Suppose $f$ is analytic in an open set that contains $\overline{B_r(z_0)}$, and $\left|\, f(z) \,\right| \le m$ holds on $\partial B_r(z_0)$ for some constant m. Then for all $k \ge 0$,

$$\begin{aligned} \left|\, f^{(k)}(z_0) \,\right| \le \frac{k!\,m}{r^k} \\ \end{aligned}$$

which turns out to be remarkably not hard to prove based on the Cauchy's Integral Formula. If I understand the notation correctly, $\overline{B_r(z_0)}$ is the disk with radius $r$ centered at $z_0$, whereas $\partial B_r(z_0)$ is the boundary of that disk.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 2:20 PM