Originally Posted by

**artvandalay11** I was thinking about this some more, and the function should have a pole at $\displaystyle z=2\pi\imath$ since $\displaystyle \lim_{z \to 2\pi\imath}|f(z)|=\infty$ (also at any multiple of $\displaystyle z=2\pi\imath$)

The series expansion is correct, but the function still tends to infinity. To show that $\displaystyle z=2\pi\imath$ is a pole we have to be able to show $\displaystyle \lim_{z \to 2\pi\imath}(z-2\pi\imath)f(z)=0$

I can't seem to be able to do that