The first shock in complex analysis is that differentiability becomes rigid. A real function can be differentiable once, twice, or infinitely often while still behaving locally in strange ways. A complex function does not have that freedom: once the complex derivative exists in a neighbourhood, the function is controlled by a convergent [power series](/page/Power%20Series). Analytic functions are the language for that control. The motivating question is local. If we know a function near a point $z_0$, can we reconstruct it from the numbers that play the role of its derivatives at $z_0$? For analytic functions the answer is yes: near $z_0$, the function is not merely approximated by a polynomial; it is equal to an infinite polynomial with a positive [radius of convergence](/theorems/262). Before defining the general concept, it is worth seeing the model example. The exponential function is not analytic because we decree it to be nice; it is analytic because its values are literally assembled from powers. [example: The Exponential Function] Let $f:\mathbb{C}\to\mathbb{C}$ be defined by \begin{align*} f(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}. \end{align*} For $|z|\le R$ and $N\le M$, \begin{align*} \left|\sum_{n=N}^{M}\frac{z^n}{n!}\right|\le \sum_{n=N}^{M}\frac{|z|^n}{n!}\le \sum_{n=N}^{M}\frac{R^n}{n!}. \end{align*} The real series $\sum_{n=0}^{\infty} R^n/n!$ converges, so its tails tend to $0$. Therefore the partial sums of $\sum z^n/n!$ are uniformly Cauchy on $\overline{B}(0,R)$, and the series converges uniformly there. Since $R>0$ was arbitrary, this defines a convergent power series on every bounded disc about $0$. Now fix $z_0\in\mathbb{C}$ and write $w=z-z_0$. The two series \begin{align*} \sum_{j=0}^{\infty}\frac{z_0^j}{j!} \end{align*} and \begin{align*} \sum_{k=0}^{\infty}\frac{w^k}{k!} \end{align*} converge absolutely, so their product is the Cauchy product: \begin{align*} f(z_0)\sum_{k=0}^{\infty}\frac{w^k}{k!}=\sum_{m=0}^{\infty}\sum_{j=0}^{m}\frac{z_0^j w^{m-j}}{j!(m-j)!}. \end{align*} For each fixed $m$, \begin{align*} \sum_{j=0}^{m}\frac{z_0^j w^{m-j}}{j!(m-j)!}=\frac{1}{m!}\sum_{j=0}^{m}\binom{m}{j}z_0^j w^{m-j}. \end{align*} By the [binomial theorem](/theorems/750), \begin{align*} \frac{1}{m!}\sum_{j=0}^{m}\binom{m}{j}z_0^j w^{m-j}=\frac{(z_0+w)^m}{m!}. \end{align*} Since $z_0+w=z$, it follows that \begin{align*} f(z_0)\sum_{k=0}^{\infty}\frac{(z-z_0)^k}{k!}=\sum_{m=0}^{\infty}\frac{z^m}{m!}=f(z). \end{align*} Thus the exponential series is not only a power series at $0$; around every centre $z_0$ it has the local expansion with coefficients $f(z_0)/k!$. [/example] This example hints at the main theme of the page. Analyticity is not a global formula written once and for all; it is the demand that every point has a neighbourhood on which the function is governed by powers of the local coordinate $z-z_0$. ## Definition A local expansion is only useful if the series actually converges on a neighbourhood, not just formally. The definition therefore asks for a positive radius around each point and a sequence of complex coefficients depending on that point. [definition: Analytic Function] Let $U \subset \mathbb{C}$ be open. A function $f: U \to \mathbb{C}$ is analytic on $U$ if for every $z_0 \in U$ there exist $r > 0$ and coefficients $(a_n)_{n=0}^{\infty}$ in $\mathbb{C}$ such that $B(z_0,r) \subset U$ and \begin{align*} f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n \end{align*} for every $z \in B(z_0,r)$. [/definition] The coefficients and radius are local data. A function may have one expansion near $z_0$ and a different-looking expansion near another point, but the overlap between the two expansions is forced to agree wherever both represent the same function. [remark: Locality of Analyticity] Analyticity is a local property. If $(U_i)_{i \in I}$ is an open cover of $U$ and $f|_{U_i}: U_i \to \mathbb{C}$ is analytic for every $i \in I$, then $f: U \to \mathbb{C}$ is analytic. [/remark] The word "analytic" is sometimes used for real power series as well, but on this page it means complex analytic unless stated otherwise. The complex case is special because it coincides with [complex differentiability](/page/Complex%20Differentiability) on open sets, a fact that has no real-variable analogue. ## Power Series and Radius of Convergence ### Discs of Convergence The definition of analytic function rests on the behaviour of power series. A formal infinite expression is not enough; the radius on which it converges determines where it defines an honest function. [definition: Complex Power Series] Let $z_0 \in \mathbb{C}$. A complex power series centred at $z_0$ is an expression of the form \begin{align*} \sum_{n=0}^{\infty} a_n (z-z_0)^n, \end{align*} where $(a_n)_{n=0}^{\infty}$ is a sequence in $\mathbb{C}$. [/definition] A power series behaves unlike a general series of functions: its convergence set is a disc, possibly degenerate or all of $\mathbb{C}$. To use a power series as a local description of a function, we need to know exactly how far from the centre that description is valid. The [radius of convergence](/theorems/265) records the open disc where the series is guaranteed to define a function, separating the reliable local theory from the boundary phenomena that require extra work. [definition: Radius of Convergence] Let $\sum_{n=0}^{\infty} a_n (z-z_0)^n$ be a complex power series. Its [radius of convergence](/theorems/273) is the number $R \in [0,\infty]$ such that the series converges absolutely for $|z-z_0| < R$ and diverges for $|z-z_0| > R$. [/definition] The boundary $|z-z_0|=R$ is deliberately absent from the definition because anything can happen there: all points may converge, no points may converge, or some may converge and others may not. [example: Boundary Behaviour of Power Series] For the geometric series \begin{align*} \sum_{n=1}^{\infty} z^n, \end{align*} if $|z|<1$, then the partial sums satisfy \begin{align*} \sum_{n=1}^{N} z^n=\frac{z(1-z^N)}{1-z}. \end{align*} Since $|z^N|=|z|^N\to 0$, the partial sums converge to $z/(1-z)$. If $|z|>1$, then $|z^n|=|z|^n$ does not tend to $0$, so the series diverges by the term test. Hence the radius of convergence is $1$. On the boundary $|z|=1$, every term has absolute value \begin{align*} |z^n|=|z|^n=1, \end{align*} so $z^n$ does not tend to $0$; therefore the series diverges at every point of the unit circle. Now consider \begin{align*} \sum_{n=1}^{\infty}\frac{z^n}{n^2}. \end{align*} If $|z|<1$, then \begin{align*} \left|\frac{z^n}{n^2}\right|=\frac{|z|^n}{n^2}\le |z|^n, \end{align*} and the geometric series $\sum_{n=1}^{\infty}|z|^n$ converges, so the given series converges absolutely. If $|z|>1$, then \begin{align*} \frac{|z|^{n+1}/(n+1)^2}{|z|^n/n^2}=|z|\left(\frac{n}{n+1}\right)^2\to |z|>1, \end{align*} so the terms $|z|^n/n^2$ cannot tend to $0$; the series diverges by the term test. Thus this second series also has radius of convergence $1$. But when $|z|=1$, \begin{align*} \left|\frac{z^n}{n^2}\right|=\frac{1}{n^2}, \end{align*} and $\sum_{n=1}^{\infty}1/n^2$ converges by the $p$-series test. Therefore the second series converges absolutely at every boundary point, while the first diverges at every boundary point; the radius controls the open disc of convergence, but not the behaviour on its boundary. [/example] Once a power series converges inside its disc, it may be differentiated term by term. This turns a representation into a calculus. ### Calculus of Power Series For analytic functions, a power series is meant to be more than a way of computing values. It should support the usual operations of calculus without leaving the class of analytic functions. The next theorem is the mechanism behind that stability: inside the disc of convergence, differentiation can be pushed through the infinite sum and the radius does not shrink. [quotetheorem:8249] This theorem explains why analytic functions automatically have derivatives of all orders. But it leaves a second question: if the same function admits a power series expansion, are the coefficients part of the data or are they forced by the function? Repeating termwise differentiation at the centre answers that question and identifies the Taylor coefficients. [quotetheorem:8250] The formula shows that an analytic expansion, if it exists, is unique. The coefficients are not choices; they are encoded in the function. This uniqueness creates a practical need for a name. When we later compare expansions, estimate coefficients, or ask how far the local formula extends, we should not have to carry an arbitrary sequence $(a_n)_{n=0}^{\infty}$ as extra data. The function itself has already selected the only possible coefficients, so the next definition packages that selected expansion as the Taylor series at the centre. [definition: Taylor Series of an Analytic Function] Let $U \subset \mathbb{C}$ be open, let $f: U\to\mathbb{C}$ be analytic, and let $z_0\in U$. The Taylor series of $f$ at $z_0$ is the power series \begin{align*} \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n. \end{align*} [/definition] For analytic functions, the Taylor series is not merely a formal record of derivatives. It converges to the original function on some neighbourhood of the centre. More precisely, once an [analytic continuation](/page/Analytic%20Continuation) domain for $f$ has been chosen, the Taylor expansion about $z_0$ can persist until it meets the boundary of that continuation or a singular obstruction inside the attempted extension. Analytic functions would be fragile if they stopped being analytic after ordinary algebraic operations. In applications, functions are built by adding, multiplying, and composing simpler functions, but each operation also raises a local convergence question: after combining two power series, or substituting one local expansion into another, is there still a neighbourhood on which the resulting expression is represented by a convergent power series? The closure theorem answers this obstruction and identifies the domains where these constructions remain analytic. [quotetheorem:8251] These closure properties are what make analytic functions usable in practice. Once the basic building blocks are analytic, finite algebraic combinations and compositions remain analytic wherever the expressions are defined. ## Holomorphicity and Complex Differentiability ### Derivative First, Series Later The analytic definition begins with power series, but complex analysis often begins with the derivative. The surprising fact is that these two starting points lead to the same class of functions on open subsets of $\mathbb{C}$. [definition: Holomorphic Function] Let $U \subset \mathbb{C}$ be open. A function $f: U \to \mathbb{C}$ is holomorphic on $U$ if the complex derivative \begin{align*} f'(z_0)=\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} \end{align*} exists for every $z_0 \in U$. [/definition] At first glance, holomorphicity looks weaker than analyticity. It asks only for a first derivative at each point, while analyticity asks for infinitely many coefficients. [Cauchy's theorem](/theorems/797) and the [Cauchy integral formula](/theorems/345) bridge that gap. In the theorem below, $\mathcal{O}(\Omega)$ denotes the space of holomorphic functions on the [open set](/page/Open%20Set) $\Omega \subset \mathbb{C}$. This notation lets the result state the key conversion compactly: a derivative-based hypothesis on $f$ forces the local power series expansions required for analyticity. [quotetheorem:3354] The reverse question is needed for the definition of analytic function to fit the rest of complex analysis. If a local power series representation did not guarantee complex differentiability, then analyticity would be a separate formal theory rather than a source of holomorphic functions. Termwise differentiation supplies exactly the missing bridge: a convergent local expansion produces the complex derivative at every point of its disc. [quotetheorem:8252] The equivalence is one of the central compression mechanisms in complex analysis. A derivative hypothesis can be converted into a power series expansion, and a power series expansion can be manipulated by algebra and calculus. ### A Real-Smooth Failure The equivalence between holomorphic and analytic functions also clarifies what complex differentiability excludes. A function can be smooth as a real map of two variables and still fail to respect complex directions. The conjugation map is the standard warning sign. [example: A Function That Is Smooth but Not Complex Analytic] Let $f:\mathbb{C}\to\mathbb{C}$ be defined by $f(z)=\bar z$. Writing $z=x+iy$ with $x,y\in\mathbb{R}$, complex conjugation gives \begin{align*} f(x+iy)=\overline{x+iy}=x-iy. \end{align*} As a real map $\mathbb{R}^2\to\mathbb{R}^2$, this is \begin{align*} (x,y)\mapsto (x,-y), \end{align*} so its coordinate functions are linear polynomials in $x$ and $y$, hence smooth as real functions. We test the complex derivative at $0$. If $h=t$ with $t\in\mathbb{R}$ and $t\ne 0$, then \begin{align*} \frac{f(0+t)-f(0)}{t}=\frac{\bar t-0}{t}=\frac{t}{t}=1. \end{align*} If instead $h=it$ with $t\in\mathbb{R}$ and $t\ne 0$, then \begin{align*} \frac{f(0+it)-f(0)}{it}=\frac{\overline{it}-0}{it}=\frac{-it}{it}=-1. \end{align*} The same difference quotient has limit $1$ along the real axis and limit $-1$ along the imaginary axis, so the limit as $h\to 0$ in $\mathbb{C}$ does not exist. Therefore $f$ is smooth as a real two-variable map but is not complex differentiable at $0$, and hence is not complex analytic on $\mathbb{C}$. [/example] The failure of $z \mapsto \bar{z}$ is not a technical annoyance; it is the dividing line between real two-dimensional calculus and complex one-dimensional calculus. Analytic functions respect the complex direction structure, while arbitrary smooth maps $\mathbb{R}^2 \to \mathbb{R}^2$ need not. ## Local Expansion and Global Consequences ### Agreement Sets Analyticity is defined point by point, but its consequences propagate. If two analytic functions agree on a set with an accumulation point, then their local power series force them to agree everywhere on a connected domain. [definition: Accumulation Point] Let $A \subset \mathbb{C}$. A point $z_0 \in \mathbb{C}$ is an accumulation point of $A$ if every $r>0$ satisfies \begin{align*} \bigl(B(z_0,r)\setminus\{z_0\}\bigr)\cap A \ne \varnothing. \end{align*} [/definition] The accumulation condition prevents agreement at a sparse list of isolated points from being mistaken for genuine equality. This is exactly the threshold needed for complex analytic functions: isolated coincidences may occur by accident, while coincidences that cluster inside the domain force the local power series to share all their coefficients. On a connected open set $U\subset\mathbb{C}$, the complex [identity principle](/theorems/3357) says that if two analytic functions $f,g:U\to\mathbb{C}$ agree on a subset of $U$ with an accumulation point in $U$, then $f=g$ on all of $U$. This identity principle is why analytic continuation is possible and dangerous. A small amount of data can determine an analytic function on a much larger connected region, provided a continuation exists. [example: Agreement on a Sequence] Let $U=\mathbb{C}$, and let $f,g:\mathbb{C}\to\mathbb{C}$ be analytic. Suppose that for every $n\in\mathbb{N}$, \begin{align*} f\left(\frac{1}{n}\right)=g\left(\frac{1}{n}\right). \end{align*} Let \begin{align*} A=\{z\in\mathbb{C}:f(z)=g(z)\}. \end{align*} Then $1/n\in A$ for every $n\in\mathbb{N}$. To see that $0$ is an accumulation point of $A$, fix $r>0$. Choose $n\in\mathbb{N}$ with $n>1/r$. Then \begin{align*} 0<\frac{1}{n}<r. \end{align*} Hence $1/n\in B(0,r)\setminus\{0\}$ and $1/n\in A$, so \begin{align*} \bigl(B(0,r)\setminus\{0\}\bigr)\cap A\ne\varnothing. \end{align*} Thus the agreement set has the accumulation point $0$, and $0\in\mathbb{C}$. Since $\mathbb{C}$ is connected and open, the *[Identity Theorem for Analytic Functions](/theorems/208)* applies to $f$ and $g$. Therefore \begin{align*} f(z)=g(z) \end{align*} for every $z\in\mathbb{C}$. Agreement on the sequence $1,1/2,1/3,\ldots$ is therefore enough to force global equality, because the sequence accumulates inside the domain. [/example] The connectedness hypothesis matters. If the domain has two components, data on one component cannot control the other. [example: Why Connectedness Is Needed] Let \begin{align*} U=B(-2,1)\cup B(2,1). \end{align*} The two discs are disjoint: if $z$ belonged to both, then $|z+2|<1$ and $|z-2|<1$, so the triangle inequality would give \begin{align*} 4=|2-(-2)|\le |2-z|+|z-(-2)|<1+1=2, \end{align*} which is impossible. Define $f:U\to\mathbb{C}$ by $f(z)=0$ for every $z\in U$, and define $g:U\to\mathbb{C}$ by \begin{align*} g(z)=0 \quad \text{for } z\in B(-2,1) \end{align*} and \begin{align*} g(z)=1 \quad \text{for } z\in B(2,1). \end{align*} The function $f$ is analytic because near every point it is represented by the constant power series $\sum_{n=0}^{\infty}a_n(z-z_0)^n$ with $a_0=0$ and $a_n=0$ for $n\ge 1$. The function $g$ is also analytic: if $z_0\in B(-2,1)$, then some small disc around $z_0$ lies inside $B(-2,1)$ and $g$ is represented there by the constant series with value $0$; if $z_0\in B(2,1)$, then some small disc around $z_0$ lies inside $B(2,1)$ and $g$ is represented there by the constant series with value $1$. The functions agree on $B(-2,1)$, because for every $z\in B(-2,1)$, \begin{align*} f(z)=0=g(z). \end{align*} Thus the agreement set contains $B(-2,1)$. In particular, $-2$ is an accumulation point of the agreement set: given $r>0$, choose \begin{align*} \varepsilon=\min\{r/2,1/2\}. \end{align*} Then $0<\varepsilon<r$ and $|-2+\varepsilon+2|=\varepsilon<1$, so $-2+\varepsilon\in B(-2,1)\setminus\{-2\}$ and hence belongs to the agreement set. However, the functions do not agree on the other component, since \begin{align*} f(2)=0 \end{align*} while \begin{align*} g(2)=1. \end{align*} So an agreement set can have an accumulation point inside a disconnected domain without forcing equality on the whole domain. Connectedness is exactly what prevents information from being trapped on only one component. [/example] ### Zeros and Multiplicity Zeros have a corresponding rigidity. A nonzero analytic function can vanish, but the identity theorem forbids its zeros from clustering inside a connected domain. This turns zero sets into discrete local objects rather than arbitrary closed sets, and it is the first step toward counting zeros with multiplicity. [quotetheorem:8253] This theorem turns local Taylor data into geometric information about the zero set. Once a zero is known to be isolated, the next question is how strongly the function vanishes there. The order of the zero records the first nonzero term in the local expansion, which is the analytic analogue of the multiplicity of a polynomial root. [definition: Order of a Zero] Let $U \subset \mathbb{C}$ be open, let $f: U\to\mathbb{C}$ be analytic, and let $z_0\in U$ satisfy $f(z_0)=0$. The zero of $f$ at $z_0$ has order $m\in\mathbb{N}$ if there exist $r>0$ and an analytic function $h:B(z_0,r)\to\mathbb{C}$ such that $h(z_0)\ne 0$ and \begin{align*} f(z)=(z-z_0)^m h(z) \end{align*} for every $z\in B(z_0,r)$. [/definition] This definition isolates the first nonvanishing term in the local expansion. It lets analytic functions behave like polynomials near a zero, even when the global function is not a polynomial. ## Cauchy Estimates and Rigidity ### Derivative Bounds Power series expansions do more than describe values. They give quantitative bounds on derivatives from bounds on the function. This is the beginning of the analytic estimates that make complex analysis so efficient. [quotetheorem:2571] The estimate says that a bound on a circle controls every derivative at the centre. This is impossible in ordinary real differentiability and is one of the reasons complex analytic functions resist local oscillation. The natural stress test is an entire function, where circles around a point can be taken with arbitrarily large radius. If the function is bounded on the whole plane, [Cauchy estimates](/theorems/2571) leave no room for a nonzero first derivative, and this motivates [Liouville's theorem](/theorems/38). [quotetheorem:346] [Liouville's theorem](/theorems/346) is a global rigidity statement from a local estimate. Entire bounded analytic functions do not merely have limited growth; they lose all variation. [example: The Boundedness Hypothesis in Liouville's Theorem] Define $f:\mathbb{C}\to\mathbb{C}$ by $f(z)=z$. Around any centre $z_0\in\mathbb{C}$, we have \begin{align*}f(z)=z=z_0+(z-z_0)\end{align*} so $f$ is represented near $z_0$ by the power series with coefficients $a_0=z_0$, $a_1=1$, and $a_n=0$ for $n\ge 2$. Hence $f$ is analytic on all of $\mathbb{C}$. It is not constant because \begin{align*}f(0)=0\end{align*} and \begin{align*}f(1)=1\end{align*} so $f(0)\ne f(1)$. The function is not bounded on $\mathbb{C}$. If $M>0$ is any proposed bound, choose a real number $R>M$. Then $R\in\mathbb{C}$ and \begin{align*}|f(R)|=|R|=R>M.\end{align*} Thus no single $M$ can satisfy $|f(z)|\le M$ for every $z\in\mathbb{C}$. This example shows that the boundedness assumption in *Liouville's Theorem* is essential: an entire analytic function need not be constant unless it is bounded. [/example] ### Boundary Control Another expression of rigidity is the maximum principle. Liouville's theorem uses control at infinity; on a bounded region the analogous question is whether the largest value of $|f|$ can occur in the interior. In many real-variable problems, a differentiable function can have an interior maximum without being constant. For analytic functions, an interior maximum of the modulus would contradict the local open behaviour encoded by the power series, so the next theorem pushes maxima toward the boundary unless the function has no genuine variation. [quotetheorem:3345] The theorem concerns $|f|$, not the real or imaginary part separately. It says that the size of a nonconstant analytic function is pushed toward the boundary of a region. ## Singularities and Failure of Analyticity Analyticity is local, so the natural way it fails is also local: a function may be analytic everywhere near a point except at the point itself. Such punctures lead to removable singularities, poles, and essential singularities. [definition: Isolated Singularity] Let $U \subset \mathbb{C}$ be open, let $z_0\in U$, and let $f: U\setminus\{z_0\}\to\mathbb{C}$ be analytic. The point $z_0$ is an isolated singularity of $f$. [/definition] Some isolated singularities are artificial. The definition intentionally includes missing points that may later be filled in, because removable singularities are still singularities before the extension is made. In one complex variable, the local question is whether boundedness near the puncture rules out any genuine obstruction. The removable singularity principle answers yes: if $f$ is analytic on $U\setminus\{z_0\}$ and is bounded on some punctured neighbourhood of $z_0$, then $f$ has a unique analytic extension to $z_0$. The boundedness assumption distinguishes a hole from a genuine blow-up. [example: A Removable Singularity] Define $f:\mathbb{C}\setminus\{0\}\to\mathbb{C}$ by \begin{align*} f(z)=\frac{\sin z}{z}. \end{align*} The sine power series is \begin{align*} \sin z=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}. \end{align*} For $z\ne 0$, scalar multiplication of this convergent series by $1/z$ gives \begin{align*} \frac{\sin z}{z}=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!z}. \end{align*} Since $z^{2n+1}/z=z^{2n}$ for $z\ne 0$, this becomes \begin{align*} \frac{\sin z}{z}=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n+1)!}. \end{align*} Now define $F:\mathbb{C}\to\mathbb{C}$ by \begin{align*} F(z)=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n+1)!}. \end{align*} For a fixed $R>0$, the absolute-value series on $|z|\le R$ is bounded by \begin{align*} \sum_{n=0}^{\infty}\left|\frac{(-1)^n z^{2n}}{(2n+1)!}\right|\le \sum_{n=0}^{\infty}\frac{R^{2n}}{(2n+1)!}. \end{align*} The ratio of consecutive terms on the right is \begin{align*} \frac{R^{2n+2}/(2n+3)!}{R^{2n}/(2n+1)!}=\frac{R^2}{(2n+3)(2n+2)}. \end{align*} This ratio tends to $0$, so $\sum_{n=0}^{\infty}R^{2n}/(2n+1)!$ converges by the [ratio test](/theorems/174). Hence $F$ is represented by a convergent power series on every bounded disc about $0$, so $F$ is analytic at $0$ and, in fact, on all of $\mathbb{C}$. At $0$, the constant term of the series for $F$ is \begin{align*} \frac{(-1)^0 0^0}{1!}=1, \end{align*} and every term with $n\ge 1$ contains the factor $0^{2n}$, so \begin{align*} F(0)=1. \end{align*} For every $z\ne 0$, the computation above gives $F(z)=\sin z/z=f(z)$. Thus $F$ is an analytic extension of $f$ across $0$, and the isolated singularity at $0$ is removable. [/example] Not every failure is removable. The function $z\mapsto 1/z$ has a genuine pole at the origin, while $z\mapsto e^{1/z}$ has much wilder behaviour. These examples belong to the broader theory of [Laurent series](/page/Laurent%20Series) and residues. ## Beyond and Connected Topics Analytic functions are a central topic in [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). The next layer is Cauchy's theorem and the Cauchy integral formula, which explain why complex differentiability produces power series and why contour integrals vanish under the right hypotheses. The local expansion viewpoint leads naturally to Laurent series and [Residue](/page/Residue) theory. Once negative powers are allowed around isolated singularities, contour integrals can be computed from a single coefficient, and analytic functions become a practical tool for evaluating real integrals. Analytic continuation studies how far a locally defined analytic function can be extended. The identity theorem makes continuation unique on connected domains, but singularities and monodromy decide whether the continuation exists globally. In several complex variables, analyticity becomes subtler. Power series still matter, but domains of holomorphy, separate analyticity, and higher-dimensional singular sets show that the one-variable picture is only the beginning. ## References Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge II Linear Analysis](/page/Cambridge%20II%20Linear%20Analysis). Androma, [Residue](/page/Residue). Lars Ahlfors, *Complex Analysis* (1979). John B. Conway, *Functions of One Complex Variable I* (1978). Elias M. Stein and Rami Shakarchi, *Complex Analysis* (2003).