Laurent series problem on $f(z)=\frac{ z }{ z^2-z-2 }$

Question

I have problems with computing Laurent series of the function $f(z)=\frac{ z }{ z^2-z-2 }\quad$ in the ring centered in $0$ containing point $1+i$. I also have to find the radius of convergence of this series.

My idea is:

$$f(z)=\frac{z}{2}\left(\frac{ 1 }{ \cfrac{ z^2-z }{ 2 }-1}\right)$$

We can treat $\frac{ z^2-z }{ 2 }$ as a ratio in geometric series, and later on a group by $z^n$ and find coefficients of the Laurent series. Unfortunately, it doesn't look good after computations.

What should I do?

Answer 1

Decompose using partial fractions: $f(z)=\frac{z}{(z-2)(z+1)}=\frac{z}3(\frac1{z-2}-\frac1{z+1})$.

Now use the geometric series: $\frac1{1-z}=\sum_{n=0}^\infty z^n$.

For instance, $\frac1{z-2}=-\frac12\frac1{1-\frac z2}=-\frac12\sum_{n=0}^\infty (\frac z2)^n$, for $\mid z\mid\lt2$.

For the other one, we have $\frac1{1+z}=\frac1z\frac1{1-(-\frac 1z)}=\sum_{n=0}^\infty (-1)^n(\frac1z)^{n+1}=\sum_{n=-\infty}^{-1}(-1)^{n+1}z^n$ for $\mid z\mid\gt1$.

The annulus of convergence is $1\lt \mid z\mid\lt 2$.

$\boxed{f(z)=\frac13\sum_{n=-\infty}^{0}(-1)^{n+1}z^n-\frac1{12}\sum_{n=1}^\infty {(\frac z2)}^n}$.