Radius of convergence of $f(z)$

Question

Radius of convergence of $f(z)=\frac{\sin z}{(z-\pi)^2}$. I find using Laurent series that $0<\vert z-\pi\vert<\pi$ but the answer is $0<\vert z-\pi\vert<\infty$. Why so?

Answer 1

$$f(z)=\frac{\sin(z)}{z-\pi}\cdot\frac1{z-\pi}$$

Note that $\frac{\sin(z)}{z-\pi}$ has one and the only singularity at $z=\pi$ which is removable, therefore this function converges on the whole complex plane.

$\frac1{z-\pi}$ has a singularity at $\pi$.

I think $|z|<\pi$ is correct.

For the Laurent series at $\pi$, the radius of convergence is of course infinite except at $\pi$ because there are no other singularities on the whole complex plane.

So, $0<|z-\pi|<\infty$.

Answer 2

We have $ \sin z= - \sin(z- \pi)= -(z- \pi)+\frac{(z- \pi)^3}{3!}-\frac{(z- \pi)^5}{5!}+- ....$ for all (!) $z$.

Hence $f(z)=-\frac{1}{z- \pi}+\frac{z- \pi}{3!}-\frac{(z- \pi)^3}{5!}+- ....$ for all $z \ne \pi$.