Let $f: U \rightarrow \mathbb{C}$ be analytic and let $\{|z - z_0| < r \subset U\}$. Then in this disk, $f$ has a power series representation:

$$\begin{aligned} f(z) = \sum_{k=0}^\infty a_k (z - z_0)^k, \quad |z - z_0| < r, \quad \text{where } a_k = \frac{f^{(k)}(z_0)}{k!} \\ \end{aligned}$$

The radius of convergence of this power series is $R \ge r$. Note this means an analytic function is determined entirely in a disk by all it's derivatives $f^{(k)}(z_0)$ at the center $z_0$ of the disk!

In other words, if $f$ and $g$ are analytic in $\{|z - z_0| < r\}$ and if $f^{(k)}(z_0) = g^{(k)}(z_0)$ for all k, then $f(z) = g(z)$ for all $z$ in $\{|z - z_0| < r\}$ !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 4:34 PM