Radius of convergence of Maclaurin series for $\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$

Question

What is the radius of convergence of the Taylor series about $z=0$ for $h(z)=\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$? Here's a plot

http://www.wolframalpha.com/input/?i=2z%2F%28z^2-pi^2%29%2B1%2Fsin+z+-+1%2Fz

Remark that $h$ is analytic in the disk $|z|<2\pi$ except for removable singularities at $z=0,\pm\pi$.

What are the first few coefficients (with positive and negative indices) in the laurent series expansion for $\csc z$, valid in the annulus $\pi<|z|<2\pi$? I'm supposed to use the above result to answer this. Thank you.

Answer 1

The radius of convergence is the distance to the closest non-removable singularity. You've already shown that the singularities at $0$ and $\pm \pi$ are removable. So where are the other singularities?