In real analysis, the equation $e^y = x$ has a unique solution $y = \ln x$ for each $x > 0$: the real exponential is a bijection from $\mathbb{R}$ onto $(0, \infty)$, and its inverse is the natural logarithm. In complex analysis, the situation is fundamentally different. The complex exponential $\exp : \mathbb{C} \to \mathbb{C} \setminus \{0\}$ is surjective but not injective — every nonzero complex number has infinitely many preimages, differing by integer multiples of $2\pi i$. The "complex logarithm" is therefore not a function in the classical sense but a *multivalued* object, and the same problem afflicts square roots, arbitrary powers, and inverse trigonometric [functions](/page/Function). The theory of branch cuts provides a systematic way to extract single-valued holomorphic functions from these multivalued objects. The idea is geometric: by removing a carefully chosen curve from the domain — the branch cut — one destroys the closed paths that would force the function to take multiple values, and the resulting restricted domain admits a well-defined holomorphic "branch." This interplay between topology (the fundamental group of the punctured domain) and analysis (holomorphicity of the selected branch) is one of the distinctive features of complex function theory. ## Motivation [motivation] ### Inverting the Exponential The complex exponential is the entire function \begin{align*} \exp : \mathbb{C} &\to \mathbb{C} \setminus \{0\} \\ w &\mapsto e^w. \end{align*} It is periodic with period $2\pi i$: for every $w \in \mathbb{C}$ and every $k \in \mathbb{Z}$, $\exp(w + 2\pi i k) = \exp(w)$. In particular, $\exp$ is not injective — each value in $\mathbb{C} \setminus \{0\}$ has infinitely many preimages. If $z = r e^{i\theta}$ with $r > 0$ and $\theta \in \mathbb{R}$, then every solution of $e^w = z$ has the form \begin{align*} w = \ln r + i(\theta + 2\pi k), \qquad k \in \mathbb{Z}, \end{align*} where $\ln r$ denotes the ordinary real logarithm of the positive number $r$. The [set](/page/Set) of all "logarithms of $z$" is the infinite discrete set $\{\ln r + i(\theta + 2\pi k) : k \in \mathbb{Z}\}$ — a vertical arithmetic progression in the complex plane with common difference $2\pi i$. To define a "logarithm function," we must select exactly one of these infinitely many values for each $z \in \mathbb{C} \setminus \{0\}$. The question is whether this selection can be made continuously — and, ideally, holomorphically. ### The [Continuity](/page/Continuity) Obstruction Suppose, seeking a contradiction, that there exists a continuous function $L : \mathbb{C} \setminus \{0\} \to \mathbb{C}$ satisfying $\exp(L(z)) = z$ for all $z \neq 0$. Consider the unit circle parametrised by $\gamma(t) = e^{it}$ for $t \in [0, 2\pi]$. The composition $t \mapsto L(e^{it})$ is a continuous function with \begin{align*} \exp(L(e^{it})) = e^{it} \qquad \text{for all } t \in [0, 2\pi]. \end{align*} Since $L(e^{it})$ is a continuous logarithm of $e^{it}$, it must have the form $L(e^{it}) = it + 2\pi i k(t)$ for some function $k : [0, 2\pi] \to \mathbb{Z}$. But $L$ is continuous and $k(t)$ is integer-valued and continuous, so $k$ is constant: $k(t) = k_0$ for all $t$. Evaluating at the endpoints: \begin{align*} L(e^{i \cdot 0}) &= L(1) = 0 + 2\pi i k_0 = 2\pi i k_0, \\ L(e^{i \cdot 2\pi}) &= L(1) = 2\pi i + 2\pi i k_0. \end{align*} The first equation gives $L(1) = 2\pi i k_0$ and the second gives $L(1) = 2\pi i(k_0 + 1)$. These are contradictory. The [topological](/page/Topology) explanation is that $\mathbb{C} \setminus \{0\}$ has nontrivial fundamental group $\pi_1(\mathbb{C} \setminus \{0\}) \cong \mathbb{Z}$: there exist closed curves (such as the unit circle) that cannot be continuously contracted to a point. The logarithm "detects" these curves via the winding number: each trip around the origin shifts the imaginary part of $\log z$ by $2\pi$. ### The Resolution: Cutting the Domain The obstruction above relies on the existence of closed curves encircling the origin. If we remove a curve connecting the origin to infinity — say, the closed ray $(-\infty, 0]$ — the resulting domain $\mathbb{C} \setminus (-\infty, 0]$ is simply connected: every closed curve in this domain can be continuously deformed to a point without leaving the domain. In particular, no closed curve in $\mathbb{C} \setminus (-\infty, 0]$ winds around the origin, so the obstruction vanishes. On this simply connected domain, we can make a continuous — in fact holomorphic — selection of $\log z$ by restricting the argument to the interval $(-\pi, \pi)$. The removed ray $(-\infty, 0]$ is the *branch cut*, the origin is the *branch point*, and the resulting single-valued holomorphic function is a *branch* of the logarithm. [/motivation] ## Definition Before giving formal definitions, we need to make precise the three concepts that appeared in the motivating discussion: the point around which values change (the branch point), the curve we remove (the branch cut), and the single-valued function that results (the branch). These are logically distinct notions — the branch point is a property of the multivalued function, the branch cut is a choice of surgery on the domain, and the branch is the holomorphic function that lives on the surgered domain. A branch point is characterised by what happens when one analytically continues a function element along a closed curve. If the function returns to a different value after a full circuit, the enclosed singularity is a branch point. The formal definition requires the notion of analytic continuation along a path, which we take as given. [definition: Branch Point] Let $f$ be a multivalued analytic function on $\Omega \subseteq \mathbb{C}$ (that is, a collection of function elements connected by analytic continuation). A point $z_0 \in \mathbb{C}$ is a **branch point** of $f$ if there exists a closed curve $\gamma : [0,1] \to \Omega \setminus \{z_0\}$ with $\gamma(0) = \gamma(1)$ and $n(\gamma, z_0) \neq 0$ (where $n(\gamma, z_0)$ denotes the winding number of $\gamma$ around $z_0$) such that analytic continuation of some function element of $f$ along $\gamma$ returns a different function element at $\gamma(1) = \gamma(0)$. [/definition] The winding number condition $n(\gamma, z_0) \neq 0$ ensures that $\gamma$ genuinely encircles $z_0$; curves that avoid the neighborhood of $z_0$ entirely will not detect the branching. For the complex logarithm, the origin is a branch point: continuing $\log z$ once around the origin shifts the value by $2\pi i$. For $z^{1/n}$, the origin is also a branch point, but the continuation is periodic — after $n$ circuits, the function returns to its original value. Such a branch point is called a *branch point of order $n - 1$* (or equivalently, the function has $n$ sheets). The logarithm, by contrast, has an *infinite-order* branch point (or *logarithmic branch point*), since no finite number of circuits returns the function to its starting value. The next step is to remove enough of the domain to eliminate the problematic closed curves. A branch cut is a curve whose removal makes it impossible to wind around the branch point. [definition: Branch Cut] Let $f$ be a multivalued analytic function with a branch point at $z_0 \in \mathbb{C}$. A **branch cut** for $f$ at $z_0$ is a simple arc $\Gamma \subset \mathbb{C}$ with one endpoint at $z_0$ and extending to another branch point or to infinity, such that the complement $\Omega \setminus \Gamma$ (where $\Omega$ is the domain of multivaluedness) is a domain on which a single-valued branch of $f$ can be defined. [/definition] The branch cut is not unique — any arc from $z_0$ to another branch point or to infinity will do, provided the complement remains connected. For the logarithm, the standard choice is the ray $(-\infty, 0]$, but the positive real axis, a spiral, or even a wiggly curve from $0$ to $\infty$ would serve equally well. The choice is a matter of convention and computational convenience, not of mathematical necessity. Finally, a branch is the holomorphic function that results from making the cut and selecting a value. [definition: Branch] Let $f$ be a multivalued analytic function on $\Omega$ and let $\Gamma$ be a branch cut for $f$. A **branch** of $f$ on $\Omega \setminus \Gamma$ is a holomorphic function $F : \Omega \setminus \Gamma \to \mathbb{C}$ such that $F(z)$ is one of the values of $f$ at each point $z \in \Omega \setminus \Gamma$. [/definition] Different branches on the same cut domain differ by the "sheet" they lie on. For the logarithm, the branches on $\mathbb{C} \setminus (-\infty, 0]$ are the functions $\log_k(z) = \ln|z| + i(\operatorname{Arg}(z) + 2\pi k)$ for $k \in \mathbb{Z}$, where $\operatorname{Arg}(z) \in (-\pi, \pi)$ is the principal argument. Each branch is holomorphic on the cut plane, and any two branches differ by the constant $2\pi i k$. ## The Complex Logarithm ### The Principal Branch The most important multivalued function in complex analysis is the logarithm. To extract a holomorphic branch, we must choose a branch cut (removing a ray from the origin) and then select a consistent value of the argument on the resulting domain. The most common convention is to cut along the negative real axis $(-\infty, 0]$ and restrict the argument to the interval $(-\pi, \pi)$. [definition: Principal Argument] The **principal argument** is the function \begin{align*} \operatorname{Arg} : \mathbb{C} \setminus (-\infty, 0] &\to (-\pi, \pi) \\ z = re^{i\theta} &\mapsto \theta, \qquad \text{where } \theta \in (-\pi, \pi) \text{ and } r > 0. \end{align*} [/definition] The restriction to $(-\pi, \pi)$ rather than $(-\pi, \pi]$ ensures that $\operatorname{Arg}$ is continuous on $\mathbb{C} \setminus (-\infty, 0]$: approaching the negative real axis from above gives arguments near $\pi$, while approaching from below gives arguments near $-\pi$. Including the endpoint $\pi$ would create a discontinuity at the cut. With this convention in hand, the principal logarithm is defined by combining the real logarithm of the modulus with the principal argument. [definition: Principal Logarithm] The **principal logarithm** (or **principal branch of the logarithm**) is the function \begin{align*} \operatorname{Log} : \mathbb{C} \setminus (-\infty, 0] &\to \{w \in \mathbb{C} : -\pi < \operatorname{Im}(w) < \pi\} \\ z &\mapsto \ln|z| + i\operatorname{Arg}(z). \end{align*} [/definition] The principal logarithm satisfies $\exp(\operatorname{Log}(z)) = z$ for every $z \in \mathbb{C} \setminus (-\infty, 0]$, and its image is the horizontal strip $\{w \in \mathbb{C} : -\pi < \operatorname{Im}(w) < \pi\}$. On the positive real axis, $\operatorname{Log}(x) = \ln x$ for $x > 0$, recovering the real logarithm. The key analytic property is holomorphicity, which we now establish. [theorem: Holomorphicity Of The Principal Logarithm] The principal logarithm $\operatorname{Log} : \mathbb{C} \setminus (-\infty, 0] \to \mathbb{C}$ is holomorphic, with derivative \begin{align*} \frac{d}{dz} \operatorname{Log}(z) = \frac{1}{z} \qquad \text{for all } z \in \mathbb{C} \setminus (-\infty, 0]. \end{align*} [/theorem] The proof proceeds by verifying that $\operatorname{Log}$ is a local inverse of $\exp$. Since $\exp : \{w : -\pi < \operatorname{Im}(w) < \pi\} \to \mathbb{C} \setminus (-\infty, 0]$ is a bijection with $\exp'(w) = e^w \neq 0$, the [inverse function theorem](/page/Inverse%20Function%20Theorem) for holomorphic maps guarantees that $\operatorname{Log} = \exp^{-1}$ is holomorphic with $\operatorname{Log}'(z) = 1/\exp'(\operatorname{Log}(z)) = 1/z$. Alternatively, one can verify the Cauchy–Riemann equations directly: writing $z = re^{i\theta}$ and $\operatorname{Log}(z) = u(r, \theta) + iv(r, \theta)$ with $u = \ln r$ and $v = \theta$, the polar Cauchy–Riemann equations $\partial_r u = (1/r)\partial_\theta v$ and $(1/r)\partial_\theta u = -\partial_r v$ reduce to $1/r = 1/r$ and $0 = 0$. The [derivative](/page/Derivative) formula $\operatorname{Log}'(z) = 1/z$ is exactly what one would expect from the real case, and it holds on the entire cut plane. The formula is independent of the branch: every branch of $\log z$ has derivative $1/z$, since any two branches differ by a constant. This is why the antiderivative of $1/z$ is "the logarithm" — the derivative does not see the branch ambiguity. ### Discontinuity Across the Branch Cut The branch cut $(-\infty, 0]$ is not merely a formal device — the principal logarithm has a genuine discontinuity there. To see this, approach a point $x_0 < 0$ on the negative real axis from above and below. Writing $x_0 = |x_0| e^{i\pi}$ (approached from above, $\theta \to \pi^-$) and $x_0 = |x_0| e^{-i\pi}$ (approached from below, $\theta \to -\pi^+$): \begin{align*} \lim_{\epsilon \to 0^+} \operatorname{Log}(x_0 + i\epsilon) &= \ln|x_0| + i\pi, \\ \lim_{\epsilon \to 0^+} \operatorname{Log}(x_0 - i\epsilon) &= \ln|x_0| - i\pi. \end{align*} The jump across the cut is \begin{align*} \lim_{\epsilon \to 0^+} \left[\operatorname{Log}(x_0 + i\epsilon) - \operatorname{Log}(x_0 - i\epsilon)\right] = 2\pi i. \end{align*} This constant jump of $2\pi i$ is a direct manifestation of the monodromy of $\log z$: crossing the branch cut is equivalent to completing one circuit around the origin, which shifts the argument by $2\pi$. The jump is the same at every point of the cut — it does not depend on $x_0$. ## The Theta Method The construction of the principal logarithm above is an instance of a general technique for building branches of multivalued functions that we call the **theta method**: one removes a ray emanating from the branch point and parametrises the resulting domain by restricting the angular coordinate to an open interval of length $2\pi$. This approach is direct and computational — it reduces the construction of a branch to an explicit choice of angular range — and it applies uniformly to logarithms, roots, and general power functions. ### The Heuristic Idea Consider a multivalued function $f$ with a branch point at the origin whose values at $z = re^{i\theta}$ depend on $\theta$ modulo $2\pi$ (such as $\log z = \ln r + i\theta$ or $z^{1/n} = r^{1/n} e^{i\theta/n}$). The multivaluedness arises because $\theta$ is only defined modulo $2\pi$: the same point $z$ can be written as $re^{i\theta}$ or $re^{i(\theta + 2\pi)}$, and $f$ may assign different values to these two representations. To eliminate the ambiguity, fix a real number $\alpha$ and restrict to arguments $\theta \in (\alpha, \alpha + 2\pi)$. This restriction amounts to removing the ray \begin{align*} R_\alpha := \{re^{i\alpha} : r \ge 0\} \end{align*} from the domain. On the resulting slit plane $\mathbb{C} \setminus R_\alpha$, every nonzero point has a unique representation $z = re^{i\theta}$ with $r > 0$ and $\theta \in (\alpha, \alpha + 2\pi)$, so the function $f$ inherits a single, well-defined value. For example, with $\alpha = -\pi$ (removing the ray $R_{-\pi} = (-\infty, 0]$) and using the angular range $\theta \in (-\pi, \pi)$, we recover the principal logarithm: $\operatorname{Log}(z) = \ln r + i\theta$ with $\theta \in (-\pi, \pi)$. With $\alpha = 0$ (removing $R_0 = [0, \infty)$) and using $\theta \in (0, 2\pi)$, we obtain a different branch of the logarithm that is continuous across the negative real axis but discontinuous across the positive real axis. ### Formalisation The heuristic description above — "restrict the angle to $(\alpha, \alpha + 2\pi)$" — can be made precise using the definitions of the previous sections. The key step is to verify that the resulting function is not merely continuous but holomorphic. Given $\alpha \in \mathbb{R}$, define the ray $R_\alpha = \{re^{i\alpha} : r \ge 0\}$. The set $\mathbb{C} \setminus R_\alpha$ is a simply connected domain (it is the image of the half-strip $\{(s, t) : s \in \mathbb{R}, \, t \in (\alpha, \alpha + 2\pi)\}$ under the map $(s,t) \mapsto e^{s+it}$, which is a conformal equivalence). Define the angular function associated to $\alpha$: \begin{align*} \operatorname{arg}_\alpha : \mathbb{C} \setminus R_\alpha &\to (\alpha, \alpha + 2\pi) \\ z = re^{i\theta} &\mapsto \theta, \qquad \text{where } \theta \text{ is the unique value in } (\alpha, \alpha + 2\pi) \text{ with } z = re^{i\theta}. \end{align*} This function is well-defined because every $z \in \mathbb{C} \setminus R_\alpha$ with $z \neq 0$ has a unique polar representation with angle in the open interval $(\alpha, \alpha + 2\pi)$, and $0 \notin \mathbb{C} \setminus R_\alpha$. The $\alpha$-branch of the logarithm is then the function \begin{align*} \log_\alpha : \mathbb{C} \setminus R_\alpha &\to \mathbb{C} \\ z &\mapsto \ln|z| + i \operatorname{arg}_\alpha(z). \end{align*} The subscript $\alpha$ records the choice of cut; when $\alpha = -\pi$, we recover $\log_{-\pi} = \operatorname{Log}$. The following theorem confirms that every branch produced by the theta method is holomorphic and is a legitimate branch of $\log z$ in the sense of our earlier definition. [theorem: Theta Method Produces Holomorphic Branches] For every $\alpha \in \mathbb{R}$, the function $\log_\alpha : \mathbb{C} \setminus R_\alpha \to \mathbb{C}$ defined by $\log_\alpha(z) = \ln|z| + i\operatorname{arg}_\alpha(z)$ satisfies: (i) $\exp(\log_\alpha(z)) = z$ for every $z \in \mathbb{C} \setminus R_\alpha$. (ii) $\log_\alpha$ is holomorphic on $\mathbb{C} \setminus R_\alpha$, with $\log_\alpha'(z) = 1/z$. (iii) The image of $\log_\alpha$ is the horizontal strip $\{w \in \mathbb{C} : \alpha < \operatorname{Im}(w) < \alpha + 2\pi\}$. (iv) Any two theta-method branches are related by a constant: if $\alpha, \beta \in \mathbb{R}$ and $z \in (\mathbb{C} \setminus R_\alpha) \cap (\mathbb{C} \setminus R_\beta)$, then $\log_\alpha(z) - \log_\beta(z) = 2\pi i k$ for some $k \in \mathbb{Z}$ (depending on $z$, $\alpha$, and $\beta$, but constant on each connected component of the intersection). [/theorem] Part (i) is immediate from the definition: $\exp(\ln|z| + i\operatorname{arg}_\alpha(z)) = |z| e^{i\operatorname{arg}_\alpha(z)} = z$. Part (ii) follows from the same inverse function theorem argument used for the principal logarithm: on $\mathbb{C} \setminus R_\alpha$, the exponential $\exp$ restricts to a bijection from the strip $\{w : \alpha < \operatorname{Im}(w) < \alpha + 2\pi\}$ with nonvanishing derivative, so its inverse $\log_\alpha$ is holomorphic with derivative $1/\exp'(\log_\alpha(z)) = 1/z$. Part (iii) follows from the range of $\operatorname{arg}_\alpha$. Part (iv) reflects the fact that any two logarithms of the same number differ by an integer multiple of $2\pi i$. The theta method thus reduces the problem of constructing branches to the elementary task of choosing an angle $\alpha$. Different choices of $\alpha$ give different branches, each holomorphic on a different slit plane, and together they "tile" the full multivalued logarithm. [example: Branch With Cut Along The Positive Real Axis] Take $\alpha = 0$, so the branch cut is $R_0 = [0, \infty)$ and the angular range is $\theta \in (0, 2\pi)$. The resulting branch is \begin{align*} \log_0(z) = \ln|z| + i\operatorname{arg}_0(z), \qquad \operatorname{arg}_0(z) \in (0, 2\pi). \end{align*} We compute its values at several points: At $z = -1$: we write $-1 = e^{i\pi}$, and $\pi \in (0, 2\pi)$, so $\log_0(-1) = \ln 1 + i\pi = i\pi$. At $z = i$: we write $i = e^{i\pi/2}$, and $\pi/2 \in (0, 2\pi)$, so $\log_0(i) = \ln 1 + i\pi/2 = i\pi/2$. At $z = -i$: we write $-i = e^{i \cdot 3\pi/2}$, and $3\pi/2 \in (0, 2\pi)$, so $\log_0(-i) = \ln 1 + i \cdot 3\pi/2 = 3\pi i/2$. Compare with the principal logarithm: $\operatorname{Log}(-i) = \ln 1 + i(-\pi/2) = -i\pi/2$. The difference is $\log_0(-i) - \operatorname{Log}(-i) = 3\pi i/2 - (-\pi i/2) = 2\pi i$, confirming part (iv) of the theorem with $k = 1$. The discontinuity of $\log_0$ occurs across the positive real axis. Approaching $x_0 > 0$ from above (argument near $0^+$) and below (argument near $2\pi^-$): \begin{align*} \lim_{\epsilon \to 0^+} \log_0(x_0 + i\epsilon) &= \ln x_0 + i \cdot 0^+ \to \ln x_0, \\ \lim_{\epsilon \to 0^+} \log_0(x_0 - i\epsilon) &= \ln x_0 + i(2\pi)^- \to \ln x_0 + 2\pi i. \end{align*} The jump is $-2\pi i$ (from above to below), the same magnitude as for the principal logarithm but occurring on a different ray. [/example] [example: Branch With Cut Along The Positive Imaginary Axis] Take $\alpha = \pi/2$, so the branch cut is $R_{\pi/2} = \{ri : r \ge 0\}$ (the positive imaginary axis including the origin) and the angular range is $\theta \in (\pi/2, \pi/2 + 2\pi) = (\pi/2, 5\pi/2)$. This is a less standard choice but perfectly valid. At $z = 1$: write $1 = e^{i \cdot 2\pi}$, and $2\pi \in (\pi/2, 5\pi/2)$, so $\log_{\pi/2}(1) = \ln 1 + i \cdot 2\pi = 2\pi i$. This differs from $\operatorname{Log}(1) = 0$ by $2\pi i$, which is expected: the cut along the positive imaginary axis forces the argument of points on the positive real axis to be $2\pi$ rather than $0$ (since $0 \notin (\pi/2, 5\pi/2)$). At $z = -1$: write $-1 = e^{i\pi}$, and $\pi \in (\pi/2, 5\pi/2)$, so $\log_{\pi/2}(-1) = i\pi$. This agrees with $\operatorname{Log}(-1)$. At $z = -i$: write $-i = e^{i \cdot 3\pi/2}$, and $3\pi/2 \in (\pi/2, 5\pi/2)$, so $\log_{\pi/2}(-i) = i \cdot 3\pi/2$. This differs from $\operatorname{Log}(-i) = -i\pi/2$ by $2\pi i$. This example illustrates that the theta method does not require the branch cut to lie along a coordinate axis. Any ray from the origin — along a coordinate direction, at an irrational angle, or anywhere else — produces a valid holomorphic branch of the logarithm. [/example] ## Power Functions and $n$-th Roots The complex logarithm is the engine behind all complex power functions: to define $z^\alpha$ for $\alpha \in \mathbb{C}$, one writes $z^\alpha = \exp(\alpha \log z)$, and the multivaluedness of $\log z$ propagates to $z^\alpha$. However, the nature of the multivaluedness depends on whether $\alpha$ is an integer, a rational number, or irrational/complex. When $\alpha = 1/n$ for a positive integer $n$, the function $z^{1/n}$ takes exactly $n$ distinct values at each point (since changing the branch of $\log$ by $2\pi i$ changes $z^{1/n}$ by a factor of $e^{2\pi i/n}$, and $n$ such changes exhaust all possibilities). When $\alpha$ is irrational or non-real, $z^\alpha$ takes infinitely many values, just like the logarithm itself. The theta method extends directly to all of these cases. [definition: Complex Power] Let $\alpha \in \mathbb{C}$. For $z \in \mathbb{C} \setminus \{0\}$, the **(multivalued) complex power** $z^\alpha$ is the set \begin{align*} z^\alpha := \left\{ \exp(\alpha w) : w \in \mathbb{C}, \, \exp(w) = z \right\} = \left\{ \exp\left(\alpha(\ln|z| + i\operatorname{arg}(z) + 2\pi i k)\right) : k \in \mathbb{Z} \right\}. \end{align*} Given a branch $\log_\alpha$ of the logarithm on a slit plane $\mathbb{C} \setminus R_\alpha$, the **corresponding branch of $z^\alpha$** is the function $z \mapsto \exp(\alpha \log_\alpha(z))$. [/definition] The branch inherits holomorphicity from the logarithm: since $\log_\alpha$ is holomorphic and $\exp$ is entire, the composition $\exp \circ (\alpha \cdot \log_\alpha)$ is holomorphic on $\mathbb{C} \setminus R_\alpha$. [example: The Square Root Via The Theta Method] We construct the principal square root using $\alpha = -\pi$ (cut along $(-\infty, 0]$, arguments in $(-\pi, \pi)$). For $z = re^{i\theta}$ with $r > 0$ and $\theta \in (-\pi, \pi)$: \begin{align*} z^{1/2} := \exp\left(\frac{1}{2} \operatorname{Log}(z)\right) = \exp\left(\frac{1}{2}(\ln r + i\theta)\right) = \sqrt{r} \, e^{i\theta/2}, \end{align*} where $\sqrt{r}$ denotes the positive real square root. Since $\theta \in (-\pi, \pi)$, the half-angle satisfies $\theta/2 \in (-\pi/2, \pi/2)$, so this branch maps into the right half-plane $\{\operatorname{Re}(w) > 0\} \cup \{0\}$ (with values on the positive imaginary axis excluded). In particular, $z^{1/2}$ has positive real part for every $z \in \mathbb{C} \setminus (-\infty, 0]$. We verify at several points: At $z = 4$: $r = 4$, $\theta = 0$, so $z^{1/2} = \sqrt{4} \, e^0 = 2$. At $z = -1 + i\epsilon$ for small $\epsilon > 0$: $r \approx 1$, $\theta \approx \pi - \epsilon$, so $z^{1/2} \approx e^{i(\pi - \epsilon)/2} = e^{i\pi/2 - i\epsilon/2} \approx i$ (approaching $i$ from the fourth-to-first quadrant [boundary](/page/Boundary)). At $z = -1 - i\epsilon$ for small $\epsilon > 0$: $r \approx 1$, $\theta \approx -\pi + \epsilon$, so $z^{1/2} \approx e^{-i\pi/2 + i\epsilon/2} \approx -i$. The jump across the branch cut is dramatic: approaching $x_0 < 0$ from above gives $z^{1/2} \to \sqrt{|x_0|} \, e^{i\pi/2} = i\sqrt{|x_0|}$, while approaching from below gives $z^{1/2} \to \sqrt{|x_0|} \, e^{-i\pi/2} = -i\sqrt{|x_0|}$. The jump is $2i\sqrt{|x_0|}$, which is *not* constant along the cut (unlike the logarithm's constant jump of $2\pi i$). There is a second branch of the square root on the same cut plane, obtained by replacing $k = 0$ with $k = 1$ (or equivalently, multiplying by $e^{2\pi i / 2} = -1$): \begin{align*} z^{1/2}_{\text{other}} = -\sqrt{r} \, e^{i\theta/2}. \end{align*} This branch maps into the left half-plane and satisfies $z^{1/2}_{\text{other}} = -z^{1/2}$. Together, the two branches cover all nonzero complex numbers: for every $w \neq 0$, exactly one of $w$ and $-w$ is in the image of the principal branch. [/example] The general $n$-th root exhibits the same structure, with $n$ branches separated by factors of $e^{2\pi i/n}$. The following theorem makes this precise and explains why the branch cut of $z^{1/n}$ is identical to that of $\log z$: the multivaluedness comes entirely from the logarithm. [theorem: Branches Of The N-th Root] Let $n \ge 2$ be an integer and let $\alpha \in \mathbb{R}$. On the slit plane $\mathbb{C} \setminus R_\alpha$, the function $z^{1/n}$ admits exactly $n$ holomorphic branches, given by \begin{align*} f_k(z) = |z|^{1/n} \exp\left(\frac{i(\operatorname{arg}_\alpha(z) + 2\pi k)}{n}\right), \qquad k = 0, 1, \ldots, n-1. \end{align*} These branches satisfy: (i) $f_k(z)^n = z$ for all $z \in \mathbb{C} \setminus R_\alpha$ and all $k$. (ii) $f_k(z) = e^{2\pi i k / n} f_0(z)$ for all $z$ and $k$. (iii) For each $z \in \mathbb{C} \setminus R_\alpha$, the values $\{f_0(z), f_1(z), \ldots, f_{n-1}(z)\}$ are the $n$ distinct $n$-th roots of $z$, equally spaced on a circle of radius $|z|^{1/n}$. (iv) Each $f_k$ is holomorphic on $\mathbb{C} \setminus R_\alpha$ with derivative $f_k'(z) = \frac{1}{n} z^{(1/n) - 1} = \frac{f_k(z)}{nz}$. [/theorem] Part (ii) shows that the $n$ branches are related by multiplication by $n$-th roots of unity. This is specific to the $n$-th root — for irrational powers $z^\alpha$, there are infinitely many branches, and no finite [group](/page/Group) relates them. Part (iii) confirms the geometric picture: at each point, the $n$ branch values form a regular $n$-gon centered at the origin. Part (iv) gives the derivative, which again takes the "expected" form from real calculus. ## The Monodromy Theorem The theta method and the explicit constructions above all rely on the same underlying principle: branch cuts work because they make the domain simply connected, and on a simply connected domain, analytic continuation cannot produce ambiguity. This principle is made precise by the monodromy theorem, which is the theoretical foundation of the entire theory of branch cuts. To state the theorem, we need the concept of analytic continuation along a path. Given a holomorphic function element $(f, D)$ — a holomorphic function $f$ defined on a disc $D$ — and a path $\gamma : [0,1] \to \Omega$, an **analytic continuation of $(f, D)$ along $\gamma$** is a family of function elements $(f_t, D_t)_{t \in [0,1]}$ such that $(f_0, D_0) = (f, D)$, each $\gamma(t) \in D_t$, and consecutive elements agree on their overlap. The question is whether the result at $\gamma(1)$ depends on the choice of path. [theorem: Monodromy Theorem] Let $\Omega \subseteq \mathbb{C}$ be a simply connected domain, let $z_0 \in \Omega$, and let $(f, D)$ be a holomorphic function element with $z_0 \in D \subseteq \Omega$. Suppose that $(f, D)$ can be analytically continued along every path in $\Omega$ starting at $z_0$. Then the analytic continuation is independent of the path: if $\gamma_1, \gamma_2 : [0,1] \to \Omega$ are any two paths with $\gamma_1(0) = \gamma_2(0) = z_0$ and $\gamma_1(1) = \gamma_2(1) = z_1$, then the function elements obtained at $z_1$ by continuing along $\gamma_1$ and $\gamma_2$ are equal. In particular, the analytic continuation defines a single-valued holomorphic function $F : \Omega \to \mathbb{C}$ extending $f$. [/theorem] The hypothesis that $\Omega$ is simply connected is essential. On a non-simply connected domain like $\mathbb{C} \setminus \{0\}$, two paths from $z_0$ to $z_1$ that wind around the origin a different number of times will produce different continuations of $\log z$ — this is exactly the monodromy phenomenon that branch cuts are designed to eliminate. The proof relies on the fact that in a simply connected domain, any two paths with the same endpoints are homotopic (continuously deformable into each other while fixing the endpoints). One then shows that analytic continuation is invariant under homotopy: if $\gamma_1$ and $\gamma_2$ are homotopic, the continuations along them agree. The key step is a compactness argument — the continuous deformation is covered by finitely many discs on which the function elements agree — combined with the identity theorem for holomorphic functions. The monodromy theorem explains *why* branch cuts work at a structural level. Removing a branch cut from $\mathbb{C} \setminus \{0\}$ produces a simply connected domain. The monodromy theorem then guarantees that analytic continuation of any function element on this domain is path-independent, yielding a single-valued holomorphic function — a branch. Different branch cuts produce different simply connected domains, and hence potentially different branches, but the monodromy theorem ensures that each cut leads to a well-defined result. [remark: Branch Cuts Are Not Unique] The monodromy theorem shows that *any* branch cut that renders the domain simply connected will work — there is nothing canonical about the standard choice $(-\infty, 0]$ for the logarithm. In applications, one chooses the cut to avoid the region of interest. For instance, when computing $\int_0^\infty f(x) \, dx$ using contour methods, one often places the cut along $[0, \infty)$ rather than $(-\infty, 0]$, because the contour must wrap around the cut and the computation requires knowledge of the function's discontinuity across it. When two or more branch points are present (as for $(z^2 - 1)^{1/2}$, which has branch points at $\pm 1$), the cut is typically a curve connecting them (e.g. the segment $[-1, 1]$), rather than a ray to infinity from each. [/remark] ## Contour Integration with Branch Cuts One of the most striking applications of branch cuts is in the evaluation of real [integrals](/page/Integral) by contour integration. The technique exploits the discontinuity of a multivalued function across its branch cut: by integrating around a contour that hugs the cut, the integral picks up the jump, which can be related to the desired real integral. [example: Keyhole Contour For A Fractional Power Integral] We evaluate the integral \begin{align*} I := \int_0^\infty \frac{x^{\alpha - 1}}{1 + x} \, d\mathcal{L}^1(x), \qquad 0 < \alpha < 1. \end{align*} The integrand involves $x^{\alpha - 1}$, which is a power function, so we extend it to the complex plane using a branch of $z^{\alpha - 1}$. We place the branch cut along $[0, \infty)$ and use the theta method with $\alpha_{\text{cut}} = 0$: define $z^{\alpha - 1} = |z|^{\alpha - 1} e^{i(\alpha - 1)\operatorname{arg}_0(z)}$ with $\operatorname{arg}_0(z) \in (0, 2\pi)$. This branch is holomorphic on $\mathbb{C} \setminus [0, \infty)$. Consider the "keyhole" contour $\Gamma$ consisting of four pieces: (1) the segment from $\epsilon$ to $R$ just above the positive real axis (argument $\theta = 0^+$), (2) a large circle $C_R$ of radius $R$ traversed counterclockwise, (3) the segment from $R$ to $\epsilon$ just below the positive real axis (argument $\theta = 2\pi^-$), and (4) a small circle $C_\epsilon$ of radius $\epsilon$ traversed clockwise. The function \begin{align*} g : \mathbb{C} \setminus [0, \infty) &\to \mathbb{C} \\ z &\mapsto \frac{z^{\alpha - 1}}{1 + z} \end{align*} is meromorphic on $\mathbb{C} \setminus [0, \infty)$ with a single simple pole at $z = -1$ (which lies off the branch cut since $-1 \notin [0, \infty)$). By the residue theorem: \begin{align*} \oint_\Gamma g(z) \, dz = 2\pi i \operatorname{Res}(g, -1). \end{align*} **Computing the residue.** At $z = -1 = e^{i\pi}$ (with $\pi \in (0, 2\pi)$): \begin{align*} \operatorname{Res}(g, -1) = \lim_{z \to -1} (z + 1) \cdot \frac{z^{\alpha - 1}}{1 + z} = (-1)^{\alpha - 1} = e^{i\pi(\alpha - 1)}. \end{align*} **The contribution above the cut** (segment (1)). Just above the positive real axis, $z = x e^{i \cdot 0^+}$ for $x \in (\epsilon, R)$, so $z^{\alpha - 1} = x^{\alpha - 1} e^{i(\alpha - 1) \cdot 0} = x^{\alpha - 1}$. Thus: \begin{align*} \int_{\text{above}} g(z) \, dz = \int_\epsilon^R \frac{x^{\alpha - 1}}{1 + x} \, dx. \end{align*} **The contribution below the cut** (segment (3)). Just below the positive real axis, $z = x e^{i \cdot 2\pi^-}$ for $x \in (\epsilon, R)$, so $z^{\alpha - 1} = x^{\alpha - 1} e^{i(\alpha - 1) \cdot 2\pi} = x^{\alpha - 1} e^{2\pi i(\alpha - 1)}$. The direction of integration is from $R$ to $\epsilon$ (reversed), so: \begin{align*} \int_{\text{below}} g(z) \, dz = -\int_\epsilon^R \frac{x^{\alpha - 1} e^{2\pi i(\alpha - 1)}}{1 + x} \, dx. \end{align*} **The circular arcs vanish.** On $C_R$: $|g(z)| \le R^{\alpha - 1} / (R - 1) \sim R^{\alpha - 2}$ as $R \to \infty$, and the length of $C_R$ is $2\pi R$, so $|\int_{C_R} g \, dz| \le 2\pi R \cdot R^{\alpha - 2} = 2\pi R^{\alpha - 1} \to 0$ since $\alpha < 1$. On $C_\epsilon$: $|g(z)| \le \epsilon^{\alpha - 1} / (1 - \epsilon)$ for small $\epsilon$, and the length of $C_\epsilon$ is $2\pi\epsilon$, so $|\int_{C_\epsilon} g \, dz| \le 2\pi \epsilon^{\alpha} / (1 - \epsilon) \to 0$ since $\alpha > 0$. **Assembling the result.** Taking $\epsilon \to 0$ and $R \to \infty$: \begin{align*} \int_0^\infty \frac{x^{\alpha - 1}}{1 + x} \, dx - e^{2\pi i(\alpha - 1)} \int_0^\infty \frac{x^{\alpha - 1}}{1 + x} \, dx = 2\pi i \, e^{i\pi(\alpha - 1)}. \end{align*} Factoring: \begin{align*} I \left(1 - e^{2\pi i(\alpha - 1)}\right) = 2\pi i \, e^{i\pi(\alpha - 1)}. \end{align*} Now $e^{2\pi i(\alpha - 1)} = e^{2\pi i \alpha} e^{-2\pi i} = e^{2\pi i \alpha}$, so: \begin{align*} I = \frac{2\pi i \, e^{i\pi(\alpha - 1)}}{1 - e^{2\pi i \alpha}}. \end{align*} To simplify, multiply numerator and denominator by $e^{-i\pi\alpha}$: \begin{align*} I = \frac{2\pi i \, e^{i\pi(\alpha - 1)} \cdot e^{-i\pi\alpha}}{e^{-i\pi\alpha} - e^{i\pi\alpha}} = \frac{2\pi i \, e^{-i\pi}}{e^{-i\pi\alpha} - e^{i\pi\alpha}} = \frac{2\pi i \cdot (-1)}{-2i\sin(\pi\alpha)} = \frac{-2\pi i}{-2i\sin(\pi\alpha)} = \frac{\pi}{\sin(\pi\alpha)}. \end{align*} Thus: \begin{align*} \int_0^\infty \frac{x^{\alpha - 1}}{1 + x} \, d\mathcal{L}^1(x) = \frac{\pi}{\sin(\pi\alpha)}, \qquad 0 < \alpha < 1. \end{align*} This formula is a special case of Euler's reflection formula for the Gamma function: $\Gamma(\alpha)\Gamma(1 - \alpha) = \pi / \sin(\pi\alpha)$, and the integral equals $\Gamma(\alpha)\Gamma(1 - \alpha)$ by the Beta function identity. [/example] The computation above depends critically on the branch cut being along $[0, \infty)$: this is what creates the "seam" in the keyhole contour where the function takes different values above and below. If we had placed the cut along $(-\infty, 0]$ instead, the contour would have to wrap around the negative real axis, and the pole at $z = -1$ would sit on the cut — making the computation impossible. ## Problems [problem] Let $f(z) = (z^2 - 1)^{1/2}$. This function has branch points at $z = \pm 1$. (a) Verify that $z = 1$ and $z = -1$ are branch points by showing that analytic continuation of $f$ around a small circle enclosing exactly one of these points changes the sign of $f$. (b) Show that $z = \infty$ is *not* a branch point: analytic continuation around a large circle enclosing both $\pm 1$ returns $f$ to its original value. (c) Define a holomorphic branch of $f$ on $\mathbb{C} \setminus [-1, 1]$ (cutting along the segment connecting the two branch points) by specifying the value at $z = 2$ to be $f(2) = \sqrt{3}$. [/problem] [solution] **Step 1: Branch point at $z = 1$.** Write $f(z) = (z - 1)^{1/2}(z + 1)^{1/2}$. Consider a small circle $\gamma_1$ of radius $\delta < 1$ centered at $z = 1$, traversed counterclockwise. On this circle, $z - 1 = \delta e^{i\phi}$ for $\phi \in [0, 2\pi]$, while $z + 1 = 2 + \delta e^{i\phi}$, which is a small perturbation of $2$ and does not wind around the origin. After one full circuit ($\phi : 0 \to 2\pi$), the factor $(z - 1)^{1/2} = \delta^{1/2} e^{i\phi/2}$ picks up a phase $e^{i\pi} = -1$ (the half-angle $\phi/2$ increases by $\pi$). The factor $(z + 1)^{1/2}$ returns to its original value because $z + 1$ does not wind around the origin. Therefore $f(z)$ changes sign after the circuit, confirming that $z = 1$ is a branch point. **Step 2: Branch point at $z = -1$.** The argument is identical with the roles of $(z - 1)$ and $(z + 1)$ reversed. A small circle $\gamma_{-1}$ around $z = -1$ causes $(z + 1)^{1/2}$ to pick up a factor of $-1$, while $(z - 1)^{1/2}$ returns to its original value. **Step 3: Infinity is not a branch point.** Consider a large circle $\gamma_R$ of radius $R > 1$ centered at the origin, traversed counterclockwise. Both $z - 1$ and $z + 1$ wind once around the origin as $z$ traverses $\gamma_R$. Each square root factor picks up a factor of $-1$, so $f(z)$ picks up $(-1)(-1) = 1$: the function returns to its original value. Thus $\infty$ is not a branch point. (This is why we can connect the two finite branch points by a finite segment, rather than extending rays to infinity.) **Step 4: Construction of a branch on $\mathbb{C} \setminus [-1, 1]$.** The complement $\mathbb{C} \setminus [-1, 1]$ is simply connected (the segment $[-1, 1]$ does not disconnect the plane but prevents paths from separating the two branch points). By the monodromy theorem, $f$ admits a single-valued holomorphic branch on this domain. To pin down the branch, we use the theta method adapted to each factor. For $z \in \mathbb{C} \setminus [-1, 1]$, write \begin{align*} z - 1 = r_1 e^{i\theta_1}, \qquad z + 1 = r_2 e^{i\theta_2}, \end{align*} where $r_1 = |z - 1|$, $r_2 = |z + 1|$, and we choose $\theta_1, \theta_2 \in (-\pi, \pi)$ (i.e., we use the principal argument with respect to each branch point). This is well-defined because neither $z - 1$ nor $z + 1$ lies on $(-\infty, 0]$ when $z \in \mathbb{C} \setminus [-1, 1]$: the point $z - 1$ lies on $(-\infty, 0]$ precisely when $z \in (-\infty, 1]$, but the only part of $(-\infty, 1]$ in our domain is $(-\infty, -1)$ (the interval $[-1, 1]$ is removed); and on $(-\infty, -1)$, $z - 1 < -2 < 0$ does lie on $(-\infty, 0]$. So the principal argument is *not* directly applicable in this naive way. Instead, define the branch by the explicit formula: \begin{align*} F(z) = \exp\left(\frac{1}{2}\operatorname{Log}(z^2 - 1)\right), \end{align*} where $\operatorname{Log}$ is the principal logarithm. For $z \in \mathbb{C} \setminus [-1, 1]$, the value $z^2 - 1$ avoids $(-\infty, 0]$: when $z$ is real and $|z| > 1$, $z^2 - 1 > 0$; and one can verify that $z^2 - 1 \in (-\infty, 0]$ implies $z \in [-1, 1]$ (the excluded set). Therefore $\operatorname{Log}(z^2 - 1)$ is well-defined and holomorphic, and so is $F(z)$. **Step 5: Verification at $z = 2$.** \begin{align*} F(2) = \exp\left(\frac{1}{2}\operatorname{Log}(4 - 1)\right) = \exp\left(\frac{1}{2}\operatorname{Log}(3)\right) = \exp\left(\frac{1}{2}\ln 3\right) = 3^{1/2} = \sqrt{3}. \end{align*} This confirms the normalisation $F(2) = \sqrt{3}$. **Step 6: Verification that $F(z)^2 = z^2 - 1$.** \begin{align*} F(z)^2 = \exp\left(\operatorname{Log}(z^2 - 1)\right) = z^2 - 1, \end{align*} since $\exp \circ \operatorname{Log}$ is the identity on $\mathbb{C} \setminus (-\infty, 0]$. [/solution] [problem] Evaluate $\displaystyle \int_0^\infty \frac{\ln x}{1 + x^2} \, d\mathcal{L}^1(x)$ using a contour integral with an appropriate branch cut. [/problem] [solution] **Step 1: Set up the branch and contour.** Consider the function \begin{align*} g : \mathbb{C} \setminus [0, \infty) &\to \mathbb{C} \\ z &\mapsto \frac{\log_0(z)}{1 + z^2}, \end{align*} where $\log_0(z) = \ln|z| + i\operatorname{arg}_0(z)$ with $\operatorname{arg}_0(z) \in (0, 2\pi)$ is the branch with cut along $[0, \infty)$. The function $g$ is meromorphic on $\mathbb{C} \setminus [0, \infty)$ with simple poles at $z = \pm i$. Both poles lie in the domain since $\pm i \notin [0, \infty)$. Use the keyhole contour $\Gamma$ from the previous example: above the cut from $\epsilon$ to $R$, the large circle $C_R$, below the cut from $R$ to $\epsilon$, and the small circle $C_\epsilon$. **Step 2: Compute the residues.** At $z = i = e^{i\pi/2}$ (with $\pi/2 \in (0, 2\pi)$): \begin{align*} \operatorname{Res}(g, i) = \frac{\log_0(i)}{2i} = \frac{i\pi/2}{2i} = \frac{\pi}{4}. \end{align*} At $z = -i = e^{i \cdot 3\pi/2}$ (with $3\pi/2 \in (0, 2\pi)$): \begin{align*} \operatorname{Res}(g, -i) = \frac{\log_0(-i)}{-2i} = \frac{i \cdot 3\pi/2}{-2i} = -\frac{3\pi}{4}. \end{align*} The sum of residues is $\pi/4 - 3\pi/4 = -\pi/2$, so: \begin{align*} \oint_\Gamma g(z) \, dz = 2\pi i \cdot \left(-\frac{\pi}{2}\right) = -\pi^2 i. \end{align*} **Step 3: Evaluate the contributions from the contour segments.** Above the cut ($z = x e^{i \cdot 0^+}$, $x \in (\epsilon, R)$): $\log_0(z) = \ln x + i \cdot 0 = \ln x$, so: \begin{align*} \int_{\text{above}} g(z) \, dz = \int_\epsilon^R \frac{\ln x}{1 + x^2} \, dx. \end{align*} Below the cut ($z = x e^{i \cdot 2\pi^-}$, $x \in (\epsilon, R)$, direction reversed): $\log_0(z) = \ln x + 2\pi i$, so: \begin{align*} \int_{\text{below}} g(z) \, dz = -\int_\epsilon^R \frac{\ln x + 2\pi i}{1 + x^2} \, dx. \end{align*} **Step 4: Show the circular arcs vanish.** On $C_R$: $|g(z)| \le (\ln R + 2\pi)/(R^2 - 1)$ for $R$ large, and the length is $2\pi R$, so the integral is $O(R \ln R / R^2) = O(\ln R / R) \to 0$. On $C_\epsilon$: $|g(z)| \le (|\ln \epsilon| + 2\pi)/(1 - \epsilon^2)$, and the length is $2\pi\epsilon$, so the integral is $O(\epsilon |\ln \epsilon|) \to 0$ (since $\epsilon \ln \epsilon \to 0$ as $\epsilon \to 0^+$). **Step 5: Combine and solve.** Taking $\epsilon \to 0$ and $R \to \infty$: \begin{align*} \int_0^\infty \frac{\ln x}{1 + x^2} \, dx - \int_0^\infty \frac{\ln x + 2\pi i}{1 + x^2} \, dx &= -\pi^2 i. \end{align*} The left side simplifies: \begin{align*} -2\pi i \int_0^\infty \frac{1}{1 + x^2} \, dx = -\pi^2 i. \end{align*} Since $\int_0^\infty \frac{1}{1+x^2} \, dx = \pi/2$, we get $-2\pi i \cdot \pi/2 = -\pi^2 i$, which is consistent but does not yet determine $\int_0^\infty \frac{\ln x}{1 + x^2} \, dx$. To extract the real integral, equate real parts. The full contour integral gives: \begin{align*} \int_0^\infty \frac{\ln x}{1 + x^2} \, dx - \int_0^\infty \frac{\ln x}{1 + x^2} \, dx - 2\pi i \int_0^\infty \frac{1}{1 + x^2} \, dx = -\pi^2 i. \end{align*} The $\ln x$ terms cancel, leaving $-\pi^2 i = -\pi^2 i$, which is an identity. This means the imaginary equation is automatically satisfied and the real equation is $0 = 0$. **Step 6: Alternative approach — use a semicircular contour.** The keyhole contour above does not isolate $\int_0^\infty \frac{\ln x}{1+x^2} \, dx$ because $\ln x$ appears on both sides of the cut and cancels. Instead, use the function $g(z) = (\operatorname{Log} z)^2 / (1 + z^2)$ on the branch cut $(-\infty, 0]$ with a contour consisting of the real axis from $\epsilon$ to $R$, the semicircle $C_R^+$ in the upper half-plane, the real axis from $-R$ to $-\epsilon$, and the semicircle $C_\epsilon^+$ around the origin. By the residue theorem (only the pole at $z = i$ is enclosed): \begin{align*} \oint g(z) \, dz = 2\pi i \cdot \frac{(\operatorname{Log}(i))^2}{2i} = 2\pi i \cdot \frac{(i\pi/2)^2}{2i} = 2\pi i \cdot \frac{-\pi^2/4}{2i} = -\frac{\pi^3}{4}. \end{align*} On the positive real axis: $\operatorname{Log}(x) = \ln x$ for $x > 0$, contributing $\int_\epsilon^R \frac{(\ln x)^2}{1 + x^2} \, dx$. On the negative real axis ($z = -t$, $t > 0$): $\operatorname{Log}(-t) = \ln t + i\pi$, contributing \begin{align*} \int_\epsilon^R \frac{(\ln t + i\pi)^2}{1 + t^2} \, dt = \int_\epsilon^R \frac{(\ln t)^2 + 2\pi i \ln t - \pi^2}{1 + t^2} \, dt. \end{align*} As $\epsilon \to 0$ and $R \to \infty$, and verifying that the semicircular contributions vanish (by the same estimates as before), equating real parts: \begin{align*} \int_0^\infty \frac{(\ln x)^2}{1 + x^2} \, dx + \int_0^\infty \frac{(\ln t)^2 - \pi^2}{1 + t^2} \, dt = -\frac{\pi^3}{4}. \end{align*} This gives: \begin{align*} 2\int_0^\infty \frac{(\ln x)^2}{1 + x^2} \, dx - \pi^2 \cdot \frac{\pi}{2} = -\frac{\pi^3}{4}, \end{align*} so $\int_0^\infty \frac{(\ln x)^2}{1+x^2} \, dx = \pi^3/8$. Equating imaginary parts: \begin{align*} 2\pi \int_0^\infty \frac{\ln t}{1 + t^2} \, dt = 0. \end{align*} Since $\pi \neq 0$: \begin{align*} \int_0^\infty \frac{\ln x}{1 + x^2} \, d\mathcal{L}^1(x) = 0. \end{align*} [/solution] ## References - Ahlfors, L. V., *Complex Analysis* (1979). - Conway, J. B., *Functions of One Complex Variable I* (1978). - Stein, E. M. and Shakarchi, R., *Complex Analysis* (2003). - Needham, T., *Visual Complex Analysis* (1997).