Power series of $f(z) = \frac{z}{1-z}$

Question

Find power series of $f(z) = \frac{z}{1-z}$ in point $z_0 = i$ and find radius of convergence this power series.

Of course, I can find $f^{(n)}(z_0)$ and then I will have $$f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n$$ where $a_n = \frac{f^{(n)}(z_0)}{n!}$ but maybe there is faster way to get this power series?

Answer 1

A geometric series is known without thee need for differentiation:

$$u=z-i$$

$$f(u)=\frac{u+i}{1-i-u}=\frac{(u+i)/(1-i)}{1-u/(1-i)}=\frac{1+i}{2}\frac{u+i}{1-u(1+i)/2}$$

Now expand

$$\frac{1}{1-u(1+i)/2}=\sum_{n=0}^\infty \left(\frac{1+i}{2}\right)^n u^n$$

This also sets the radius of convergence as $|u(1+i)/2|<1$.

The multiplication by $u+i$ just means you have to add two series where one has all powers increased by 1.