Question

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

Answer 1

Statement of Cauchy's Residue Theorem: 

If f (z) is analytic in a closed curve C, except at a finite number of poles within C, then

\( \begin {aligned} \int_{c} f(z) dz &= 2 \pi i(\text{ Sum of all Residues}) \end{aligned} \)

 

Solution :

The poles of the integrand are given by putting the denominator equal to zero.

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

Singular points are z =1, -1

The integrand is analytic on | z | = 2 & all points is inside the circle | z | = 2.

We have to find residues at each points.

 

1)Residue at \( z_o \)=1, with pole of order 2

\( \begin{aligned} &=\frac{1}{(n-1)!} \lim _{z \rightarrow z_{0}} \frac{d^{n-1}}{d x^{n-1}}\left[\left(z-z_{0}\right)^{n} f(z)\right] \\ &=\lim _{z \rightarrow 1} \frac{d}{d z}\left[(z-1)^{2} \frac{z^{2}}{(z-1)^{2}(z+1)}\right] \\ &=\lim _{z \rightarrow 1} \frac{d}{d z}\left[\frac{z^{2}}{z+1}\right] \\ &=\lim _{z \rightarrow 1}\left(\frac{ z^{2}+2 z}{(z+1)^{2}}\right) \\ &=\frac{3}{4} \end{aligned} \)

2) Residue at \( z_o \)=-1, with simple pole 

\( \begin{aligned} &=\frac{1}{(n-1) !} \lim _{z \rightarrow z_{0}} \frac{d^{n-1}}{d x^{n-1}}\left[\left(z-z_{0}\right)^{n} f(z)\right] \\ &=\lim _{z \rightarrow -1} \left[(z+1) \frac{z^{2}}{(z-1)^{2}(z+1)}\right] \\ &=\lim _{z \rightarrow -1}\left(\frac{ z^{2}}{(z-1)^{2}}\right) \\ &=\frac{1}{4} \end{aligned} \)

3) By Cauchy's Residue theorem 

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

Hence, The sum of all residues is \( 2\pi i \)

This is solved by Cauchy's Residue Theorem.