Finding residues with Laurent series.

[Terrell]

Homework Statement

Use an appropriate Laurent series to find the indicated residue for $f(z)=\frac{4z-6}{z(2-z)}$ ; $\operatorname{Res}(f(z),0)$

Homework Equations

n/a

The Attempt at a Solution

Computations are done such that $0 \lt \vert z\vert \lt 2$.
$\frac{4z}{z(z-2)}=\frac{2}{1-z/2}$ and $\frac{6}{z(z-2)}=\frac{6}{z}\frac{1}{1-z/2}$.
\begin{align}
\frac{4z}{z(2-z)}=2\sum_{k=0}^{\infty}(\frac{z}{2})^k=2[1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\cdots]=2+z+\frac{z^2}{2}+\frac{z^3}{4}\\
\frac{6}{z}\frac{1}{1-z/2}=\frac{6}{z}\sum_{k=0}^{\infty}(\frac{z}{2})^k=\frac{6}{z}[1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\cdots]=\frac{6}{z}+3+\frac{3}{2}z+\frac{3}{4}z^2\\
f(z)=\frac{4z}{z(2-z)}-\frac{6}{z(2-z)}=-\frac{6}{z}-1-\frac{z}{2}-\frac{1}{4}z^2-\cdots
\end{align}
What am I doing wrong? The solutions manual gave an answer of -3 while according to my solution, it must be -6.

 

[George Jones]

$\frac{4z}{z(z-2)}=\frac{2}{1-z/2}$ and $\frac{6}{z(z-2)}=\frac{6}{z}\frac{1}{1-z/2}$

Neither of these equalities is correct. The second "equality" is particularly problematic.