1. ## Power series (Complex)

Find the power series about the origin for the given function.

(1) $\frac{z^3}{1-z^3}$ , |z| < 1

(2) $z^2cosz$

I will try to attempt the first one:

(1) $\frac{1}{(1-z)^3} = (\frac{1}{1-z})'$

$\frac{1}{1-z} = 1 + z + z^2 + z^3 + ....$

$\frac{z}{1-z} = z + z^2 + z^3 + z^4 + ...$

$\frac{z^3}{1-z} = z^3 + z^6 + z^9 + z^{12} + ...$

$= \sum_{n = 1}^{\infty} z^{3n}$

Somehow, I think my reasoning is wrong. The answer is correct, but I am not sure I took the correct steps.

I have no idea how to approach (2). Thanks for looking and any help provided!

(1) $\frac{z^3}{1-z^3}$ , |z| < 1
$\frac{1}{1-z^3} = 1 + z^3+z^6+...$ now multiply by $z^3$
(2) $z^2cosz$
$z^2 \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$