finding a möbius transformation for power series

Question

Let f be the function with domain $D = {{z \in C : |z| < 8}}$ given by f(z)= $\sum (i^nz^n)/(8^n) $ find a mobius transformation g such that f(z)=g(z) for all $z\in D$. Calculate f(i) and f'(0)

I literally do not know where to start so any help would be greatly appreciated thanks

Answer 1

$\sum_{n≥0} \frac{i^nz^n}{8^n} = \sum_{n≥0}(\frac{iz}{8})^n$

Let $w=\frac{iz}{8}$

Then $\sum_{n≥0}(\frac{iz}{8})^n = \sum_{n≥0}w^n$

This is now just a geometric progression so we have that

$\sum_{n≥0}w^n = \frac{1}{1-w} = \frac{1}{1-\frac{iz}{8}} = \frac{8}{8-iz}$

This is a fractional linear transformation with a=0, b=8, c=-i, d=8.

A fractional linear transformation is a Möbius transformation if $ad-bc≠0$.

$ ad-bc = 8i≠0$ so this is a Möbius transformation, as required.