Laurent series of $f(z)=\frac{1}{1-z}-\frac{1}{2-z}$ I have to find the Laurent series of $f(z)=\frac{1}{1-z}-\frac{1}{2-z}$ on $D_1(0)=\{z\mid |z|<1\}$, on $C_{1,2}(0)=\{z\mid 1<|z|<2\}$ and on $C_{2,\infty }(0)=\{z\mid |z|>2\}$.
For $D_1(0)$ no problem. For $C_{1,2}(0)$ I don't understand why 
$$f(z)=\frac{-1}{z\left(1-\frac{1}{z}\right)}-\frac{2}{\left(1-\frac{z}{2}\right)}$$
Why do we have to use this form ?
In the same way, why on $C_{2,\infty }(0)$ we use
$$f(z)=\frac{-1}{z\left(1-\frac{1}{z}\right)}+\frac{1}{z\left(1-\frac{2}{z}\right)}\ ?$$
 A: The geometric series converges when $\lvert r\rvert<1$. For Laurent series in the second domain, you have $1<\lvert z\rvert<2$ so we need $1/\lvert z\rvert<1$ and $\lvert z\rvert/2<1$ for the series to converge.
In the third domain, you have$\lvert z\rvert>2>1$ so you need $1/\lvert z\rvert<1$ and $2/\lvert z\rvert<1$.
A: The Taylor series for $\frac{1}{1-z}$, is $1+z+z^2+z^3+\cdots$.  This converges for $|z|<1$ and diverges otherwise.  Now, if you want a series for $\frac{1}{1-z}$ outside the disk of radius 1, this expansion won't work because it diverges (this makes sense because when passing outside the disk of radius 1, you're including the pole of $\frac{1}{1-z}$.
For outside the disk of radius 1, you can consider the Taylor expansion around infinity.  More precisely, it is easy to see that $\frac{1}{-z\left(1-\frac{1}{z}\right)}$ is equal to $\frac{1}{1-z}$, and, for $|z|>1$, $\left|\frac{1}{z}\right|<1$.  Therefore, the Taylor expansion from above applies (by replacing $z$ with $\frac{1}{z}$).  This gives the power series expansion that you're searching for.  
