Laurent series for $\frac{1}{(z-1)^2(z+2)}$

Question

Find Laurent expansions for $$\frac{1}{(z-1)^2(z+2)}$$

in $A_1 = D(0,1)$, $A_2 = {z : 1 < |z| < 2}$, $A_3 = {{z : \sqrt{2} < |z-i| < \sqrt{5}}}$

I've split it into partial fractions and am trying to apply Taylor's expansion to the square term and convergent sum of geometric to the other two. I'm confused because I'm not sure how you're meant to know the expansion point (like $z$ or $z-1$ etc). Please can someone tell me how to do the $A_2$ scenario? Thanks in advance.

Answer 1

Hint

For $\dfrac 1{(z-1)^2}$, note that it equals $-(\dfrac 1{z-1})'$.

So since (geometric series) $\dfrac 1{z-1}=\dfrac 1z\dfrac 1{1-\frac 1z}=\sum_{n\ge0}z^{-n}$, differentiate term by term.

Get $\dfrac 1{(z-1)^2}=-\sum_{n\ge0}-nz^{-(n+1)}$, on $\lvert z\rvert \gt1$. This is about $z_0=0$.

Since $\dfrac 1{z+2}=\dfrac 12\dfrac 1{1-(-\frac z2)}=\dfrac 12\sum_{n\ge0}(-\dfrac z2)^n$ on $\lvert z\rvert \lt2$, now you just combine.