Expand using Laurent series

Question

Could anyone help me expanding the following by Laurent series: 1/(z+1)(z+2)^2 Range:0<|z-1|<2 I'm particularly getting confused with that square term.

Answer 1

Note that$$\frac1{(z+1)(z+2)^2}=-\frac1{z+2}-\frac1{(z+2)^2}+\frac1{z+1}.$$If $0<\lvert z-1\rvert<2$, then\begin{align}\frac1{z+1}&=\frac1{2+(z-1)}\\&=\frac12\cdot\frac1{1+\frac{z-1}2}\\&=\frac12\sum_{n=0}^\infty\left(-\frac{z-1}2\right)^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-1)^n\end{align}and\begin{align}-\frac1{z+2}&=-\frac1{3+(z-1)}\\&=-\frac13\cdot\frac1{1+\frac{z-1}3}\\&=\sum_{n=0}^\infty\frac{(-1)^{n+1}}{3^{n+1}}(z-1)^n.\end{align}Finally, use the fact that$$\frac1{(z+2)^2}=\left(-\frac1{z+2}\right)'$$to get the third series.