy of Paths (Fixed Endpoints)] Let $U \subseteq \mathbb{C}$ be open, and let $\gamma_0, \gamma_1: [0, 1] \to U$ be paths with $\gamma_0(0) = \gamma_1(0) = a$ and $\gamma_0(1) = \gamma_1(1) = b$. A **homotopy with fixed endpoints** from $\gamma_0$ to $\gamma_1$ in $U$ is a continuous map $H: [0, 1] \times [0, 1] \to U$ satisfying: \begin{align*} H(t, 0) &= \gamma_0(t) \quad \text{for all } t \in [0,1], \\ H(t, 1) &= \gamma_1(t) \quad \text{for all } t \in [0,1], \\ H(0, s) &= a \quad \text{for all } s \in [0,1], \\ H(1, s) &= b \quad \text{for all } s \in [0,1]. \end{align*} If such an $H$ exists, we say $\gamma_0$ and $\gamma_1$ are **homotopic with fixed endpoints** in $U$, written $\gamma_0 \simeq \gamma_1 \operatorname{rel} \{0, 1\}$. [/definition] The map $H$ should be thought of as a "movie" deforming $\gamma_0$ into $\gamma_1$: at each "time" $s \in [0, 1]$, $H(\cdot, s)$ is an intermediate path from $a$ to $b$, starting at $\gamma_0$ (when $s = 0$) and ending at $\gamma_1$ (when $s = 1$). The conditions $H(0, s) = a$ and $H(1, s) = b$ ensure the endpoints stay fixed throughout the deformation. Crucially, *every intermediate path must lie entirely in $U$* — the deformation cannot pass through holes in the domain. [definition: Free Homotopy of Closed Paths] Let $\gamma_0, \gamma_1: [0, 1] \to U$ be closed paths. A **free homotopy** from $\gamma_0$ to $\gamma_1$ in $U$ is a continuous map $H: [0, 1] \times [0, 1] \to U$ with $H(t, 0) = \gamma_0(t)$, $H(t, 1) = \gamma_1(t)$, and $H(0, s) = H(1, s)$ for all $s$ (the intermediate curves are closed, but the basepoint may move). [/definition] [definition: Null-Homotopic Path] A closed path $\gamma$ in $U$ is **null-homotopic** (or **contractible**) if it is freely homotopic to a constant path in $U$ — i.e., there exists a continuous $H: [0,1]^2 \to U$ deforming $\gamma$ to a single point without leaving $U$. [/definition] [definition: Simply Connected Domain] A domain $U \subseteq \mathbb{C}$ is **simply connected** if every closed path in $U$ is null-homotopic in $U$. Equivalently (in the plane), $U$ is simply connected if it has no "holes": $\mathbb{C} \setminus U$ is connected. [/definition] The equivalence of these two definitions (topological: all loops contractible; planar: complement connected) is a non-trivial result from algebraic topology. Convex domains, star-shaped domains, and the slit plane $\mathbb{C} \setminus (-\infty, 0]$ are all simply connected. The punctured plane $\mathbb{C}^*$ and annuli $\{r < |z| < R\}$ are *not* simply connected — the loop $\gamma(\theta) = e^{i\theta}$ winds around the missing point and cannot be contracted to a point. ### The Homotopy Invariance Theorem [quotetheorem:343] This is the theorem that justifies *every* contour deformation argument in complex analysis. Whenever a textbook or lecture says "deform the contour from $\gamma_0$ to $\gamma_1$," the rigorous justification is: (1) construct a homotopy $H$ from $\gamma_0$ to $\gamma_1$ in the domain where $f$ is holomorphic, (2) invoke homotopy invariance to conclude the integrals are equal. The proof (which we record but do not develop in full here) proceeds as follows. The continuous image $H([0,1]^2)$ is a compact subset of the open set $U$, so it has positive distance to $\mathbb{C} \setminus U$. Subdivide $[0,1]^2$ into a fine grid of small rectangles, each mapping into a disc $B \subseteq U$ (possible by uniform continuity of $H$). On each small disc, $f$ has an antiderivative (since discs are convex), and the integral around the boundary of each rectangle vanishes by the [star-shaped Cauchy theorem](/theorems/342). Summing over all rectangles, interior edges cancel in pairs (they are traversed in opposite directions by adjacent cells), and the surviving boundary edges telescopically connect $\gamma_0$ to $\gamma_1$. The conclusion is $\int_{\gamma_0} f = \int_{\gamma_1} f$. The free-homotopy version for closed curves follows by a minor adaptation: the boundary conditions $H(0, s) = H(1, s)$ (rather than fixed endpoints) create a cylindrical rather than rectangular boundary, but the telescoping argument is identical. [example: Deforming a Square to a Circle] Let $f(z) = 1/(z - 2)$, holomorphic on $\mathbb{C} \setminus \{2\}$. Let $\gamma_0$ be the unit circle $|z| = 1$ (traversed counter-clockwise) and $\gamma_1$ the boundary of the square $[-1, 1]^2$. Both enclose the same region and neither encloses $z = 2$. The paths are freely homotopic in $\mathbb{C} \setminus \{2\}$ (deform one to the other through intermediate curves staying away from $z = 2$). By [homotopy invariance](/theorems/343), $\int_{\gamma_0} f \, dz = \int_{\gamma_1} f \, dz$. In fact, both integrals equal $0$ since the loop is null-homotopic in $\mathbb{C} \setminus \{2\}$ (neither encloses the puncture). [/example] [example: Non-Homotopic Paths Give Different Integrals] Let $f(z) = 1/z$ on $\mathbb{C}^*$. The circle $\gamma_1(\theta) = e^{i\theta}$ winds once around the origin; the constant path $\gamma_0 \equiv 1$ does not wind at all. These paths are *not* homotopic in $\mathbb{C}^*$ (any deformation would have to cross $z = 0$, which is not in the domain). Indeed, $\int_{\gamma_1} dz/z = 2\pi i$ while $\int_{\gamma_0} dz/z = 0$. The circle $\gamma_2(\theta) = e^{2i\theta}$ (winding twice) gives $\int_{\gamma_2} dz/z = 4\pi i$ — a different homotopy class, a different integral. [/example] ### Cauchy's Theorem for Simply Connected Domains [quotetheorem:344] The proof is immediate from homotopy invariance: in a simply connected domain, every closed loop is null-homotopic, so its integral equals that of a constant loop, which is $0$. The [Antiderivative Existence Characterisation](/theorems/340) then provides a global antiderivative. [citeproof:344] This is the strongest general form of Cauchy's theorem. It subsumes the star-shaped case (every star-shaped domain is simply connected) and the convex case (every convex domain is star-shaped). More importantly, it applies to domains like the slit plane $\mathbb{C} \setminus (-\infty, 0]$ that are simply connected but not star-shaped, and it *correctly fails* on non-simply-connected domains like $\mathbb{C}^*$. ## The Cauchy Integral Formula The most important single result in complex analysis: [quotetheorem:345] The Cauchy Integral Formula says that the values of a holomorphic function inside a disc are completely determined by its values on the boundary circle — an extraordinary rigidity with no real analogue. No real $C^\infty$ function is determined by its values on a circle. The proof defines $g(w) = (f(w) - f(z))/(w - z)$ for $w \neq z$ and $g(z) = f'(z)$. This function is continuous on the whole disc (continuity at $w = z$ is precisely the statement that $f'(z)$ exists) and holomorphic on the punctured disc $B(z_0, r) \setminus \{z\}$. The key step is showing $\int_{\partial T} g = 0$ for every triangle $T$ in the disc, including those containing $z$. For triangles avoiding $z$, this is Goursat's theorem. For triangles containing $z$, the same bisection argument works: the nested sub-triangles shrink to a limit point, and if that limit point is $z$, then *continuity* of $g$ at $z$ provides the bound needed to kill the integral via the ML inequality (no appeal to holomorphicity at $z$ is needed — continuity suffices). With all triangle integrals vanishing, the [star-shaped Cauchy theorem](/theorems/342) gives $\int_{\partial B} g = 0$, which expands to $\int_{\partial B} f(w)/(w-z) \, dw = f(z) \int_{\partial B} dw/(w-z) = 2\pi i \cdot f(z)$. [citeproof:345] The formula for derivatives $f^{(n)}(z) = \frac{n!}{2\pi i} \int_{\partial B} \frac{f(w)}{(w-z)^{n+1}} \, dw$ is obtained by differentiating under the integral sign (justified by [uniform convergence](/page/Uniform%20Convergence) of the difference quotients). An important corollary is **Cauchy's estimate**: for $f$ holomorphic on $B(a, R)$ with $|f| \leq M$ on $\partial B(a, R)$, \begin{align*} |f^{(n)}(a)| \leq \frac{n! \, M}{R^n}. \end{align*} [example: Computing an Integral via CIF] Compute $\int_{|z| = 2} \frac{e^z}{z - 1} \, dz$. The function $f(z) = e^z$ is entire, and $z = 1$ lies inside $|z| = 2$. By the [Cauchy Integral Formula](/theorems/345): \begin{align*} \int_{|z| = 2} \frac{e^z}{z - 1} \, dz = 2\pi i \cdot f(1) = 2\pi i \cdot e. \end{align*} [/example] [example: Computing a Derivative via CIF] Compute $\int_{|z| = 1} \frac{\sin z}{z^3} \, dz$. Write this as $\int_{|z|=1} \frac{f(z)}{(z - 0)^{2+1}} \, dz$ with $f(z) = \sin z$ and $n = 2$. By the derivative formula: \begin{align*} \int_{|z| = 1} \frac{\sin z}{z^3} \, dz = \frac{2\pi i}{2!} f''(0) = \pi i \cdot (-\sin 0) = 0. \end{align*} [/example] ## Consequences of the Cauchy Integral Formula ### Liouville's Theorem [quotetheorem:346] The proof is a one-line application of Cauchy's estimate: $|f'(z)| \leq M/R$ for any $R > 0$, so $f' \equiv 0$. [citeproof:346] This result has no real analogue: the function $\sin x$ is bounded and infinitely differentiable on $\mathbb{R}$ but non-constant. The rigidity of holomorphic functions — captured by the Cauchy integral representation — is what makes Liouville's theorem work. ### The Fundamental Theorem of Algebra [quotetheorem:347] The proof is by contradiction: if $p$ has no roots, then $1/p$ is entire and bounded (since $|p(z)| \to \infty$ ensures $1/p$ is small at infinity, and $1/p$ is continuous hence bounded on any compact set). [Liouville's theorem](/theorems/346) forces $1/p$ constant, contradicting $\deg p \geq 1$. [citeproof:347] That the algebraic closure of $\mathbb{C}$ — an algebraic statement — is proved using complex analysis is one of the most striking applications of the theory. ### Taylor's Theorem [quotetheorem:348] The proof expands the Cauchy kernel $1/(w - z)$ as a geometric series $\sum (z-a)^n/(w-a)^{n+1}$, which converges uniformly for $|w - a| = \rho$ and $|z - a| < \rho$, and interchanges sum and integral. [citeproof:348] Combined with the [Holomorphicity of Power Series](/theorems/335) from the previous section, this establishes the fundamental equivalence: **a function is holomorphic on an open set if and only if it is locally representable by a convergent power series**. In complex analysis, holomorphic = analytic. ### Morera's Theorem [quotetheorem:349] The proof is a partial converse to Cauchy's theorem: if all triangle integrals vanish, construct a local antiderivative $F$ (by integrating from a basepoint in a small disc). Then $F$ is holomorphic with $F' = f$, and since holomorphic functions are infinitely differentiable, $f = F'$ is itself holomorphic. [citeproof:349] Morera's theorem is the standard tool for proving that limits of holomorphic functions are holomorphic: if $f_n \to f$ uniformly on compact subsets and each $f_n$ is holomorphic, one shows $\int_{\partial T} f = \lim \int_{\partial T} f_n = 0$ for all triangles, and Morera gives holomorphicity of $f$. ## Singularities and Laurent Series ### Classification of Isolated Singularities When a function is holomorphic on a punctured neighbourhood $B(a, r) \setminus \{a\}$ but not at $a$ itself, the nature of the singularity is classified by the behaviour of $f$ near $a$. Taylor series represent holomorphic functions on discs, but many important functions — like $1/z$, $\cot z$, or $e^{1/z}$ — are only holomorphic on punctured discs or annuli. To represent such functions by series, we need to allow negative powers of $(z - a)$. This motivates the Laurent series. [definition: Isolated Singularity] Let $U \subseteq \mathbb{C}$ be open, $a \in U$, and $f: U \setminus \{a\} \to \mathbb{C}$ holomorphic. Then $a$ is an **isolated singularity** of $f$. [/definition] [definition: Removable Singularity] An isolated singularity $a$ of $f$ is **removable** if $f$ is bounded in some punctured neighbourhood $B(a, r) \setminus \{a\}$. Equivalently (by Riemann's removable singularity theorem), $f$ extends to a holomorphic function on $B(a, r)$. [/definition] [definition: Pole] An isolated singularity $a$ is a **pole of order $k \geq 1$** if $(z - a)^k f(z)$ extends to a holomorphic function $g$ on $B(a, r)$ with $g(a) \neq 0$. Equivalently, $f(z) = g(z)/(z-a)^k$ near $a$ with $g$ holomorphic and $g(a) \neq 0$. A pole of order $1$ is called **simple**. [/definition] [definition: Essential Singularity] An isolated singularity that is neither removable nor a pole is an **essential singularity**. By the Casorati–Weierstrass theorem, the image of any punctured neighbourhood under $f$ is dense in $\mathbb{C}$. [/definition] ### The Laurent Series [quotetheorem:350] The Laurent series generalises the Taylor series to annular domains, allowing negative powers of $(z - a)$. The **principal part** $\sum_{n < 0} c_n(z - a)^n$ encodes the singularity: it vanishes for removable singularities, is a finite sum for poles (the order of the pole is the largest $|n|$ with $c_n \neq 0$), and is an infinite series for essential singularities. [example: Laurent Series of $e^{1/z}$] The function $e^{1/z} = \sum_{n=0}^\infty \frac{1}{n! \, z^n}$ has infinitely many nonzero negative-power coefficients ($c_{-n} = 1/n!$ for all $n \geq 0$), so $z = 0$ is an essential singularity. Near $z = 0$, the function takes values arbitrarily close to any complex number (by Casorati–Weierstrass) — wildly different behaviour from a pole, where $|f(z)| \to \infty$. [/example] [example: Laurent Series of a Rational Function] The function $f(z) = \frac{1}{z(z-1)}$ has simple poles at $z = 0$ and $z = 1$. Using partial fractions: $f(z) = \frac{-1}{z} + \frac{1}{z - 1}$. On the annulus $0 < |z| < 1$, expand $\frac{1}{z-1} = \frac{-1}{1-z} = -\sum_{n=0}^\infty z^n$, giving: \begin{align*} f(z) = -\frac{1}{z} - 1 - z - z^2 - \cdots = -\frac{1}{z} - \sum_{n=0}^\infty z^n. \end{align*} The principal part is $-1/z$ (a single term), confirming a simple pole at $z = 0$. On the annulus $|z| > 1$, the expansion is different: $\frac{1}{z-1} = \frac{1}{z} \cdot \frac{1}{1 - 1/z} = \sum_{n=1}^\infty z^{-n}$, giving $f(z) = \sum_{n=2}^\infty z^{-n}$ — the Laurent series depends on the annulus. [/example] ## Worked Example [problem] Compute $\int_{|z| = 3} \frac{z^2 e^z}{(z-1)(z+2)^2} \, dz$. [/problem] [solution] **Step 1: Identify the singularities inside the contour.** The integrand has a simple pole at $z = 1$ and a double pole at $z = -2$. Both lie inside $|z| = 3$. **Step 2: Partial fraction decomposition.** Write $\frac{z^2 e^z}{(z-1)(z+2)^2} = \frac{A}{z - 1} + \frac{B}{z + 2} + \frac{C}{(z+2)^2}$, or more efficiently, use the Cauchy integral formula directly. **Step 3: Apply CIF to each singularity.** By linearity and the [Cauchy Integral Formula](/theorems/345), the integral equals $2\pi i$ times the sum of residues at the poles inside the contour. At $z = 1$ (simple pole): the residue is $\lim_{z \to 1} (z-1) \cdot \frac{z^2 e^z}{(z-1)(z+2)^2} = \frac{1 \cdot e}{(1+2)^2} = \frac{e}{9}$. At $z = -2$ (double pole): the residue is $\lim_{z \to -2} \frac{d}{dz}\left[(z+2)^2 \cdot \frac{z^2 e^z}{(z-1)(z+2)^2}\right] = \lim_{z \to -2} \frac{d}{dz}\left[\frac{z^2 e^z}{z-1}\right]$. Computing the derivative via the quotient rule: \begin{align*} \frac{d}{dz}\left[\frac{z^2 e^z}{z-1}\right] = \frac{(2z e^z + z^2 e^z)(z-1) - z^2 e^z}{(z-1)^2} = \frac{e^z(z^2 + 2z)(z-1) - z^2 e^z}{(z-1)^2}. \end{align*} At $z = -2$: numerator $= e^{-2}(4 - 4)(-3) - 4e^{-2} = 0 - 4e^{-2} = -4e^{-2}$, denominator $= (-3)^2 = 9$. So the residue at $z = -2$ is $-4e^{-2}/9$. **Step 4: Assemble.** \begin{align*} \int_{|z| = 3} \frac{z^2 e^z}{(z-1)(z+2)^2} \, dz = 2\pi i \left(\frac{e}{9} - \frac{4e^{-2}}{9}\right) = \frac{2\pi i}{9}(e - 4e^{-2}). \end{align*} This computation uses the residue theorem (developed fully in §3), but the underlying justification is the Cauchy Integral Formula: each residue computation is an application of CIF or its derivative version to the holomorphic part of $f$ after removing the singular factor. [/solution] ## References 1. Sheratt, N., *Cambridge Part IB — Complex Analysis*, Lecture Notes. # Residue Calculus The Cauchy Integral Formula tells us that a holomorphic function is completely determined by its boundary values. But what happens when the function has singularities — poles or essential singularities — inside the contour? The integral no longer vanishes; instead, it captures precisely the "singular content" of $f$ at each bad point. The *residue theorem* makes this quantitative: the integral of a meromorphic function around a closed contour equals $2\pi i$ times the sum of the residues at the enclosed singularities, each weighted by the winding number of the contour around that point. This result is the central computational tool of complex analysis. It reduces difficult contour integrals to the algebraic problem of computing residues (often by inspection or differentiation), and it provides the foundation for evaluating real definite integrals that resist elementary methods. The section then develops the *Argument Principle* — an application of the residue theorem to the logarithmic derivative $f'/f$ — which counts zeros and poles of meromorphic functions, and *Rouché's Theorem*, which compares zero counts under perturbation. These lead to the *Open Mapping Theorem*: non-constant holomorphic functions map [open sets](/page/Open%20Set) to open sets. [motivation] ### From Cauchy's Theorem to the Residue Theorem Cauchy's theorem says $\int_\gamma f \, dz = 0$ when $f$ is holomorphic inside $\gamma$. What if $f$ has isolated singularities? The Laurent series at a singularity $a$ splits into a convergent "Taylor part" $\sum_{n \geq 0} c_n(z-a)^n$ and a "principal part" $\sum_{n < 0} c_n(z-a)^n$. The Taylor part is holomorphic, so its integral vanishes by Cauchy. The principal part's terms $(z-a)^n$ for $n \leq -2$ all have antiderivatives, so they too integrate to zero over closed paths. The *only* term that survives is $c_{-1}/(z-a)$, whose integral over a loop winding once around $a$ is $2\pi i \cdot c_{-1}$. The coefficient $c_{-1}$ — the *residue* — is thus the entire contribution of the singularity to the contour integral. The residue theorem simply sums these contributions over all singularities. ### Why the Winding Number? The integral $\int_\gamma dz/(z-a)$ does not just detect *whether* $\gamma$ goes around $a$, but *how many times* and in which direction. A contour winding twice around $a$ gives $4\pi i$, not $2\pi i$. This "winding count" is formalised by the *winding number* $I(\gamma, a)$, which enters the residue theorem as a weight: the contribution of each singularity is $I(\gamma, z_j) \operatorname{Res}(f, z_j)$. For simple contours (winding number $0$ or $1$), the formula reduces to the familiar "sum of residues inside." For more complicated contours, the winding numbers are essential. ### The Argument Principle: Residues Count Zeros The logarithmic derivative $f'/f$ has a remarkable property: its residue at a zero of order $m$ is $+m$, and at a pole of order $p$ is $-p$. The residue theorem applied to $f'/f$ therefore *counts* zeros minus poles, weighted by winding number. This is the Argument Principle — the name comes from the geometric interpretation: the integral \begin{align*} \frac{1}{2\pi i}\int_\gamma \frac{f'}{f} \, dz \end{align*} measures the net change in $\arg f(z)$ as $z$ traverses $\gamma$, i.e., the winding number of the image curve $f \circ \gamma$ around the origin. [/motivation] ## Residues and Winding Numbers ### The Residue [definition: Residue] Let $f$ be holomorphic on a punctured neighbourhood $B(a, r) \setminus \{a\}$, with Laurent expansion $f(z) = \sum_{n=-\infty}^\infty c_n(z-a)^n$. The **residue** of $f$ at $a$ is: \begin{align*} \operatorname{Res}(f, a) = c_{-1} = \frac{1}{2\pi i} \int_{|z-a| = \rho} f(z) \, dz \end{align*} for any $0 < \rho < r$. [/definition] The integral formula follows immediately from integrating the Laurent series term by term: all terms except $c_{-1}/(z-a)$ integrate to zero (those with $n \neq -1$ have antiderivatives $(z-a)^{n+1}/(n+1)$), and $\int_{|z-a|=\rho} (z-a)^{-1} dz = 2\pi i$. ### The Winding Number [definition: Winding Number] Let $\gamma: [a,b] \to \mathbb{C}$ be a continuous closed path and $w \notin \gamma^*$. By the lifting lemma, there exist continuous functions $r: [a,b] \to \mathbb{R}_{> 0}$ and $\theta: [a,b] \to \mathbb{R}$ with $\gamma(t) = w + r(t)e^{i\theta(t)}$. The **winding number** (or **index**) of $\gamma$ about $w$ is: \begin{align*} I(\gamma, w) = \frac{\theta(b) - \theta(a)}{2\pi} \in \mathbb{Z}. \end{align*} [/definition] The winding number counts the net number of counter-clockwise revolutions of $\gamma$ around $w$. It is always an integer (since $\gamma$ is closed, $\gamma(b) = \gamma(a)$, so $\theta(b) - \theta(a)$ must be a multiple of $2\pi$). It is a topological invariant: $I(\gamma, w)$ is constant on each connected component of $\mathbb{C} \setminus \gamma^*$, and equals $0$ on the unbounded component. For piecewise $C^1$ paths, the winding number has an integral representation: [quotetheorem:351] The proof writes $\gamma(t) - w = r(t)e^{i\theta(t)}$ and computes \begin{align*} \frac{\gamma'}{\gamma - w} = \frac{\dot{r}}{r} + i\dot{\theta}. \end{align*} Integrating and using closedness ($r(a) = r(b)$) to kill the real part gives \begin{align*} \int \frac{dz}{z-w} = i(\theta(b) - \theta(a)) = 2\pi i \cdot I(\gamma, w). \end{align*} [citeproof:351] [example: Winding Numbers of Circles] For $\gamma(\theta) = e^{in\theta}$, $\theta \in [0, 2\pi]$, winding $n$ times around the origin: $I(\gamma, 0) = n$. For $w = 2$ (outside the unit circle): $I(\gamma, 2) = 0$. The [integral formula](/theorems/351) confirms: \begin{align*} \frac{1}{2\pi i}\int_\gamma \frac{dz}{z} = \frac{1}{2\pi i} \cdot 2\pi i n = n. \end{align*} [/example] ### Meromorphic Functions [definition: Meromorphic Function] A function $f: U \to \mathbb{C} \cup \{\infty\}$ is **meromorphic on $U$** if it is holomorphic on $U$ except for isolated singularities, all of which are poles. Equivalently, $f$ can locally be written as a quotient $g/h$ of holomorphic functions with $h \not\equiv 0$. [/definition] The key example: rational functions $p(z)/q(z)$ are meromorphic on $\mathbb{C}$ (with poles at the zeros of $q$). The function $e^{1/z}$ is *not* meromorphic at $z = 0$ (essential singularity). ## The Residue Theorem [quotetheorem:352] The proof subtracts the principal part of the Laurent expansion at each singularity: \begin{align*} P_j(z) = \sum_{n < 0} c_n^{(j)}(z - z_j)^n. \end{align*} The remainder $f - \sum P_j$ is holomorphic on all of $U$ (removing the principal parts heals each singularity), so its integral vanishes by [Cauchy's theorem for simply connected domains](/theorems/344). Each principal part contributes only through its $c_{-1}/(z - z_j)$ term (higher negative powers have antiderivatives), giving $2\pi i \cdot \operatorname{Res}(f, z_j) \cdot I(\gamma, z_j)$. [citeproof:352] For simple contours winding once around each singularity (the most common case in practice), the theorem simplifies to: \begin{align*} \int_\gamma f(z) \, dz = 2\pi i \sum_{z_j \text{ inside } \gamma} \operatorname{Res}(f, z_j). \end{align*} ## Computing Residues [quotetheorem:353] These formulas are the workhorses of residue calculus. The simple pole formula (1) is used most frequently; the quotient form (2) handles rational functions efficiently; the higher-order formula (3) is a last resort when partial fractions are impractical. [citeproof:353] [example: Residues of $1/(z^2 + 1)$] The function $f(z) = 1/(z^2 + 1)$ has simple poles at $z = \pm i$, since \begin{align*} f(z) = \frac{1}{(z-i)(z+i)}. \end{align*} By the [quotient formula](/theorems/353) with $g(z) = 1$ and $h(z) = z^2 + 1$: \begin{align*} \operatorname{Res}(f, i) = \frac{1}{h'(i)} = \frac{1}{2i}, \qquad \operatorname{Res}(f, -i) = \frac{1}{h'(-i)} = \frac{1}{-2i}. \end{align*} [/example] [example: Residue at a Double Pole] For $f(z) = e^z/(z-1)^2$, the point $z = 1$ is a pole of order $2$. By [formula (3)](/theorems/353): \begin{align*} \operatorname{Res}(f, 1) = \frac{1}{1!} \lim_{z \to 1} \frac{d}{dz}\bigl[(z-1)^2 f(z)\bigr] = \lim_{z \to 1} \frac{d}{dz} e^z = e. \end{align*} [/example] [example: Residue at an Essential Singularity] For $f(z) = e^{1/z}$, the Laurent series is \begin{align*} e^{1/z} = \sum_{n=0}^\infty \frac{z^{-n}}{n!}. \end{align*} The coefficient of $z^{-1}$ is $c_{-1} = 1$. No shortcut formula applies — one must read the residue directly from the series expansion. [/example] ## Techniques for Evaluating Real Integrals The residue theorem becomes a powerful computational tool when combined with carefully chosen contours. The general strategy is: embed a real integral as part of a complex contour integral, close the contour in a region where the integrand is holomorphic except for isolated singularities, show the added pieces contribute zero (or a known amount), and read off the real integral from the residue sum. The choice of contour depends on the integrand: - **Semicircular contours** for rational integrals $\int_{-\infty}^\infty R(x) \, dx$ where $R$ decays at infinity: close with a large semicircle in the upper or lower half-plane. - **Semicircular + Jordan's lemma** for oscillatory integrals $\int_{-\infty}^\infty R(x) e^{i\alpha x} dx$ ($\alpha > 0$): the exponential factor $e^{i\alpha z}$ decays in the upper half-plane, so close upward. Jordan's lemma handles the case where $R$ decays only like $1/|z|$. - **Indented contours** when a pole lies on the real axis: detour around it with a small semicircular arc. - **Keyhole contours** for integrals involving $x^\alpha$ or $\log x$ on $(0, \infty)$: the branch cut along $[0, \infty)$ means the integrand takes different values above and below the cut, and the difference produces the real integral. ### Semicircular Contours and Jordan's Lemma For integrals of the form $\int_{-\infty}^\infty f(x) \, dx$ where $f$ is a rational function decaying at infinity, close the contour with a large semicircle in the upper (or lower) half-plane. The contribution of the semicircular arc is controlled by: [quotetheorem:354] The key estimate is Jordan's inequality $\sin\theta \geq 2\theta/\pi$ for $\theta \in [0, \pi/2]$, which gives the exponential decay $e^{-\alpha R \sin\theta}$ enough [integrability](/page/Integral) to force the arc contribution to zero. [citeproof:354] [example: $\int_{-\infty}^\infty \frac{dx}{1 + x^2} = \pi$] Close with the upper semicircle $\gamma_R$. The function $1/(z^2 + 1)$ has a simple pole at $z = i$ (in the upper half-plane) with $\operatorname{Res}(f, i) = 1/(2i)$. On the arc, $|f| \leq 1/(R^2 - 1)$, so by the [ML inequality](/theorems/338) the arc integral tends to $0$. The [residue theorem](/theorems/352) gives: \begin{align*} \int_{-\infty}^\infty \frac{dx}{1 + x^2} = 2\pi i \cdot \frac{1}{2i} = \pi. \end{align*} [/example] [example: $\int_{-\infty}^\infty \frac{x \sin x}{x^2 + 1} \, dx = \frac{\pi}{e}$ via Jordan's Lemma] Write $\sin x = \operatorname{Im}(e^{ix})$ and consider the integral of $ze^{iz}/(z^2 + 1)$ over the closed contour. Close with the upper semicircle. The function $f(z) = z/(z^2 + 1)$ satisfies $|zf(z)| \to 1$ as $|z| \to \infty$, so [Jordan's lemma](/theorems/354) gives $\int_{\gamma_R} f(z) e^{iz} dz \to 0$. The pole at $z = i$ gives: \begin{align*} \operatorname{Res}\left(\frac{ze^{iz}}{z^2+1}, i\right) = \frac{ie^{-1}}{2i} = \frac{1}{2e}. \end{align*} So \begin{align*} \int_{-\infty}^\infty \frac{xe^{ix}}{x^2 + 1} dx = 2\pi i \cdot \frac{1}{2e} = \frac{\pi i}{e}. \end{align*} Taking imaginary parts: \begin{align*} \int_{-\infty}^\infty \frac{x \sin x}{x^2 + 1} dx = \frac{\pi}{e}. \end{align*} [/example] ### Indented Contours and the Small Arc Lemma When a pole lies *on* the real axis (or on the contour boundary), the standard strategy is to indent the contour with a small semicircular detour: [quotetheorem:355] [citeproof:355] [example: $\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2}$ via Indentation] The function $e^{iz}/z$ has a simple pole at $z = 0$ with residue $1$. Use the contour: real axis from $\varepsilon$ to $R$, upper semicircle $\gamma_R$ of radius $R$, real axis from $-R$ to $-\varepsilon$, upper semicircular indentation $\gamma_\varepsilon$ from $-\varepsilon$ to $\varepsilon$ (traversed clockwise, i.e., from angle $\pi$ to $0$). There are no poles inside this contour, so the total integral is $0$. Jordan's lemma gives $\int_{\gamma_R} \to 0$. The [small arc lemma](/theorems/355) gives \begin{align*} \int_{\gamma_\varepsilon} \frac{e^{iz}}{z} \, dz \to -i\pi \cdot 1 = -i\pi \end{align*} (the minus sign because we traverse from $\pi$ to $0$, giving angle $0 - \pi = -\pi$). The real axis integrals combine to give \begin{align*} \int_\varepsilon^R \frac{e^{ix} - e^{-ix}}{x} \, dx = 2i\int_\varepsilon^R \frac{\sin x}{x} \, dx. \end{align*} Taking limits: \begin{align*} 0 = 2i\int_0^\infty \frac{\sin x}{x} dx - i\pi, \qquad \text{so} \qquad \int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}{2}. \end{align*} [/example] ### Keyhole Contours for [Branch Cuts](/page/Branch%20Cuts) For integrals involving multi-valued functions like $x^\alpha$ or $\log x$, a keyhole contour avoids the branch cut along $[0, \infty)$. The integrand takes different values above and below the cut, and the difference produces the desired real integral. [example: $\int_0^\infty \frac{x^{s-1}}{1 + x} \, dx = \frac{\pi}{\sin(\pi s)}$ for $0 < s < 1$] Use the branch $z^{s-1} = e^{(s-1)\operatorname{Log} z}$ with cut along $[0, \infty)$, where $\operatorname{Log}$ denotes the logarithm with argument in $[0, 2\pi)$ (not the principal logarithm — we choose this branch so that $z^{s-1}$ is continuous on $\mathbb{C} \setminus [0, \infty)$). The keyhole contour consists of four pieces: the real axis above the cut from $\varepsilon$ to $R$, a large circle of radius $R$ traversed counter-clockwise, the real axis below the cut from $R$ to $\varepsilon$, and a small circle of radius $\varepsilon$ traversed clockwise. The large and small circles contribute $\to 0$ (by ML estimates using $|z^{s-1}/(1+z)| \sim R^{s-2}$ and $\varepsilon^{s-1}$ respectively, with $0 < s < 1$). Above the cut ($\arg z = 0^+$): $z^{s-1} = x^{s-1}$, giving the integral $\int_\varepsilon^R x^{s-1}/(1+x) \, dx$. Below the cut ($\arg z = 2\pi^-$): $z^{s-1} = x^{s-1} e^{2\pi i(s-1)}$, but this integral runs from $R$ to $\varepsilon$ (reversed orientation), contributing \begin{align*} -e^{2\pi i(s-1)}\int_\varepsilon^R \frac{x^{s-1}}{1+x} \, dx. \end{align*} The only singularity inside the keyhole is a simple pole at $z = -1$ (which has $\arg z = \pi$ in our branch, so $(-1)^{s-1} = e^{i\pi(s-1)}$). By the residue theorem: \begin{align*} (1 - e^{2\pi i(s-1)}) \int_0^\infty \frac{x^{s-1}}{1+x} dx = 2\pi i \cdot e^{i\pi(s-1)}. \end{align*} Simplifying the prefactor: \begin{align*} 1 - e^{2\pi i(s-1)} &= -e^{i\pi(s-1)}\bigl(e^{i\pi(s-1)} - e^{-i\pi(s-1)}\bigr) \\ &= -e^{i\pi(s-1)} \cdot 2i\sin(\pi(s-1)) \\ &= 2ie^{i\pi(s-1)}\sin(\pi s), \end{align*} where the last step uses $\sin(\pi(s-1)) = -\sin(\pi s)$. Dividing: \begin{align*} \int_0^\infty \frac{x^{s-1}}{1+x} dx = \frac{2\pi i \cdot e^{i\pi(s-1)}}{2i e^{i\pi(s-1)} \sin(\pi s)} = \frac{\pi}{\sin(\pi s)}. \end{align*} [/example] ## The Argument Principle [quotetheorem:356] The proof computes the residue of the logarithmic derivative $f'/f$ at each zero and pole. At a zero of order $m$, writing $f(z) = (z-a)^m g(z)$ with $g(a) \neq 0$ gives \begin{align*} \frac{f'}{f} = \frac{m}{z-a} + \frac{g'}{g}, \end{align*} so the residue is $+m$. At a pole of order $p$, the same argument gives residue $-p$. The [residue theorem](/theorems/352) assembles the count. [citeproof:356] The geometric content: the integral \begin{align*} \frac{1}{2\pi i}\int_\gamma \frac{f'}{f} \, dz = I(f \circ \gamma, 0) \end{align*} equals the winding number of the image curve around the origin. As $z$ traverses $\gamma$, the argument of $f(z)$ changes by $2\pi$ times (zeros minus poles inside). [example: Counting Zeros of $z^4 + 4z + 2$ Inside $|z| = 2$] Let $f(z) = z^4$ and $g(z) = 4z + 2$. On $|z| = 2$: \begin{align*} |f(z)| = |z|^4 = 16 \quad \text{and} \quad |g(z)| \leq 4 \cdot 2 + 2 = 10 < 16. \end{align*} By [Rouché's theorem](/theorems/357), $z^4 + 4z + 2$ has the same number of zeros inside $|z| < 2$ as $z^4$, namely $4$ (counted with multiplicity). On $|z| = 1$: \begin{align*} |4z| = 4 \quad \text{and} \quad |z^4 + 2| \leq 1 + 2 = 3 < 4. \end{align*} With $f(z) = 4z$ and $g(z) = z^4 + 2$: $z^4 + 4z + 2$ has $1$ zero inside $|z| < 1$ (same as $4z$). Therefore exactly $4 - 1 = 3$ zeros lie in the annulus $1 \leq |z| < 2$. (None lie on $|z| = 1$ itself, since $|z^4 + 4z + 2| \geq |4z| - |z^4 + 2| \geq 4 - 3 = 1 > 0$ there.) [/example] ## Rouché's Theorem [quotetheorem:357] The proof defines the homotopy $F_t = f + tg$ for $t \in [0,1]$. The condition $|f| > |g|$ on $\gamma$ ensures $F_t \neq 0$ on $\gamma$ for all $t$, so the zero-count $N(t) = I(F_t \circ \gamma, 0)$ is a continuous integer-valued function of $t$, hence constant. At $t = 0$ it counts zeros of $f$; at $t = 1$, zeros of $f + g$. [citeproof:357] [example: Roots of $z^4 - 6z + 3$ in $|z| < 2$] On $|z| = 2$: \begin{align*} |z^4| = 16 \quad \text{and} \quad |-6z + 3| \leq 12 + 3 = 15 < 16. \end{align*} By [Rouché](/theorems/357) with $f(z) = z^4$ and $g(z) = -6z + 3$: $z^4 - 6z + 3$ has $4$ zeros inside $|z| < 2$ (same as $z^4$). On $|z| = 1$: \begin{align*} |-6z| = 6 \quad \text{and} \quad |z^4 + 3| \leq 1 + 3 = 4 < 6. \end{align*} With $f(z) = -6z$ and $g(z) = z^4 + 3$: $z^4 - 6z + 3$ has $1$ zero inside $|z| < 1$ (same as $-6z$). Therefore exactly $4 - 1 = 3$ zeros lie in the annulus $1 \leq |z| < 2$. [/example] [example: The Fundamental Theorem of Algebra via Rouché] Let $p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_0$ with $a_n \neq 0$. For $R$ large enough, on $|z| = R$: \begin{align*} |a_nz^n| = |a_n|R^n \quad \text{dominates} \quad |a_{n-1}z^{n-1} + \cdots + a_0| \leq |a_{n-1}|R^{n-1} + \cdots + |a_0| = o(R^n). \end{align*} By [Rouché](/theorems/357), $p$ has $n$ zeros inside $|z| = R$ (same as $a_nz^n$), counted with multiplicity. This gives an alternative proof of the [Fundamental Theorem of Algebra](/theorems/347). [/example] ## The Open Mapping Theorem and Local Degree Rouché's theorem doesn't just count zeros; it also controls how zeros *move* under perturbation. If $f(z) = w_0$ has a zero of order $k$ at $z = a$, then for $w$ close to $w_0$, the equation $f(z) = w$ has exactly $k$ solutions near $a$ (by applying Rouché to $f - w_0$ and the perturbation $w_0 - w$). This stability of the zero count is what drives both the Local Degree Theorem and the Open Mapping Theorem. ### Local Degree [definition: Local Degree] Let $f: U \to \mathbb{C}$ be holomorphic and non-constant at $a \in U$. The **local degree** (or **multiplicity**) of $f$ at $a$ is $\deg(f, a) = \operatorname{ord}(f - f(a), a)$, i.e., the order of the zero of $f(z) - f(a)$ at $z = a$. If $f'(a) \neq 0$ then $\deg(f, a) = 1$; if $f'(a) = 0$ and $f''(a) \neq 0$ then $\deg(f, a) = 2$; and so on. [/definition] The local degree describes the "branching" behaviour: near $a$, the function $f$ looks like $z \mapsto z^k$ (after suitable changes of coordinate), mapping $k$ distinct points near $a$ to each nearby value $w \neq f(a)$. This is made precise by: [quotetheorem:359] The proof uses [Rouché's theorem](/theorems/357): on a small circle around $a$ where $f - f(a)$ has no other zeros, the perturbation $f - w$ (for $w$ close to $f(a)$) has the same number of zeros as $f - f(a)$, namely $k$. For $w \neq f(a)$, these zeros are simple (they avoid the isolated zeros of $f'$), hence distinct. [citeproof:359] ### The Open Mapping Theorem [quotetheorem:358] The proof is a direct corollary of the Local Degree Theorem: for any $a \in U$ and $w_0 = f(a)$, the equation $f(z) = w$ has solutions near $a$ for all $w$ near $w_0$. Thus every point of $f(U)$ is an interior point, and $f$ maps open sets to open sets. [citeproof:358] An immediate corollary is the **[Maximum Modulus Principle](/theorems/491)**: if $f$ is holomorphic and non-constant on a domain $U$, then $|f|$ has no local maximum in $U$. (If $|f(a)|$ were a local maximum, then $f$ would map a neighbourhood of $a$ into the closed disc $\overline{B(0, |f(a)|)}$, whose interior is an open set not contained in $\{|w| \leq |f(a)|\}$ — contradicting the open mapping theorem.) ## Worked Examples [problem] Evaluate $\int_0^\infty \frac{\cos x}{(x^2 + 1)^2} \, dx$. [/problem] [solution] **Step 1: Set up.** Write $\cos x = \operatorname{Re}(e^{ix})$ and consider \begin{align*} I = \int_{-\infty}^\infty \frac{e^{iz}}{(z^2+1)^2} \, dz. \end{align*} The desired integral is $\frac{1}{2}\operatorname{Re}(I)$ (by evenness of $\cos x / (x^2+1)^2$). **Step 2: Close the contour.** Use the upper semicircle $\gamma_R$ of radius $R$. By [Jordan's lemma](/theorems/354) (with $f(z) = 1/(z^2+1)^2$ satisfying $|zf(z)| \to 0$), the arc integral $\int_{\gamma_R} f(z)e^{iz} dz \to 0$. **Step 3: Find the residue.** The only singularity in the upper half-plane is a double pole at $z = i$. By [formula (3)](/theorems/353): \begin{align*} \operatorname{Res}\left(\frac{e^{iz}}{(z^2+1)^2}, i\right) = \lim_{z \to i} \frac{d}{dz}\left[\frac{e^{iz}}{(z+i)^2}\right]. \end{align*} Computing: \begin{align*} \frac{d}{dz}\left[\frac{e^{iz}}{(z+i)^2}\right] = \frac{ie^{iz}(z+i)^2 - 2(z+i)e^{iz}}{(z+i)^4} = \frac{e^{iz}(i(z+i) - 2)}{(z+i)^3}. \end{align*} At $z = i$: \begin{align*} \frac{e^{-1}(i \cdot 2i - 2)}{(2i)^3} = \frac{e^{-1}(-2-2)}{-8i} = \frac{-4e^{-1}}{-8i} = \frac{1}{2ie}. \end{align*} **Step 4: Assemble.** \begin{align*} I = 2\pi i \cdot \frac{1}{2ie} = \frac{\pi}{e}. \end{align*} Therefore \begin{align*} \int_0^\infty \frac{\cos x}{(x^2+1)^2} \, dx = \frac{1}{2}\operatorname{Re}(I) = \frac{\pi}{2e}. \end{align*} [/solution] [problem] Show that $p(z) = z^6 + 6z + 3$ has exactly $1$ root inside $|z| < 1$ and $5$ roots in $1 < |z| < 2$. [/problem] [solution] **On $|z| = 1$:** \begin{align*} |6z| = 6 > |z^6 + 3| \leq 1 + 3 = 4. \end{align*} By [Rouché](/theorems/357) with $f(z) = 6z$ (one zero at $z = 0$) and $g(z) = z^6 + 3$: $p$ has $1$ zero inside $|z| < 1$. **On $|z| = 2$:** \begin{align*} |z^6| = 64 > |6z + 3| \leq 12 + 3 = 15. \end{align*} By Rouché with $f(z) = z^6$ (six zeros at $z = 0$) and $g(z) = 6z + 3$: $p$ has $6$ zeros inside $|z| < 2$, counted with multiplicity. **On the boundary $|z| = 1$:** \begin{align*} |p(z)| \geq |6z| - |z^6 + 3| \geq 6 - 4 = 2 > 0, \end{align*} so no roots lie on $|z| = 1$. **In the annulus:** $6 - 1 = 5$ zeros lie in $1 < |z| < 2$. [/solution] ## References 1. Sheratt, N., *Cambridge Part IB — Complex Analysis*, Lecture Notes. # Transform Theory The machinery of contour integration and residue calculus has so far been directed at evaluating integrals and counting zeros. We now turn to its most powerful application domain: *transform theory*. The Fourier and Laplace transforms convert differential equations into algebraic ones and [convolutions](/page/Convolution) into products, by representing functions in terms of complex exponentials $e^{ikx}$ or $e^{pt}$. The inversion formulas that recover the original function are *contour integrals*, and the residue theorem provides the computational engine for evaluating them. This section develops the theory from the Fourier transform through the [Laplace transform](/page/Laplace%20Transform), the Bromwich inversion formula, and applications to differential equations. [motivation] ### Why Transforms? The Differential-to-Algebraic Principle The core idea is simple. Differentiation in the time domain corresponds to multiplication by a variable in the transform domain. If $\hat{f}(p) = \mathcal{L}\{f\}(p)$, then $\mathcal{L}\{f'\}(p) = p\hat{f}(p) - f(0)$. A linear ODE with constant coefficients becomes a polynomial equation in $p$, solvable by algebra. The solution in the transform domain is then a rational function of $p$, and recovering $f(t)$ requires "inverting" the transform — which turns out to be a contour integral evaluated by residues. This strategy only works because complex analysis provides both the *existence* of the inversion integral (the Bromwich formula, derived from Fourier inversion) and the *computational method* for evaluating it (close the contour, apply the residue theorem). Without these tools, transform methods would be formal manipulations without rigorous foundations. ### From Fourier to Laplace: Why the Complex Shift? The Fourier transform $\tilde{f}(k) = \int f(x) e^{-ikx} dx$ decomposes a function into oscillatory components. It works beautifully for $L^1$ functions (those with $\int |f| < \infty$), but many physically relevant functions — step functions, exponentially growing solutions of ODEs — are not in $L^1$. The Laplace transform fixes this by introducing a damping factor: $\hat{f}(p) = \int_0^\infty f(t) e^{-pt} dt$ with $\operatorname{Re} p > \sigma_0$. The factor $e^{-\operatorname{Re}(p) t}$ forces convergence for functions of exponential growth. Setting $p = c + i\omega$ reveals the relationship: the Laplace transform is the Fourier transform of the damped function $f(t)e^{-ct}$. This shift from the imaginary axis (Fourier) to a vertical line in the right half-plane (Laplace) is what makes the transform applicable to initial value problems. ### Causality and Contour Closing The most elegant feature of the Bromwich inversion formula is how it automatically enforces *causality*: for $t > 0$, close the contour to the left and pick up residues (giving the physical solution); for $t < 0$, close to the right where there are no singularities (giving zero — the function hasn't started yet). The sign of $t$ determines which half-plane the exponential $e^{pt}$ decays in, and this dictates the direction of contour closure. The entire structure of causal response — the fact that effects follow causes — is encoded in the analytic structure of the transform. [/motivation] ## The Fourier Transform ### Definitions [definition: Fourier Transform] Let $f: \mathbb{R} \to \mathbb{C}$ be absolutely integrable ($f \in L^1(\mathbb{R})$). The **Fourier transform** of $f$ is: \begin{align*} \tilde{f}(k) = \int_{-\infty}^\infty f(x) e^{-ikx} \, dx. \end{align*} [/definition] The transform decomposes $f$ into a superposition of complex exponentials $e^{ikx}$ of frequency $k$. The coefficient $\tilde{f}(k)$ measures the "amplitude and phase" of the frequency-$k$ component. The absolute integrability condition ensures the integral converges and $\tilde{f}$ is continuous and bounded ($|\tilde{f}(k)| \leq \|f\|_{L^1}$). [definition: Cauchy Principal Value] The **Cauchy principal value** of $\int_{-\infty}^\infty g(x) \, dx$ is: \begin{align*} \text{p.v.} \int_{-\infty}^\infty g(x) \, dx = \lim_{R \to \infty} \int_{-R}^R g(x) \, dx, \end{align*} provided the limit exists. This symmetric limiting process can converge even when the two-sided limit $\lim_{S \to -\infty, R \to \infty} \int_S^R g \, dx$ does not. [/definition] ### Operational Properties of the Fourier Transform The Fourier transform converts operations in the time domain to algebraic operations in the frequency domain: **Linearity:** $\widetilde{af + bg} = a\tilde{f} + b\tilde{g}$. **Frequency shift:** $\widetilde{e^{iak}f}(k) = \tilde{f}(k - a)$. Modulation by $e^{iax}$ shifts the spectrum. **Time shift:** $\widetilde{f(x - a)}(k) = e^{-iak}\tilde{f}(k)$. A delay introduces a phase shift. **Scaling:** $\widetilde{f(ax)}(k) = \frac{1}{|a|}\tilde{f}(k/a)$. Compressing in time spreads in frequency (the uncertainty principle in embryonic form). **Derivative rule:** $\widetilde{f'}(k) = ik\tilde{f}(k)$ (when $f' \in L^1$). Differentiation becomes multiplication by $ik$ — the engine for solving ODEs and PDEs. **Convolution:** $\widetilde{f * g}(k) = \tilde{f}(k)\tilde{g}(k)$, where $(f*g)(x) = \int_{-\infty}^\infty f(x-t)g(t) \, dt$. Convolution becomes pointwise multiplication. **Parseval's theorem:** $\int_{-\infty}^\infty |f(x)|^2 dx = \frac{1}{2\pi}\int_{-\infty}^\infty |\tilde{f}(k)|^2 dk$. The Fourier transform preserves energy (up to the $2\pi$ convention factor). ### The Fourier Inversion Theorem [quotetheorem:364] The inversion formula recovers $f$ from $\tilde{f}$: the Fourier transform is (up to convention) its own inverse. At points of discontinuity, the inversion returns the average of the left and right limits — the same Gibbs-type averaging that appears in [Fourier series](/page/Fourier%20Series). The principal value is necessary because $\tilde{f}$ need not be in $L^1$ even when $f$ is. [example: Fourier Transform of $e^{-|x|}$] For $f(x) = e^{-|x|}$, compute directly: \begin{align*} \tilde{f}(k) = \int_{-\infty}^0 e^{x} e^{-ikx} dx + \int_0^\infty e^{-x} e^{-ikx} dx = \frac{1}{1 - ik} + \frac{1}{1 + ik} = \frac{2}{1 + k^2}. \end{align*} Inversion gives $e^{-|x|} = \frac{1}{2\pi}\int_{-\infty}^\infty \frac{2e^{ikx}}{1 + k^2} dk$, i.e., $\int_{-\infty}^\infty \frac{e^{ikx}}{1 + k^2} dk = \pi e^{-|x|}$ — an integral we previously evaluated by residues. [/example] [example: Fourier Transform of the Gaussian] For $f(x) = e^{-x^2/2}$, complete the square in the exponent: \begin{align*} \tilde{f}(k) = \int_{-\infty}^\infty e^{-x^2/2 - ikx} dx = e^{-k^2/2} \int_{-\infty}^\infty e^{-(x + ik)^2/2} dx. \end{align*} The shifted integral $\int_{-\infty + ik}^{\infty + ik} e^{-z^2/2} dz$ equals $\int_{-\infty}^\infty e^{-x^2/2} dx = \sqrt{2\pi}$ by a rectangular contour argument: $e^{-z^2/2}$ is entire, the vertical sides of the rectangle $[-R, R] \times [0, k]$ contribute $O(e^{-R^2/2}) \to 0$, and Cauchy's theorem gives equality. Therefore $\tilde{f}(k) = \sqrt{2\pi} \, e^{-k^2/2}$ — the Gaussian is an eigenfunction of the Fourier transform. [/example] ## The Fourier–Laplace Connection Before developing the Laplace transform, we note the relationship that motivates it. If $f$ is causal (vanishing for $t < 0$) and of exponential order $\sigma_0$, then for $c > \sigma_0$: \begin{align*} \hat{f}(c + i\omega) = \int_0^\infty f(t)e^{-ct} e^{-i\omega t} \, dt = \widetilde{f_c}(\omega), \end{align*} where $f_c(t) = f(t)e^{-ct}$ is the "damped" version of $f$. The Laplace transform is the Fourier transform of the damped signal, evaluated on a vertical line shifted into the right half-plane. When $\sigma_0 < 0$ (i.e., $f$ decays exponentially), one can take $c \to 0^+$ to recover $\hat{f}(i\omega) = \tilde{f}(\omega)$. This connection explains why the Bromwich inversion formula (below) is a vertical-line integral — it is Fourier inversion for the damped signal, with the damping factor removed. ## The Laplace Transform ### Definitions [definition: Laplace Transform] Let $f: \mathbb{R} \to \mathbb{C}$ satisfy $f(t) = 0$ for $t < 0$ and $|f(t)| \leq Me^{\sigma_0 t}$ for some $M, \sigma_0$ (i.e., $f$ is **causal** and of **exponential order** $\sigma_0$). The **Laplace transform** of $f$ is: \begin{align*} \hat{f}(p) = \mathcal{L}\{f\}(p) = \int_0^\infty f(t) e^{-pt} \, dt, \end{align*} defined and holomorphic for $\operatorname{Re} p > \sigma_0$. [/definition] The Laplace transform is holomorphic in the half-plane $\operatorname{Re} p > \sigma_0$ (differentiate under the integral sign, justified by uniform convergence on compact subsets). The boundary $\operatorname{Re} p = \sigma_0$ is called the **abscissa of convergence**. [definition: Convolution (Causal)] For causal functions $f, g$ (vanishing for $t < 0$), their **convolution** is: \begin{align*} (f * g)(t) = \int_0^t f(t - t') g(t') \, dt'. \end{align*} The upper limit is $t$ (not $\infty$) because causality forces $f(t - t') = 0$ for $t' > t$. [/definition] ### Standard Transforms The following table collects the most commonly used Laplace transforms, all verifiable by direct integration: | $f(t)$ (for $t > 0$) | $\hat{f}(p)$ | Region of convergence | |---|---|---| | $1$ (Heaviside step) | $1/p$ | $\operatorname{Re} p > 0$ | | $t^n$ | $n!/p^{n+1}$ | $\operatorname{Re} p > 0$ | | $e^{at}$ | $1/(p-a)$ | $\operatorname{Re} p > \operatorname{Re} a$ | | $\cos(\omega t)$ | $p/(p^2 + \omega^2)$ | $\operatorname{Re} p > 0$ | | $\sin(\omega t)$ | $\omega/(p^2 + \omega^2)$ | $\operatorname{Re} p > 0$ | | $t e^{at}$ | $1/(p-a)^2$ | $\operatorname{Re} p > \operatorname{Re} a$ | ## Operational Properties ### The Derivative Rule [quotetheorem:362] The proof is [integration by parts](/theorems/210): the boundary term at $t = \infty$ vanishes by exponential damping, and the boundary term at $t = 0$ gives $-f(0)$. [citeproof:362] This is the key property that makes the Laplace transform useful for differential equations. An $n$-th order linear ODE with constant coefficients $\sum a_k y^{(k)} = g(t)$ becomes $\sum a_k p^k \hat{y}(p) = \hat{g}(p) + \text{(initial conditions)}$ — a polynomial equation in $p$ that can be solved algebraically. ### The Convolution Theorem [quotetheorem:363] The proof interchanges the order of integration (Fubini) and substitutes $\tau = t - t'$ to separate the double integral into a product. [citeproof:363] Physically, the convolution theorem says: if a linear time-invariant system has impulse response $h(t)$ (so the output for input $g(t)$ is $h * g$), then in the transform domain the output is $\hat{h}(p) \cdot \hat{g}(p)$. The function $\hat{h}(p)$ is the **transfer function** of the system. [example: Verification for $f = g = H$ (Heaviside)] $(H * H)(t) = \int_0^t 1 \cdot 1 \, dt' = t$ for $t > 0$. In the transform domain: $\hat{H}(p) = 1/p$, so $\hat{H}(p)^2 = 1/p^2 = \mathcal{L}\{t\}(p)$. ✓ [/example] ### Other Useful Rules Several additional properties follow from the definition by direct manipulation: **Shifting in $p$ (frequency shift):** $\mathcal{L}\{e^{at}f(t)\}(p) = \hat{f}(p - a)$. This follows by absorbing $e^{at}$ into the exponential kernel. **Shifting in $t$ (time delay):** $\mathcal{L}\{f(t - \tau)H(t - \tau)\}(p) = e^{-p\tau}\hat{f}(p)$ for $\tau > 0$. The Heaviside factor ensures causality. **Scaling:** $\mathcal{L}\{f(at)\}(p) = \frac{1}{a}\hat{f}(p/a)$ for $a > 0$. ## The Bromwich Inversion Formula [quotetheorem:360] The derivation reduces to the Fourier inversion theorem: setting $g(t) = f(t)e^{-ct}$ (which is in $L^1$ for $c > \sigma_0$), the Fourier transform of $g$ is $\hat{f}(c + i\omega)$, and Fourier inversion recovers $g(t)$. Multiplying back by $e^{ct}$ and substituting $p = c + i\omega$ gives the Bromwich integral. [citeproof:360] The formula expresses the inverse Laplace transform as an integral along a vertical line in the complex $p$-plane. The choice of $c$ does not matter (as long as $c > \sigma_0$), because shifting the vertical line does not change the integral — a consequence of Cauchy's theorem applied in the half-plane where $\hat{f}$ is holomorphic. ### Residue Inversion In practice, the Bromwich integral is evaluated by closing the contour and applying the residue theorem: [quotetheorem:361] The key insight is the direction of closure: for $t > 0$, the factor $e^{pt}$ decays in the left half-plane ($\operatorname{Re} p \to -\infty$), so we close left and pick up all singularities. For $t < 0$, $e^{pt}$ decays in the right half-plane, so we close right where there are no singularities, automatically giving $f(t) = 0$. This is causality encoded in complex analysis. [citeproof:361] [example: Inverse Laplace Transform of $1/(p-a)$] $\hat{f}(p) = 1/(p-a)$ has a simple pole at $p = a$. For $t > 0$: \begin{align*} f(t) = \operatorname{Res}_{p=a}\left(\frac{e^{pt}}{p-a}\right) = e^{at}. \end{align*} For $t < 0$: $f(t) = 0$. So $\mathcal{L}^{-1}\{1/(p-a)\} = e^{at}H(t)$. [/example] [example: Inverse Laplace Transform of $1/(p^2 + \omega^2)$] $\hat{f}(p) = 1/(p^2 + \omega^2) = 1/((p - i\omega)(p + i\omega))$ has simple poles at $p = \pm i\omega$. For $t > 0$: \begin{align*} f(t) = \operatorname{Res}_{p=i\omega}\left(\frac{e^{pt}}{(p-i\omega)(p+i\omega)}\right) + \operatorname{Res}_{p=-i\omega}\left(\cdots\right) = \frac{e^{i\omega t}}{2i\omega} + \frac{e^{-i\omega t}}{-2i\omega} = \frac{\sin(\omega t)}{\omega}. \end{align*} [/example] [example: Double Pole — $1/(p+1)^2$] $\hat{f}(p) = 1/(p+1)^2$ has a double pole at $p = -1$. For $t > 0$: \begin{align*} f(t) = \operatorname{Res}_{p=-1}\left(\frac{e^{pt}}{(p+1)^2}\right) = \lim_{p \to -1} \frac{d}{dp} e^{pt} = te^{-t}. \end{align*} [/example] ## Solving Differential Equations ### Second-Order IVPs The full power of the Laplace transform emerges when solving initial value problems. The method is: (1) transform both sides using the [derivative rule](/theorems/362), (2) solve the resulting algebraic equation for $\hat{y}(p)$, (3) invert using [residue inversion](/theorems/361). [example: Damped Oscillator $y'' + 2y' + 5y = 0$, $y(0) = 1$, $y'(0) = 0$] Transforming: $p^2\hat{y} - p - 0 + 2(p\hat{y} - 1) + 5\hat{y} = 0$, so: \begin{align*} (p^2 + 2p + 5)\hat{y} = p + 2, \qquad \hat{y}(p) = \frac{p + 2}{p^2 + 2p + 5} = \frac{p + 2}{(p+1)^2 + 4}. \end{align*} The poles are at $p = -1 \pm 2i$. Computing residues: \begin{align*} \operatorname{Res}_{p = -1+2i} \frac{(p+2)e^{pt}}{(p+1-2i)(p+1+2i)} = \frac{(1 + 2i)e^{(-1+2i)t}}{4i}. \end{align*} Adding the conjugate residue and simplifying: \begin{align*} y(t) = e^{-t}\left(\cos 2t + \tfrac{1}{2}\sin 2t\right). \end{align*} The initial conditions are automatically incorporated through the derivative rule — no need to solve for constants of integration as in the classical method. [/example] ### Systems of ODEs For $y' = My + g(t)$ with $y(0) = y_0$, transforming gives $(pI - M)\hat{y} = y_0 + \hat{g}$, so $\hat{y} = (pI - M)^{-1}(y_0 + \hat{g})$. The matrix $(pI - M)^{-1}$ is the **resolvent** of $M$ — so named because it "resolves" the algebraic equation for $\hat{y}$. Its entries are rational functions of $p$ with poles at the eigenvalues of $M$. Applying residue inversion entry-by-entry produces the **matrix exponential** $e^{Mt}$, defined as $\mathcal{L}^{-1}\{(pI - M)^{-1}\}$, which gives the general solution $y(t) = e^{Mt}y_0$ of the homogeneous system. [example: System with Repeated Eigenvalue] For $y' = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}y$ with $y(0) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$: \begin{align*} (pI - M)^{-1} = \frac{1}{(p+1)^2}\begin{pmatrix} p+1 & 1 \\ 0 & p+1 \end{pmatrix}. \end{align*} So $\hat{y} = (pI-M)^{-1}y_0 = \frac{1}{(p+1)^2}\begin{pmatrix} p+2 \\ p+1 \end{pmatrix} = \begin{pmatrix} \frac{1}{p+1} + \frac{1}{(p+1)^2} \\ \frac{1}{p+1} \end{pmatrix}$. Inverting: $y(t) = \begin{pmatrix} e^{-t} + te^{-t} \\ e^{-t} \end{pmatrix} = e^{-t}\begin{pmatrix} 1 + t \\ 1 \end{pmatrix}$. [/example] ## Branch Points and Keyhole Inversion When $\hat{f}(p)$ has branch points rather than poles, the Bromwich contour must be deformed around the branch cut. The residue theorem does not apply directly — instead, the discontinuity of the integrand across the cut produces a real integral. The prototype is: [example: Inverse Laplace of $1/\sqrt{p}$] The function $\hat{f}(p) = p^{-1/2}$ has a branch point at $p = 0$. Place the branch cut along $(-\infty, 0]$ and take $\arg p \in (-\pi, \pi)$ (the principal branch), so that $p^{-1/2}$ is holomorphic on $\mathbb{C} \setminus (-\infty, 0]$. **The contour.** Since $\hat{f}$ has no poles, we deform the Bromwich line into a keyhole contour wrapping around the branch cut: a path just above the negative real axis from $-R$ to $-\varepsilon$, a small circle $C_\varepsilon$ of radius $\varepsilon$ around the origin (counterclockwise), a path just below the negative real axis from $-\varepsilon$ to $-R$, and a large arc $C_R$ connecting back to the start. The Bromwich integral equals the keyhole integral by Cauchy's theorem (the two contours are homotopic in the region of holomorphicity). **Vanishing arcs.** On $C_R$: $|p^{-1/2}e^{pt}| \leq R^{-1/2}e^{Rt\cos\theta}$. For $t > 0$ and $\theta \in (\pi/2, 3\pi/2)$, $\cos\theta < 0$, giving exponential decay; the arc integral tends to $0$. On $C_\varepsilon$: $|p^{-1/2}e^{pt}| \leq \varepsilon^{-1/2}$ and the arc has length $2\pi\varepsilon$, giving $O(\varepsilon^{1/2}) \to 0$. **Above the cut.** Parametrise as $p = -x + i0^+$ for $x > 0$ (argument $= \pi$). Then $p^{-1/2} = x^{-1/2}e^{-i\pi/2} = -ix^{-1/2}$, and $e^{pt} = e^{-xt}$. This path runs from $-R$ to $-\varepsilon$, which in the $x$-parametrisation means $x$ from $R$ to $\varepsilon$: \begin{align*} \int_{\text{above}} = \int_R^\varepsilon (-ix^{-1/2})e^{-xt}(-dx) = \int_\varepsilon^R (-ix^{-1/2})e^{-xt} \, dx. \end{align*} **Below the cut.** Parametrise as $p = -x - i0^+$ for $x > 0$ (argument $= -\pi$). Then $p^{-1/2} = x^{-1/2}e^{i\pi/2} = ix^{-1/2}$. This path runs from $-\varepsilon$ to $-R$ (reversed direction compared to above), so $x$ goes from $\varepsilon$ to $R$: \begin{align*} \int_{\text{below}} = \int_\varepsilon^R (ix^{-1/2})e^{-xt}(-dx) = -\int_\varepsilon^R (ix^{-1/2})e^{-xt} \, dx. \end{align*} **Assembly.** Combining above and below (the two terms have the same sign because the path reversal introduces a minus that flips the $+i$ to $-i$): \begin{align*} \int_{\text{above}} + \int_{\text{below}} = \int_\varepsilon^R (-i - i)x^{-1/2}e^{-xt} \, dx = -2i\int_\varepsilon^R x^{-1/2}e^{-xt} \, dx. \end{align*} The Bromwich integral equals the keyhole integral, so: \begin{align*} f(t) = \frac{1}{2\pi i}\cdot(-2i)\int_0^\infty x^{-1/2}e^{-xt} \, dx = \frac{1}{\pi}\int_0^\infty x^{-1/2}e^{-xt} \, dx. \end{align*} Substituting $u = \sqrt{xt}$: $f(t) = \frac{1}{\pi\sqrt{t}} \int_0^\infty 2e^{-u^2} \, du = \frac{1}{\pi\sqrt{t}} \cdot \sqrt{\pi} = \frac{1}{\sqrt{\pi t}}$. [/example] ## Worked Example [problem] Solve $y'' + y = H(t)\sin(2t)$ with $y(0) = 0$, $y'(0) = 1$, where $H(t)$ is the Heaviside step function. [/problem] [solution] **Step 1: Transform.** Using the [derivative rule](/theorems/362): $\mathcal{L}\{y''\} = p^2\hat{y} - p \cdot 0 - 1 = p^2\hat{y} - 1$. The right side: $\mathcal{L}\{H(t)\sin 2t\} = 2/(p^2 + 4)$. The transformed equation: \begin{align*} p^2\hat{y} - 1 + \hat{y} = \frac{2}{p^2 + 4}, \qquad \hat{y}(p) = \frac{1}{p^2 + 1} + \frac{2}{(p^2+1)(p^2+4)}. \end{align*} **Step 2: Partial fractions.** Decompose the second term: $\frac{2}{(p^2+1)(p^2+4)} = \frac{2}{3}\left(\frac{1}{p^2+1} - \frac{1}{p^2+4}\right)$. So: \begin{align*} \hat{y}(p) = \frac{1}{p^2+1} + \frac{2}{3}\cdot\frac{1}{p^2+1} - \frac{2}{3}\cdot\frac{1}{p^2+4} = \frac{5}{3}\cdot\frac{1}{p^2+1} - \frac{2}{3}\cdot\frac{1}{p^2+4}. \end{align*} **Step 3: Invert.** Using the standard transform $\mathcal{L}^{-1}\{1/(p^2 + \omega^2)\} = \sin(\omega t)/\omega$: \begin{align*} y(t) = \frac{5}{3}\sin t - \frac{2}{3} \cdot \frac{\sin 2t}{2} = \frac{5}{3}\sin t - \frac{1}{3}\sin 2t. \end{align*} **Check:** $y(0) = 0$ ✓; $y'(t) = \frac{5}{3}\cos t - \frac{2}{3}\cos 2t$, so $y'(0) = \frac{5}{3} - \frac{2}{3} = 1$ ✓; $y'' + y = (-\frac{5}{3}\sin t + \frac{4}{3}\sin 2t) + (\frac{5}{3}\sin t - \frac{1}{3}\sin 2t) = \sin 2t$ ✓. [/solution] ## References 1. Sheratt, N., *Cambridge Part IB — Complex Analysis*, Lecture Notes. # Uniform Limits of Holomorphic Functions A recurring theme throughout this course has been the construction of holomorphic functions by limiting processes: power series (limits of partial sums), Cauchy integrals (limits of Riemann sums), Laurent series, and Laplace transforms. Each time, we used holomorphicity of the approximants and asked whether the limit inherits this property. This section addresses the question systematically: *when does a limit of holomorphic functions remain holomorphic?* The answer is one of the most striking contrasts between real and complex analysis. In real analysis, the uniform limit of smooth functions is continuous but need not be differentiable — let alone smooth. In complex analysis, the locally uniform limit of holomorphic functions is *automatically holomorphic*, and all derivatives converge as well. This is a consequence of the integral representation (Cauchy's formula), which converts convergence of function values into convergence of derivatives via a contour integral — a mechanism with no real-variable analogue. [motivation] ### Why Does Uniform Convergence Preserve Holomorphicity but Not Real Smoothness? In real analysis, differentiation and uniform convergence interact badly. The sequence $f_n(x) = \frac{1}{n}\sin(n^2 x)$ converges uniformly to $0$ on $\mathbb{R}$, but $f_n'(x) = n\cos(n^2 x)$ diverges everywhere. The derivatives are uncontrolled because differentiation amplifies high-frequency oscillations, and uniform convergence of $f_n$ gives no control over these oscillations. In complex analysis, the situation is fundamentally different. The [Cauchy integral formula for derivatives](/theorems/345) gives: \begin{align*} f_n'(z) = \frac{1}{2\pi i} \int_{|w - z| = r} \frac{f_n(w)}{(w - z)^2} \, dw. \end{align*} If $f_n \to f$ uniformly on the circle $|w - z| = r$, then $f_n'(z) \to f'(z)$ — the derivative is expressed as an *integral* of $f_n$, and integrals commute with uniform limits. This is the key: in complex analysis, derivatives are controlled by values on a surrounding curve, not by local oscillation. The Cauchy formula acts as a "smoothing operator" that converts uniform convergence of functions into convergence of all derivatives. ### Why Locally Uniform, Not Pointwise? Pointwise convergence is too weak: $f_n(z) = z^n$ converges pointwise on $B(0,1)$ to $0$ (holomorphic), but on $\overline{B(0,1)}$ the limit is $0$ for $|z| < 1$ and $1$ at $z = 1$ — discontinuous. More dramatically, it is possible to construct [sequences](/page/Sequence) of entire functions converging pointwise to a function that is nowhere analytic. The right notion is *locally uniform convergence*: uniform on every compact subset, or equivalently, uniform on some disc around each point. This is strong enough to pass limits through contour integrals (which run over compact curves), but weak enough to accommodate functions that behave differently near the boundary of their domain. [/motivation] ## Locally Uniform Convergence [definition: Locally Uniform Convergence] Let $U \subseteq \mathbb{C}$ be open and $f_n: U \to \mathbb{C}$. The sequence $f_n$ **converges locally uniformly** to $f$ on $U$ if for every $a \in U$, there exists $r > 0$ with $\overline{B(a, r)} \subseteq U$ such that $f_n \to f$ uniformly on $B(a, r)$. [/definition] The condition says: at every point, there is a *neighbourhood* on which the convergence is uniform. The neighbourhood can depend on the point, and typically shrinks near the boundary of $U$. This is weaker than uniform convergence on all of $U$ (which requires a single rate of convergence everywhere), but strong enough for all the applications we need. The following equivalence is often more convenient in practice: [quotetheorem:366] The equivalence uses compactness in both directions: a compact set is covered by finitely many of the local discs (forward direction), and a closed disc is itself compact (reverse direction). [citeproof:366] Because of this equivalence, "locally uniform convergence" and "uniform convergence on compact subsets" (sometimes called "compact convergence") are used interchangeably. The topology this defines on the space of holomorphic functions is called the **compact-open topology**. [example: $z^n \to 0$ Locally Uniformly on $B(0,1)$ but Not Uniformly] The sequence $f_n(z) = z^n$ converges pointwise to $0$ on $B(0, 1)$. For any compact $K \subseteq B(0, 1)$, there exists $\rho < 1$ with $K \subseteq \overline{B(0, \rho)}$. Then $\sup_{z \in K} |z^n| \leq \rho^n \to 0$, so $f_n \to 0$ uniformly on $K$. By the [equivalence theorem](/theorems/366), $f_n \to 0$ locally uniformly on $B(0, 1)$. However, $\sup_{z \in B(0,1)} |z^n| = 1$ for all $n$, so convergence is *not* uniform on $B(0, 1)$. The problem is at the boundary: for each $n$, there are points with $|z|$ close to $1$ where $|z^n|$ is close to $1$. Locally uniform convergence avoids this boundary issue by only requiring uniformity on compact subsets that stay away from $\partial B(0, 1)$. [/example] ## The Main Theorem [quotetheorem:365] This is the central result of the section — and arguably one of the most important structural results in complex analysis. It says that the space of holomorphic functions on $U$ is *closed* under locally uniform limits, and that this closure extends to all derivatives. **Proof of Part (1): Holomorphicity via Morera.** The strategy is to verify [Morera's criterion](/theorems/349): show that $\int_{\partial T} f \, dz = 0$ for every triangle $T \subseteq U$. Since each $f_n$ is holomorphic, [Cauchy's theorem for triangles](/theorems/341) gives $\int_{\partial T} f_n = 0$. Since $T$ is compact and $f_n \to f$ uniformly on compact subsets, we can pass the limit through the integral: \begin{align*} \int_{\partial T} f \, dz = \lim_{n \to \infty} \int_{\partial T} f_n \, dz = 0. \end{align*} The limit $f$ is continuous (as a uniform limit of continuous functions), and all triangle integrals vanish. Morera's theorem concludes that $f$ is holomorphic. **Proof of Part (2): Derivative convergence via Cauchy's formula.** Fix $a \in U$ and choose $r > 0$ with $\overline{B(a, 2r)} \subseteq U$. For any $z \in B(a, r)$, the [Cauchy integral formula](/theorems/345) gives: \begin{align*} f_n'(z) - f'(z) = \frac{1}{2\pi i} \int_{|w-z| = r} \frac{f_n(w) - f(w)}{(w-z)^2} \, dw. \end{align*} The circle $|w - z| = r$ lies inside $\overline{B(a, 2r)}$, where $f_n \to f$ uniformly. Let $\varepsilon_n = \sup_{\overline{B(a,2r)}} |f_n - f| \to 0$. The [ML inequality](/theorems/338) gives: \begin{align*} |f_n'(z) - f'(z)| \leq \frac{1}{2\pi} \cdot 2\pi r \cdot \frac{\varepsilon_n}{r^2} = \frac{\varepsilon_n}{r}, \end{align*} uniformly for $z \in B(a, r)$. Since $a$ was arbitrary, $f_n' \to f'$ locally uniformly. [citeproof:365] Iterating Part (2) — since $f_n'$ are holomorphic and $f_n' \to f'$ locally uniformly, we can apply the theorem again — gives $f_n'' \to f''$ locally uniformly, and by induction $f_n^{(k)} \to f^{(k)}$ locally uniformly for every $k \geq 0$. ### The Real-Variable Contrast The theorem fails completely for real $C^\infty$ functions. Consider $f_n(x) = \frac{1}{n}\sin(n^2 x)$ on $\mathbb{R}$. Then $f_n \to 0$ uniformly ($|f_n| \leq 1/n$), $0$ is $C^\infty$, yet $f_n'(x) = n\cos(n^2 x)$ does not converge anywhere. The Weierstrass function $\sum 2^{-n}\cos(3^n \pi x)$ is a uniform limit of smooth partial sums but is continuous and *nowhere* differentiable. In complex analysis, such pathologies cannot occur: the Cauchy integral formula provides automatic derivative control. ## Applications ### Convergence of Power Series If $f(z) = \sum_{n=0}^\infty c_n(z-a)^n$ has radius of convergence $R > 0$, the partial sums $S_N(z) = \sum_{n=0}^N c_n(z-a)^n$ are polynomials (hence entire, a fortiori holomorphic on $B(a, R)$) converging locally uniformly on $B(a, R)$. The [main theorem](/theorems/365) immediately gives: - $f$ is holomorphic on $B(a, R)$, - $f'(z) = \sum_{n=1}^\infty nc_n(z-a)^{n-1}$ (term-by-term differentiation), with locally uniform convergence. This recovers the [Holomorphicity of Power Series](/theorems/335) from §1 as a special case. ### Normally Convergent Series A series $\sum g_n(z)$ of holomorphic functions **converges normally** on $U$ if for every compact $K \subseteq U$, $\sum \sup_K |g_n| < \infty$. Normal convergence implies locally uniform convergence (by the Weierstrass M-test applied on each compact set). The main theorem then guarantees that the sum is holomorphic and can be differentiated term by term. [example: The Riemann Zeta Function for $\operatorname{Re} s > 1$] The Dirichlet series $\zeta(s) = \sum_{n=1}^\infty n^{-s}$ converges normally on every half-plane $\{\operatorname{Re} s > 1 + \delta\}$ for $\delta > 0$ (since $|n^{-s}| = n^{-\operatorname{Re} s} \leq n^{-(1+\delta)}$ and $\sum n^{-(1+\delta)} < \infty$). Each $n^{-s} = e^{-s\log n}$ is entire, so the [main theorem](/theorems/365) gives: $\zeta(s)$ is holomorphic on $\{\operatorname{Re} s > 1\}$ with $\zeta'(s) = -\sum_{n=1}^\infty \frac{\log n}{n^s}$. [/example] ### Justifying Limit Interchange in Contour Integrals Many constructions in earlier sections implicitly used the main theorem. For instance, in the proof of [Taylor's theorem](/theorems/348), we interchanged a sum and an integral: \begin{align*} f(z) = \frac{1}{2\pi i} \int_{|w-a| = \rho} \frac{f(w)}{w - z} \, dw = \frac{1}{2\pi i} \int_{|w-a|=\rho} \sum_{n=0}^\infty \frac{(z-a)^n}{(w-a)^{n+1}} f(w) \, dw = \sum_{n=0}^\infty c_n(z-a)^n. \end{align*} The interchange of sum and integral is justified by uniform convergence of the geometric series on the compact circle $|w-a| = \rho$. ### Limits of Meromorphic Functions If $f_n$ are meromorphic on $U$ (holomorphic except for isolated poles) and $f_n \to f$ locally uniformly on $U \setminus S$ (where $S$ is a discrete set), the limit $f$ is holomorphic on $U \setminus S$. Under additional conditions (e.g., if the poles of $f_n$ converge to fixed points in $S$), $f$ extends to a meromorphic function on $U$. This is used in the theory of modular forms and automorphic functions. ## Worked Example [problem] Let $f_n(z) = \sum_{k=0}^n \frac{z^k}{k!}$ be the partial sums of $e^z$. Show that $f_n \to e^z$ locally uniformly on $\mathbb{C}$, and deduce that $\frac{d}{dz}e^z = e^z$ using the main theorem (without using the power series differentiation result directly). [/problem] [solution] **Step 1: Locally uniform convergence.** Fix a compact set $K \subseteq \mathbb{C}$. Then $R = \max_{z \in K} |z| < \infty$. For $z \in K$: \begin{align*} |e^z - f_n(z)| = \left|\sum_{k=n+1}^\infty \frac{z^k}{k!}\right| \leq \sum_{k=n+1}^\infty \frac{R^k}{k!} \to 0 \quad \text{as } n \to \infty, \end{align*} since the tail of a convergent series tends to $0$. The bound is uniform in $z \in K$, so $f_n \to e^z$ uniformly on $K$. By the [equivalence theorem](/theorems/366), $f_n \to e^z$ locally uniformly on $\mathbb{C}$. **Step 2: Apply the main theorem.** Each $f_n$ is a polynomial, hence entire. By [preservation of holomorphicity](/theorems/365), the limit $e^z$ is holomorphic on $\mathbb{C}$ (which we already know), and $f_n' \to (e^z)'$ locally uniformly. But $f_n'(z) = \sum_{k=1}^n \frac{z^{k-1}}{(k-1)!} = \sum_{k=0}^{n-1} \frac{z^k}{k!} = f_{n-1}(z)$. So $f_{n-1} \to (e^z)'$ locally uniformly. Since also $f_{n-1} \to e^z$ locally uniformly (relabelling), we conclude $(e^z)' = e^z$. The argument generalises: for any power series $f(z) = \sum c_n z^n$ with positive radius of convergence, the main theorem gives $f'(z) = \sum nc_n z^{n-1}$ — term-by-term differentiation is valid because partial sums converge locally uniformly and are holomorphic. [/solution] ## References 1. Sheratt, N., *Cambridge Part IB — Complex Analysis*, Lecture Notes. # Example Sheets This section collects problems that synthesise the ideas developed across the course — complex differentiation, contour integration, residue calculus, transform theory, and uniform limits. The problems are organised by theme and chosen to highlight the key techniques and subtleties of the subject. Each problem illustrates how the abstract machinery applies in concrete settings, and several reveal surprising rigidity phenomena that distinguish complex analysis from its real counterpart. ## Cauchy–Riemann Equations and Holomorphicity These problems probe the boundary between real differentiability and complex differentiability, and the rigidity that holomorphicity imposes. [problem] Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a real-linear map. Regarding $T$ as a map $\mathbb{C} \to \mathbb{C}$, show there exist unique $A, B \in \mathbb{C}$ with $T(z) = Az + B\bar{z}$, and that $T$ is complex-differentiable if and only if $B = 0$. [/problem] [solution] Since $\{1, i\}$ is an $\mathbb{R}$-basis for $\mathbb{C}$, $T$ is determined by $T(1) = \alpha$ and $T(i) = \beta$. For any $z = x + iy$: $T(z) = xT(1) + yT(i) = x\alpha + y\beta$. Write $z = x + iy$ and $\bar{z} = x - iy$, so $x = (z + \bar{z})/2$ and $y = (z - \bar{z})/(2i)$. Then: \begin{align*} T(z) = \frac{\alpha}{2}(z + \bar{z}) + \frac{\beta}{2i}(z - \bar{z}) = \frac{\alpha - i\beta}{2}z + \frac{\alpha + i\beta}{2}\bar{z} = Az + B\bar{z}, \end{align*} where $A = (\alpha - i\beta)/2$ and $B = (\alpha + i\beta)/2$. Uniqueness: if $Az + B\bar{z} = A'z + B'\bar{z}$ for all $z$, set $z = 1$: $A + B = A' + B'$; set $z = i$: $iA - iB = iA' - iB'$, giving $A = A'$, $B = B'$. Write $T(z) = Az + B\bar{z}$, i.e., $T(x + iy) = (u(x,y), v(x,y))$ where $u + iv = Az + B\bar{z}$. The [Cauchy–Riemann equations](/theorems/333) require $u_x = v_y$ and $u_y = -v_x$. Computing: $u_x + iv_x = A + B$ and $u_y + iv_y = i(A - B)$. The CR equations $u_x = v_y$ and $u_y = -v_x$ are equivalent to $A + B = -i \cdot i(A - B) = A - B + 2B$... more directly, $T$ is $\mathbb{C}$-linear iff $T(iz) = iT(z)$, i.e., $A(iz) + B\overline{iz} = i(Az + B\bar{z})$. The left side is $iAz - iB\bar{z}$; the right is $iAz + iB\bar{z}$. These are equal for all $z$ iff $B = 0$. $\square$ [/solution] [problem] Let $f: U \to \mathbb{C}$ be holomorphic and non-constant on a domain $U$. Show that none of $\operatorname{Re}(f)$, $|f|$, or $\arg(f)$ can be constant on $U$. [/problem] [solution] If $\operatorname{Re}(f) = c$, write $f = c + iv$. The [CR equations](/theorems/333) give $0 = u_x = v_y$ and $0 = u_y = -v_x$, so $v_x = v_y = 0$, hence $v$ is constant and $f$ is constant. Contradiction. If $|f| = c$: if $c = 0$ then $f \equiv 0$. If $c > 0$, then $f$ never vanishes, so $|f|^2 = f\bar{f} = c^2$ gives $\bar{f} = c^2/f$. But $\bar{f}$ is anti-holomorphic while $c^2/f$ is holomorphic (since $f \neq 0$), so $\bar{f}$ is both holomorphic and anti-holomorphic, hence constant, so $f$ is constant. Alternatively, the [Open Mapping Theorem](/theorems/358) gives: $f(U)$ is open, but $\{|w| = c\}$ contains no open set, contradiction. If $\arg(f) = \theta_0$ (constant), then $f(U) \subseteq \{re^{i\theta_0} : r > 0\}$, a ray — but by the [Open Mapping Theorem](/theorems/358), $f(U)$ must be open, and a ray contains no open set. So $f$ is constant. $\square$ [/solution] [problem] Show that the only entire functions of the form $f(x + iy) = u(x) + iv(y)$ (with $u, v$ depending on one real variable each) are the functions $f(z) = az + b$ for $a \in \mathbb{R}$, $b \in \mathbb{C}$. [/problem] [solution] The [CR equations](/theorems/333): $u'(x) = v'(y)$. The left side depends only on $x$, the right only on $y$, so both equal a constant $a \in \mathbb{R}$. Then $u(x) = ax + c_1$ and $v(y) = ay + c_2$, giving $f(z) = a(x + iy) + (c_1 + ic_2) = az + b$ with $b = c_1 + ic_2$. [/solution] [problem] Define $f: \mathbb{C} \to \mathbb{C}$ by $f(0) = 0$ and $f(z) = \frac{(1+i)x^3 - (1-i)y^3}{x^2 + y^2}$ for $z = x + iy \neq 0$. Show that $f$ satisfies the Cauchy–Riemann equations at $z = 0$ but is not complex-differentiable there. [/problem] [solution] Write $f = u + iv$ with $u(x,y) = (x^3 - y^3)/(x^2 + y^2)$ and $v(x,y) = (x^3 + y^3)/(x^2 + y^2)$ for $(x,y) \neq (0,0)$, and $u(0,0) = v(0,0) = 0$. The partial derivatives at the origin: \begin{align*} u_x(0,0) = \lim_{h \to 0} \frac{u(h, 0)}{h} = \lim_{h \to 0} \frac{h^3/h^2}{h} = 1, \quad u_y(0,0) = \lim_{h \to 0} \frac{u(0,h)}{h} = \lim_{h \to 0} \frac{-h^3/h^2}{h} = -1. \end{align*} Similarly $v_x(0,0) = 1$ and $v_y(0,0) = 1$. The CR equations $u_x = v_y$ ($1 = 1$ ✓) and $u_y = -v_x$ ($-1 = -1$ ✓) hold. However, $f$ is not complex-differentiable at $0$. If it were, $f'(0)$ would equal $u_x + iv_x = 1 + i$. But along the line $y = x$: $f(z)/z = f(t + it)/(t + it) = \frac{(1+i)t^3 - (1-i)t^3}{2t^2} \cdot \frac{1}{t(1+i)} = \frac{2it}{2t^2(1+i)} = \frac{i}{t(1+i)}$, which diverges as $t \to 0$, so $f(z)/z$ has no limit. The CR equations hold but $f$ is not even continuous at the origin along this ray, so complex differentiability fails. This illustrates that CR equations plus real-differentiability (not just existence of partial derivatives) is the correct condition for the [Cauchy–Riemann characterisation](/theorems/333). $\square$ [/solution] ## Harmonic Functions and Conjugates [problem] Show that $u(x, y) = \log\sqrt{x^2 + y^2}$ is harmonic on $\mathbb{C}^*$ but is not the real part of any holomorphic function on $\mathbb{C}^*$. [/problem] [solution] $u = \frac{1}{2}\log(x^2 + y^2)$. Compute: $u_x = x/(x^2+y^2)$, $u_{xx} = (y^2 - x^2)/(x^2+y^2)^2$, and by symmetry $u_{yy} = (x^2 - y^2)/(x^2+y^2)^2$. So $u_{xx} + u_{yy} = 0$ — harmonic on $\mathbb{C}^*$. If $u = \operatorname{Re}(F)$ for some holomorphic $F$ on $\mathbb{C}^*$, then $F'(z) = u_x - iu_y = \frac{x - iy}{x^2 + y^2} = \frac{\bar{z}}{|z|^2} = \frac{1}{z}$. So $F$ would be an antiderivative of $1/z$ on $\mathbb{C}^*$. But $\int_{|z|=1} dz/z = 2\pi i \neq 0$, so by the [Antiderivative Existence Characterisation](/theorems/340), no such antiderivative exists on $\mathbb{C}^*$. On any simply connected subdomain of $\mathbb{C}^*$ (e.g., $\mathbb{C} \setminus (-\infty, 0]$), the antiderivative does exist: $F(z) = \operatorname{Log} z = \log|z| + i\operatorname{Arg}(z)$, and indeed $\operatorname{Re}(\operatorname{Log} z) = \log|z| = u$. The obstruction is purely topological: $\mathbb{C}^*$ is not simply connected. $\square$ [/solution] ## Antiderivatives and Path Dependence [problem] Show that neither (a) $f(z) = 1/z - 1/(z-1)$ on $\mathbb{C} \setminus \{0, 1\}$ nor (b) $g(z) = z/(1 + z^2)$ on $\{|z| > 1\}$ has an antiderivative on its domain. [/problem] [solution] **(a)** Integrate around $|z| = 1/2$ (enclosing $z = 0$ but not $z = 1$). The [residue theorem](/theorems/352) gives: \begin{align*} \int_{|z|=1/2} f(z) \, dz = 2\pi i \bigl(\operatorname{Res}(f, 0)\bigr) = 2\pi i \cdot 1 = 2\pi i \neq 0. \end{align*} (The residue of $-1/(z-1)$ at $z = 0$ is $0$ since $z = 0$ is not a singularity of $1/(z-1)$.) By the [Fundamental Theorem of Contour Integration](/theorems/339), no antiderivative exists. **(b)** Write $g(z) = z/(z^2+1) = \frac{1}{2}\left(\frac{1}{z-i} + \frac{1}{z+i}\right)$. On $|z| = 2$ (inside the domain $\{|z| > 1\}$): \begin{align*} \int_{|z|=2} g(z) \, dz = 2\pi i\bigl(\operatorname{Res}(g, i) + \operatorname{Res}(g, -i)\bigr) = 2\pi i\left(\frac{1}{2} + \frac{1}{2}\right) = 2\pi i \neq 0. \end{align*} Both poles $z = \pm i$ lie inside $|z| = 2$, which is a valid closed path in $\{|z| > 1\}$. So no antiderivative exists. $\square$ [/solution] ## Laurent Expansions [problem] Find the Laurent expansion of $f(z) = \frac{1}{(z-1)(z-2)}$ in each of the regions: (i) $|z| < 1$, (ii) $1 < |z| < 2$, (iii) $|z| > 2$. [/problem] [solution] Partial fractions: $f(z) = \frac{-1}{z - 1} + \frac{1}{z - 2}$. **(i) $|z| < 1$.** Both $|z/1| < 1$ and $|z/2| < 1$. Expand each term as a convergent geometric series: \begin{align*} \frac{-1}{z-1} = \frac{1}{1-z} = \sum_{n=0}^\infty z^n, \qquad \frac{1}{z-2} = \frac{-1}{2}\cdot\frac{1}{1 - z/2} = -\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}. \end{align*} So $f(z) = \sum_{n=0}^\infty (1 - 2^{-(n+1)})z^n$. (No negative powers — $f$ is holomorphic on $|z| < 1$.) **(ii) $1 < |z| < 2$.** Now $|1/z| < 1$ and $|z/2| < 1$: \begin{align*} \frac{-1}{z-1} = \frac{-1}{z}\cdot\frac{1}{1 - 1/z} = -\sum_{n=0}^\infty \frac{1}{z^{n+1}} = -\sum_{n=1}^\infty z^{-n}, \end{align*} and $\frac{1}{z-2} = -\sum_{n=0}^\infty z^n/2^{n+1}$ as before. So $f(z) = -\sum_{n=1}^\infty z^{-n} - \sum_{n=0}^\infty z^n/2^{n+1}$. The principal part $-\sum z^{-n}$ reflects the pole at $z = 1$ (now inside the inner boundary). **(iii) $|z| > 2$.** Both $|1/z| < 1$ and $|2/z| < 1$: \begin{align*} \frac{1}{z-2} = \frac{1}{z}\cdot\frac{1}{1-2/z} = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}} = \sum_{n=1}^\infty \frac{2^{n-1}}{z^n}. \end{align*} Combined: $f(z) = \sum_{n=1}^\infty (2^{n-1} - 1)z^{-n}$. (Only negative powers — $f$ vanishes at $\infty$.) $\square$ [/solution] ## Growth Conditions and Rigidity [quotetheorem:367] Part (1) uses Cauchy's estimate $|a_n| \leq M(1+R)^k/R^n \to 0$ for $n > k$ as $R \to \infty$, killing all high-degree Taylor coefficients. Part (2) uses the Casorati–Weierstrass theorem to rule out essential singularities at infinity. [citeproof:367] [quotetheorem:368] The proof defines $g(z) = f(z)/z$ (removable singularity at $0$), applies the maximum modulus principle on discs $|z| \leq r < 1$ to get $|g| \leq 1/r$, and sends $r \to 1^-$. Equality forces $|g|$ to attain its maximum in the interior, making $g$ constant by the maximum modulus principle. [citeproof:368] [problem] Let $f$ be entire with $|f(z)| \leq 3 + 2|z|^{3/2}$ for all $z$. Show that $f$ is a polynomial of degree at most $1$. [/problem] [solution] By Cauchy's estimate for the $n$-th Taylor coefficient on $|z| = R$: \begin{align*} |a_n| \leq \frac{\sup_{|z|=R} |f(z)|}{R^n} \leq \frac{3 + 2R^{3/2}}{R^n}. \end{align*} For $n \geq 2$: $(3 + 2R^{3/2})/R^n \leq 3/R^n + 2/R^{n - 3/2} \to 0$ as $R \to \infty$ (since $n \geq 2 > 3/2$). So $a_n = 0$ for $n \geq 2$. Therefore $f(z) = a_0 + a_1 z$. Note: the bound $|z|^{3/2}$ is *not* polynomial growth, but Cauchy's estimate still applies. The key is that $3/2 < 2$, so the growth is sub-quadratic, killing the $z^2$ coefficient and above. $\square$ [/solution] ## The Open Mapping Theorem and Maximum Principle [problem] Let $f$ be holomorphic on a domain $U$ with $\operatorname{Re}(f)$ having a local maximum at $a \in U$. Show $f$ is constant. [/problem] [solution] Consider $g(z) = e^{f(z)}$, which is holomorphic and non-vanishing on $U$. Then $|g(z)| = e^{\operatorname{Re}(f(z))}$. If $\operatorname{Re}(f)$ has a local maximum at $a$, then $|g|$ has a local maximum at $a$. By the maximum modulus principle (which follows from the [Open Mapping Theorem](/theorems/358): if $g$ is non-constant, $g(U)$ is open, so $|g(a)|$ cannot be a local maximum of $|g|$ on $U$), $g$ is constant. Since $g = e^f$ is constant and the exponential is injective on any horizontal strip of height $< 2\pi$, $f$ is constant (locally, hence globally by the identity theorem). $\square$ [/solution] ## Winding Numbers and Rouché Applications [problem] Let $\gamma, \delta: [0,1] \to \mathbb{C}$ be closed curves with $|\gamma(t) - \delta(t)| < |\gamma(t) - w|$ for all $t$ and some $w \notin \gamma^* \cup \delta^*$. Show $I(\gamma, w) = I(\delta, w)$. [/problem] [solution] Define $\sigma(t) = (\delta(t) - w)/(\gamma(t) - w)$. This is well-defined and continuous since $\gamma(t) \neq w$. For each $t$: \begin{align*} |\sigma(t) - 1| = \frac{|\delta(t) - \gamma(t)|}{|\gamma(t) - w|} < 1, \end{align*} so $\sigma(t) \in B(1, 1)$, which does not contain $0$. Therefore $I(\sigma, 0) = 0$ (the image of $\sigma$ lies in a simply connected region avoiding $0$). By the multiplicativity of winding numbers: $I(\sigma, 0) = I(\delta - w, 0) - I(\gamma - w, 0) = I(\delta, w) - I(\gamma, w)$. So $I(\delta, w) = I(\gamma, w)$. This is essentially [Rouché's theorem](/theorems/357) in disguise: the condition $|\gamma - \delta| < |\gamma - w|$ says the "perturbation" $\delta - \gamma$ is smaller than the "dominant term" $\gamma - w$, so the winding number is unchanged. $\square$ [/solution] [problem] Show that all roots of $p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0$ (monic, degree $n$) lie in $|z| < A + 1$, where $A = \max_j |a_j|$. [/problem] [solution] On $|z| = A + 1$: $|z^n| = (A+1)^n$. For the remaining terms: \begin{align*} |a_{n-1}z^{n-1} + \cdots + a_0| \leq A\sum_{k=0}^{n-1}(A+1)^k = A \cdot \frac{(A+1)^n - 1}{A} = (A+1)^n - 1 < (A+1)^n = |z^n|. \end{align*} By [Rouché's theorem](/theorems/357) with $f(z) = z^n$ and $g(z) = a_{n-1}z^{n-1} + \cdots + a_0$: $p = f + g$ has the same number of zeros as $z^n$ inside $|z| < A + 1$, namely $n$. So all $n$ roots lie in $|z| < A + 1$. $\square$ [/solution] ## Natural Boundaries [problem] Show that $f(z) = \sum_{n=0}^\infty z^{2^n}$ is holomorphic on $B(0,1)$ but cannot be analytically continued past any point of $|z| = 1$. [/problem] [solution] **Holomorphicity on $B(0,1)$:** For compact $K \subseteq B(0,1)$, let $\rho = \max_K |z| < 1$. Then $\sup_K |z^{2^n}| = \rho^{2^n}$ and $\sum \rho^{2^n} < \infty$ (the terms decrease super-exponentially). By the Weierstrass M-test, the series converges normally on $B(0,1)$, and the [main theorem on uniform limits](/theorems/365) gives holomorphicity. **Natural boundary:** The $2^k$-th roots of unity $\omega = e^{2\pi i p/2^k}$ (with $p \in \mathbb{Z}$, $k \geq 0$) are dense on $|z| = 1$. For such $\omega$, the partial sum $\sum_{n=0}^{k-1} \omega^{2^n}$ is bounded, but for $n \geq k$: $\omega^{2^n} = e^{2\pi i p \cdot 2^{n-k}} = 1$ (since $2^{n-k}$ is a positive integer). So $\sum_{n=k}^N \omega^{2^n} = N - k + 1 \to \infty$, and the series diverges at $\omega$. If $f$ could be analytically continued past some arc of $|z| = 1$, it would be continuous (hence bounded) on a compact subset of that arc. But the divergence at a dense set of points on $|z| = 1$ shows that $f(r\omega) \to \infty$ as $r \to 1^-$ for each such $\omega$, preventing bounded extension to any arc. Therefore $|z| = 1$ is a **natural boundary**: no analytic continuation exists past any point of the unit circle. $\square$ [/solution] ## Contour Integral Inversions [problem] Let $f$ be holomorphic on a domain $U$ with $f'(a) \neq 0$. Show that for $r > 0$ small enough, the function \begin{align*} g(w) = \frac{1}{2\pi i}\int_{|z - a| = r} \frac{z f'(z)}{f(z) - w} \, dz \end{align*} defines a holomorphic local inverse to $f$ near $w_0 = f(a)$. [/problem] [solution] Since $f'(a) \neq 0$, the [Local Degree Theorem](/theorems/359) gives $\deg(f, a) = 1$: for $r$ small enough, $f$ is injective on $B(a, r)$, and every $w$ near $w_0$ has exactly one preimage $z(w) \in B(a, r)$. The integrand $zf'(z)/(f(z) - w)$ has a simple pole at $z = z(w)$ (where $f(z) = w$) with residue: \begin{align*} \operatorname{Res}_{z = z(w)} \frac{zf'(z)}{f(z) - w} = \frac{z(w) \cdot f'(z(w))}{f'(z(w))} = z(w). \end{align*} (Using [residue formula (2)](/theorems/353) with $g(z) = zf'(z)$ and $h(z) = f(z) - w$, since $h$ has a simple zero at $z(w)$.) By the [residue theorem](/theorems/352): $g(w) = z(w)$, the unique preimage. Holomorphicity of $g$: for $w$ near $w_0$, the pole $z(w)$ stays inside $|z - a| = r$ (by the Local Degree Theorem), so the integrand is a continuous function of $(z, w)$ on the compact circle $|z - a| = r$. Differentiating under the integral sign (justified by uniform convergence) gives $g'(w) = \frac{1}{2\pi i}\int_{|z-a|=r} \frac{zf'(z)}{(f(z) - w)^2} dz$, so $g$ is holomorphic. Finally, $f(g(w)) = f(z(w)) = w$ by construction, so $g$ is a local inverse. $\square$ [/solution] ## References 1. Sheratt, N., *Cambridge Part IB — Complex Analysis*, Lecture Notes and Example Sheets.