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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.

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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 $z=0$: $$ \begin{align} \frac{z^2}{(z-1)^2} &=z^2\,\left(\frac{1}{1-z}\right)' =z^2\,\left(\sum_{n=0}^{\infty}z^n\right)' =z^2\,\sum_{n=0}^{\infty}\left(z^n\right)' \\[2mm] &=z^2\,\sum_{n=0}^{\infty}n\,z^{n-1} =\color{red}{\sum_{n=1}^{\infty}n\,z^{n+1}} \end{align} $$

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  • $\begingroup$ I don't understand what to do :s $\endgroup$ – Sebastian TG Feb 23 '17 at 11:41
  • $\begingroup$ Thank you for your help, but where you have expanded the taylor series at $z=0$ how would you find the residue? $\endgroup$ – Sebastian TG Feb 23 '17 at 16:01
  • $\begingroup$ @SebastianTG: Using Laurent series, the residue is the coefficient of $\,(z-1)^{-1}\,$, which is $\,2\,$. $\endgroup$ – Hazem Orabi Feb 23 '17 at 20:52
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Hint:

Note that $$ \left( \frac{z}{z-1}\right)^2=\left(1+\frac{1}{z-1} \right)^2 $$

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  • $\begingroup$ My confusion comes with dealing with the power of 2. Can I deal with the power series on its own within the brackets and then square the resulting laurent series $\endgroup$ – Sebastian TG Feb 23 '17 at 10:43

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