Computing a Laurent series Let $$f(z) = \frac{1}{(2z-1)(z-3)} $$. Compute the Laurent series about the point z = 1 in the annular domain $$ \frac{1}{2} < |z-1| < 2$$
My attempt:
I broke f(z) up into the partial fraction decomposition:
$$ -\frac{2}{5(2z-1)} + \frac{1}{5(z-3)} = -\frac{2}{5}*\frac{1}{(1-\frac{(z+\frac{1}{2})}{2})} +\frac{1}{5}*\frac{1}{1-(z-2)} = $$
$$-\frac{2}{5}\sum_{n=0}^\infty(-1)^{n}\frac{(z+1)^{n}}{2^n}-\frac{1}{5}\sum_{n=0}^\infty(z-2)^n $$
And that was my answer. But I was told I was wrong, and I'm not sure where I went wrong in there. So if someone could point out where I went wrong, it would be greatly appreciated! 

$$\frac{1}{(2z-1)(z-3)}=\frac{1}{5}\left(\frac{1}{z-3}-\frac{1}{z-\frac{1}{2}}\right)=-\frac{1}{10}\frac{1}{1-\frac{z-1}{2}}-\frac{1}{5(z-1)}\frac{1}{\left(1+\frac{1}{2(z-1)}\right)}$$
Now only check that 
$$\left|\frac{z-1}{2}\right|<1\,\,,\,\,|2(z-1)|^{-1}<1$$
and you'll be able to use the developments
$$\frac{1}{1-z}=1+z+z^2+...=\sum_{n=0}^\infty z^n\,\,,\,\,\frac{1}{1+z}=1-z+z^2-...=\sum_{n=0}^\infty(-1)^{n}z^n$$
