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!