# UMD Analysis Qualifying Exam/Jan10 Complex

## Problem 2

 The function $\sec \pi z$ has a convergent Taylor expansion $\sum _{n=0}^{\infty }a_{n}(z+i)^{n}$ . Find $\lim \sup _{n\to \infty }|a_{n}|^{1/n}$ .

### Solution

$\sec \pi z={\frac {1}{\cos \pi z}}$ . By using the definition $\cos z={\frac {e^{iz}+e^{-iz}}{2}}$ , we get that $\cos(z)=0$  if and only if $e^{-y}e^{ix}=-e^{y}e^{-ix}$ . It is not hard to show that this happens if and only if $y=0$  and $x={\frac {(2n+1)\pi }{2}}$ . Therefore, the only zeros of $\cos \pi z$  all occur on the real axis at integer distances away from 1/2. Therefore, $\sec \pi z$  is analytic everywhere except at these points.

Our Taylor series $\sum _{n=0}^{\infty }a_{n}(z+i)^{n}$  is centered at $z=-i$ . By simple geometry, the shortest distance from $z=-i$  to $z=1/2$  or $z=-1/2$  (the closest poles of $\sec \pi z$ ) is $R={\sqrt {1/2^{2}+1^{2}}}={\frac {\sqrt {5}}{2}}$ . This is the radius of convergence of the Taylor series.

From calculus (root test), we know that $\lim \sup _{n\to \infty }|a_{n}|^{1/n}=1/R$ . Therefore, $\lim \sup _{n\to \infty }|a_{n}|^{1/n}={\frac {2}{\sqrt {5}}}$ .