I need to compute the Laurent series expansion of the following function - $$f(z) = \frac{1}{z^2-z}$$ centered at $z= -1$ and converges at $z=-1/2$ .
I tried this problem using substitution $w=z+1$. My function changes to $$\frac{1}{(w-1)(w-2)}$$
Now, I used the following formula to write the Laurent expansion of the above function $$\frac{1}{1-z} = \sum_{n=0}^{\infty}z^n \space \space \space ; \space |z| < 1$$ and it came out to be $$\sum_{n=0}^{\infty}(z+1)^n \frac{2^{n+1} -1 }{2^{n+1}}$$
But this series is not convergent at $z=1/2$. I have two questions -
1) If the condition of convergence at $z=1/2$ wasn't given then is my solution right?
2) How to do such problems where such conditions of convergence are given?
Let $0<|z+1|<1$, then \begin{align} \frac{1}{z^2-z}&=\frac{1}{z(z-1)}\\ &=\frac{1}{z-1}-\frac{1}{z}\\ &=-\frac{1}{2(1-\frac{z+1}{2})}+\frac{1}{1-(1+z)}\\ &=-\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z+1}{2}\right)^n +\sum_{n=0}^{\infty} (z+1)^n, \end{align} This series converges at $z=-\frac{1}{2}$. However, $f(z)$ has simple poles at $z=0$ and $z=1$, and if domain of Laurent series contains poles, then Laurent series is not valid. For $z=\frac{1}{2}$, it is outside the annulus of convergence. Instead, you can find the Laurent series in $0< |z-1|< 1$ or $0<|z|<1$ and they converge at $z=\frac{1}{2}$. For example, the Laurent series of $f(z)$ in $0<|z-1|<1$ is $$ \frac{1}{z^2-z}=\frac{1}{z-1}\sum_{n=0}^{\infty}(-1)^n (z-1)^n $$ and it converges at $z=\frac{1}{2}$.