UMD Analysis Qualifying Exam/Jan10 Complex

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, $$\lim \sup_{n\to\infty} |a_n|^{1/n}=\frac{2}{\sqrt 5}$$.