Expand $f(z)=\frac{1}{z(z-1)}$ in a Laurent series valid for the follwing annular domains.
$a)0\lt \vert z \rvert \lt 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)1\le\lvert z \rvert\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c)0\le \lvert z-1 \rvert \lt1\,\,\,\,\,\,\,\,\,\,\,\,\,d) 1\le \lvert z-1 \rvert$
Ok, so here are some things I know, There are singularities at $z_0=0$and $z_0=1$
a) is the unit circle, shaded inside.$f(z)=\frac{1}{z(z-1)}=\frac{1}{z}*\frac{1}{z-1}=\frac{-1}{z}*\frac{1}{1-z}=\frac{-1}{z}*\sum_{n=0}^\infty Z^n=\frac{-1}{z}[z^0+z^1+z^2...]=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f(z)=\frac{-1}{z}-1-z-z^2-...\,$ which converges for $0\lt \vert z\rvert \lt 1$
b)is the unit circle shaded outside the circle. $f(z)=\frac{1}{z(z-1)}$$=\frac{1}{z}*\frac{1}{z-1}$$=\frac{1}{z}*\frac{1}{z(1-1/z)}=\frac{1}{z^2}*\frac{1}{1-1/z}=\frac{1}{z^2}*\sum_{n=0}^\infty(\frac{1}{z})^n=\frac{1}{z^2}[\frac{1}{z}^0+\frac{1}{z}^1+\frac{1}{z}^2+...]=\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+...$
c)is a circle with r=1, shaded inside. Here, Im not so sure how to manipulate the function and have gotten stuck
d)is a cirlce with r=1, shaded outside
For (c)
$$\frac{1}{z(z-1)}= \frac1{z-1}\frac{1}{1+(z-1)}= \frac1{z-1}\sum_{k=0}^{\infty}(-1)^k(z-1)^{k}$$
For (d)
$$\frac{1}{z(z-1)}= \frac{\frac1{(z-1)^2}}{1+\frac1{z-1}}=\frac1{(z-1)^2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(z-1)^k}$$