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.