Radius of convergence of complex series around 0+i Let $f(z):=\frac{\sin z}{z(z-\pi)(z+\pi)}$.  What is the radius of convergence of the series
$$
\sum_{n=0}^{\infty} \frac{f^{(n)}(i)}{n!}(z-i)^n\quad\text{?}
$$Explain.

My Attempt :

Note that the series in question is just the Taylor series of the function $f(z)$ about the point $i$.  So the solution reduces to determining the distance between the closest non-removable isolated singularity and the point $0+i$.  All of the isolated singularities are removable and we must conclude that the radius of convergence is $\infty$.  I verify that the singularities are all removable below.

$$
\lim_{z\to 0} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/\pi^2
$$
$$
\lim_{z\to \pi} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/2\pi^2
$$
$$
\lim_{z\to -\pi} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/2\pi^2
$$
Note that 


  The radius of convergence is the distance to the nearest singularity for the function $f(z)$.
After editing my attempt to change the RoC to $\infty$ from $\mathbb{C}$ I believe my attempt is correct.