Finding Laurent and Taylor series I need to find both a Laurent and a Taylor expansion. 
$$f(z)=\frac{z}{(z-1)(z-2)} = \frac{-1}{(z-1)}+\frac{2}{(z-2)}$$
If I choose $z_0=0$ 
$$f(z)=\frac{1}{(1 + z)} - \frac{4}{\left(1 - \frac{z}{4}\right)}$$
$$f(z)=\sum_{n}^{\infty}(-1)^n{z^n} - 4\sum_{n}^{\infty}(\frac{z}{4})^n$$
Which is a Taylor series.

What value of $z_0$ would you pick for a Laurent series?

It is sufficient to consider a Laurent expansion around $z=0$. Depending  on the region which you then choose you obtain a principal part of a Laurent expansion respectively an expansion as Taylor series.

The function
\begin{align*}
 f(z)&= \frac{-1}{z-1}+\frac{2}{z-2}\\
\end{align*}
   has two simple poles at $1$ and $2$.
We look at the poles $1$ and $2$ and see they determine three regions.
\begin{align*}
 |z|<1,\qquad\quad
 1<|z|<2,\qquad\quad
 2<|z|
 \end{align*} 
  
  
*
  
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $1$ and $2$  admit a representation as Taylor series at $z=0$.
  
*The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with pole $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $2$ admits a representation as power series.
  
*The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A Tayor series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
