Laurent series for $\frac{1}{z+z^2}$

Question

I need to find the Laurent series for $\frac{1}{z+z^2}$ in these domains:

i) $0<|z|<1$

ii) $1<|z|$

iii) $1<|z+1|$

and I'm not even sure how to do it in the first one, I have all the answers but no solutions to them..

So i notice that there is no singularities inside the first domain and I tried to rewrite it as following

$\frac{1}{z} - \frac{1}{1-(-z)}$

then what?

Answer 1

1) $$\frac{1}{1-(-z)}=\sum_{k=0}^\infty (-1)^nz^n$$

2) Set $u=\frac{1}{z}$, then $|u|<1$ $$\frac{1}{z+z^2}=\frac{1}{(\frac{1}{z}+1)z^2}=\frac{u^2}{1+u}=u^2\frac{1}{1+u}=u^2\sum_{k=0}^\infty (-1)^ku^k=...$$