UMD Analysis Qualifying Exam/Jan10 Complex

Problem 2

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

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 $\sum _{n=0}^{\mathrm{\infty }}{a}_n(z+i{)}^n$ 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 $\lim \underset{n\to \mathrm{\infty }}{\sup }|a_n{|}^{1/n}=1/R$. Therefore, $\lim \underset{n\to \mathrm{\infty }}{\sup }|a_n{|}^{1/n}=\frac{2}{\sqrt{5}}$.

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