Evaluate $\int _{c} \frac {z^2}{(z-1)^2(z+1)}dz$ where c is |z|=2 using Cauchy's residue theorem

Solution :

$ \text{Let} \\ \begin{array}{I} f(z) =\int_{c} \frac{z^{2} d z}{(z-1)^{2}(z+1)} d z \\ \end{array} $

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$ (z-1)^{2}(z+1) \Rightarrow z =1,-1 $

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Singular points are $z=1,-1$

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$z = 1$ is pole of order 2 and $z = -1$ is simple pole.

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$|z| = 2$ is circle with centre (0,0) radius 2.

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hence both the pole lies inside the region

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1) Residue at $z_{0}$=1 with pole of order 2

$ \begin{array}{I} &=\frac{1}{(n-1)!} \lim _{z \rightarrow z_{0}} \frac{d^{n-1}}{d x^{n-1}}\left[\left(z-z_{0}\right)^{n} f(z)\right] \\ \end{array} $

$ \begin {array}{I} &=\lim _{z \rightarrow 1} \frac{d}{d z}\left[(z-1)^{2} \frac{z^{2}}{(z-1)^{2}(z+1)}\right] \\ \end{array} $

$ \begin{array}{I} &=\lim _{z \rightarrow 1} \frac{d}{d z}\left[\frac{z^{2}}{z+1}\right] \\ \end{array} $

$ \begin{array}{I} &=\lim _{z \rightarrow 1}\left(\frac{ z^{2}+2 z}{(z+1)^{2}}\right) \\ \end{array} $

$ \begin{array}{I} &=\frac{3}{4} \end{array} $

2) Residue at $z_{0}$=-1 with simple pole

$ \begin{array}{I} &=\frac{1}{(n-1) !} \lim _{z \rightarrow z_{0}} \frac{d^{n-1}}{d x^{n-1}}\left[\left(z-z_{0}\right)^{n} f(z)\right] \\ \end{array} $ $ \begin{array}{I} &=\lim _{z \rightarrow -1} \left[(z+1) \frac{z^{2}}{(z-1)^{2}(z+1)}\right] \\ \end{array} $ $ \begin{array}{I} &=\lim _{z \rightarrow -1}\left(\frac{ z^{2}}{(z-1)^{2}}\right) \\ \end{array} $

$ \begin{array}{I} &=\frac{1}{4} \end{array} $

By Cauchy's Residue Theorem,

$ \begin {array}{I} \int_{c} f(z) dz &= 2 \pi i(\text{ Sum of all Residues}) \\ &= 2 \pi i ({ \frac{3}{4}}+{ \frac{1}{4}}) \\ &=2 \pi i \end{array} $