# UMD Analysis Qualifying Exam/Jan10 Complex

## Problem 2

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

### Solution

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

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

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