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, $\lim \sup _{n\to \infty }|a_{n}|^{1/n}={\frac {2}{\sqrt {5}}}$.