Find a Laurent series - proof check I am trying to find the Laurent expansion of:

$$f(z)=\frac{1}{(1-z)(2-z)}$$

In the region $\{z:1<|z|<2\}$. So for any $p\in (1,2)$, $f$ has a simple pole at $z=1$ in $\{z:|z|\leq p\}.$ Hence the only negative power of $z$ in the expansion is $\frac{1}{z}$. To find the coefficient:
$$c_n=\frac{1}{2\pi i}\int_{|z|=p}\frac{f(z)}{z^{n+1}}dz\Rightarrow c_{-1}=\int_{|z|=p}f(z)dz=\frac{1}{2-z}\Big|_{z=1}=1$$
So:
$$f(z)=\frac{1}{z}+h(z)$$
For some holomorphic $h$. 
First, is that correct? Also, what is $h$? I am guessing maybe just $\frac{1}{2-z}$ but not sure...
 A: As far as my experience with this goes, using the integral formula it's a no go - we only need to use the geometric series. We have: $$\begin{align}f(z) &= \frac{1}{(1-z)(2-z)} = \frac{1}{1-z} - \frac{1}{2-z}  \\ &= -\frac{1}{z}\frac{1}{1-\frac{1}{z}} - \frac{1}{2} \frac{1}{1-\frac{z}{2}}\end{align}$$In that annulus, we have $|1/z| < 1$ and $|z/2| < 1$, so: $$\begin{align} &=-\frac{1}{z}\sum_{n \geq 0}\frac{1}{z^n}-\frac{1}{2}\sum_{n \geq 0}\frac{z^n}{2^n} \\ &= -\sum_{n \geq 0} \frac{1}{z^{n+1}} -\sum_{n \geq  0}\frac{z^n}{2^{n+1}}.\end{align}$$
A: $$f(z)=\frac{1}{(1-z)(2-z)}\overset{\text{partial fractions}}{=}\underbrace{\frac{1}{z-2}}_{\text{Taylor}}-\underbrace{\frac{1}{z-1}}_{\text{Laurent}}$$
$|\frac 1 z|<1$ and $|\frac z 2|<1$
$$\frac{1}{z-2}=-\frac{1}{2-z}=-\frac 1 2 \sum_{n=0}^{\infty}\left(\frac z 2\right)^n$$
$$\frac{1}{1-z}=-\frac{1}{z-1}=-\frac 1 z \frac{1}{1-1/z}=-\frac 1 z\sum_{n=0}^{\infty}z^{-n}=-\sum_{n=0}^{\infty}z^{-n-1}$$
So
$$\boxed{-\sum_{n=0}^{\infty}z^{-n-1}-\frac 1 2 \sum_{n=0}^{\infty}\left(\frac z 2\right)^n}$$

