Find the Laurent Series Representation of the Function

Question

$ f(z)=\frac{2z+1}{(z-1)(z+3) } $

Around, $ z=0$

Inside the annulus defined by $ 1< \left | z \right |< 3$

This is what I get: $ \frac{3}{4}\sum_{n=1}^{\infty}(\frac{1}{z^n})+\frac{4}{15}\sum_{n=0}^{\infty}(\frac{-1}{3})^n(z^n) $

Is it right?

Answer 1

I get another factor in front of the second series: $$ \frac{2z+1}{(z-1)(z+3)} = \frac{3/4}{z-1} + \frac{5/4}{z+3} = \frac{3}{4}\frac{1}{z}\frac{1}{1-\frac1z} + \frac{5}{4}\frac{1}{3}\frac{1}{1+\frac z3} \\ = \frac34 \frac{1}{z} \sum_{n=0}^{\infty} \left(\frac1z\right)^n + \frac{5}{12} \sum_{n=0}^{\infty} \left(-\frac z3\right)^n \\ = \frac34 \sum_{n=1}^{\infty} \left(\frac1z\right)^n + \frac{5}{12} \sum_{n=0}^{\infty} \left(-\frac z3\right)^n \\ $$