Is there an easy way to find the radius of convergence?

Question

Question :

Describe the given function $f(z)$ below in the power series around the center of $z_0$ and determine the radius of convergence.

$$f(z)=\frac{1}{(z-i)(z+3)}\,;\quad z_0=0$$

My attempt:

$$\begin{align} f(z)&=\frac{1}{(z-i)(z+3)}=\frac{\frac{3-i}{10}}{(z-i)}+\frac{\frac{-3+i}{10}}{(z+3)}\\ &=\frac{i\frac{3-i}{10}}{i(z-i)}+\frac{\frac{-3+i}{10}}{3(1+\frac{z}{3})}\\ &=\frac{i\frac{3-i}{10}}{(1+iz)}+\frac{\frac{-3+i}{10}}{3(1+\frac{z}{3})}\\ &=\frac{3-i}{10}\sum_{n=0}^{\infty} \frac{z^n}{i^{n-1}}+\frac{-3+i}{10}\sum_{n=0}^{\infty} \frac{(-1)^nz^n}{3^{n+1}}\\ &=\frac{3-i}{10}\sum_{n=0}^{\infty}\left[\frac{1}{i^{n-1}}-\frac{(-1)^n}{3^{n+1}}\right]z^n \end{align} $$

With Cauchy-Hadamard formula :

$$\begin{align} \rho&=\lim_{n \to \infty} \frac{|a_n|}{|a_{n+1}|}\\ &=\lim_{n \to \infty} \left|\frac{3-i}{10}\left(\frac{1}{i^{n-1}}-\frac{(-1)^n}{3^{n+1}}\right)\right| \left|\frac{10}{3-i}\left(\frac{1}{i^{n}}-\frac{(-1)^{n+1}}{3^{n+2}}\right)^{-1}\right| \\ &=\lim_{n \to \infty} \left|\frac{1}{i^{n-1}}-\frac{(-1)^n}{3^{n+1}}\right| \left|\left(\frac{1}{i^{n}}-\frac{(-1)^{n+1}}{3^{n+2}}\right)^{-1}\right| \\ &=\lim_{n \to \infty} \left|\frac{3^{n+1}-(-1)^ni^{n-1}}{i^{n-1}3^{n+1}}\right| \left|\frac{i^n 3^{n+2}}{3^{n+2}-(-1)^{n+1}i^n}\right| \\ \end{align} $$

This is getting complicated.

Please help. is there an easy way to find the radius of convergence?

Answer 1

The radius of convergence is the distance from the center of the expansion to the closest singularity.

In this case, the singularities are poles, so the radius is the distance from the center to the closest pole:

$$\min\{|0-i|, |0-(-3)|\}=\min\{1,3\}=1$$