# Help needed on laurent series for a complex function

I'm struggling to understand the ways in which one could find the laurent series and there for the residuals for:

Find the Laurent series expansion and residue at $$\left(\frac{z}{z-1}\right)^2$$ for $z = 1$

Any help that could be provided as to where to start would be appreciated. I attempted differentiating the Laurent series expansion for

$$\frac{1}{z-1}$$

Aswell as trying to multiply out the coefficients of the Laurent expansion for

$$\frac{1}{z-1}$$

But have had no luck whatsoever and just get myself into a state.

Laurent series about $z=1$: \begin{align}\frac{z^2}{(z-1)^2} &=\frac{z^2\color{blue}{-2z+2+2z-2}}{(z-1)^2} =\frac{\left(z^2-2z+1\right)+2\left(z-1\right)+1}{(z-1)^2} \\[2mm] &=\frac{(z-1)^2+2\left(z-1\right)+1}{(z-1)^2} =\color{red}{1+\frac{2}{z-1}+\frac{1}{(z-1)^2}} \end{align} Taylor series at $z=0$: \begin{align} \frac{z^2}{(z-1)^2} &=z^2\,\left(\frac{1}{1-z}\right)' =z^2\,\left(\sum_{n=0}^{\infty}z^n\right)' =z^2\,\sum_{n=0}^{\infty}\left(z^n\right)' \\[2mm] &=z^2\,\sum_{n=0}^{\infty}n\,z^{n-1} =\color{red}{\sum_{n=1}^{\infty}n\,z^{n+1}} \end{align}
• Thank you for your help, but where you have expanded the taylor series at $z=0$ how would you find the residue? – Sebastian TG Feb 23 '17 at 16:01
• @SebastianTG: Using Laurent series, the residue is the coefficient of $\,(z-1)^{-1}\,$, which is $\,2\,$. – Hazem Orabi Feb 23 '17 at 20:52
Note that $$\left( \frac{z}{z-1}\right)^2=\left(1+\frac{1}{z-1} \right)^2$$