Question

To find that the integrals by computing residue at infinity.

${\mathbf{\oint }}_{\mathbf{c}}\frac{{\mathbf{z}}^{\mathbf{2}}}{\left(2z+1\right)\left(z^2+9\right)}$ around $\left|z\right|\mathbf{=}\mathbf{5}$ .

Short answer

The answer is, ${\oint }_{c}f\left(z\right)dz=2\mathrm{\pi iRes}(\mathrm{z}=\infty )=2\mathrm{\pi i}\left(\frac{-1}{2}\right)=\mathrm{\pi i}$

Step-by-step solution

Evaluate the integral by residue at  t=∞

Let $I={\oint }_c\frac{z^2}{\left(2z+1\right)\left(z^2+9\right)}$ around$\left|z\right|=5$ .

The integral by residue at  t=∞

The integral by residue is expressed as,

${\mathbf{\oint }}_{\mathbf{c}}\mathit{f}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathit{d}\mathit{z}\mathbf{=}\mathbf{2}\mathbf{\pi iRes}\mathbf{(}\mathbf{z}\mathbf{=}\mathbf{\infty }\mathbf{)}$

C  is described as counter clock wise direction

Suppose,.$\begin{array}{rcl}f\left(z\right)& =& \frac{z^2}{\left(2z+1\right)\left(z^2+9\right)}\\ & =& \frac{1}{\left(2z+1\right)\left(1+{\displaystyle \frac{9}{z^2}}\right)}\\ & =& \frac{1}{2\left(z+{\displaystyle \frac{1}{2}}\right)\left(1+{\displaystyle \frac{9}{z^2}}\right)}\end{array}$

In regular at $\begin{array}{rcl}z& =& \infty \end{array}$ is given$\begin{array}{rcl}& & \underset{z\to \infty }{\lim }(-zf(z\left)\right)\end{array}$.

$\begin{array}{rcl}& \Rightarrow & \underset{z\to \infty }{\lim }\frac{1}{2\left(z+{\displaystyle \frac{1}{2}}\right)\left(1+{\displaystyle \frac{9}{z^2}}\right)}=-\frac{1}{2}\end{array}$

Hence, $\begin{array}{rcl}{\oint }_{c}f\left(z\right)dz=2\mathrm{\pi iRes}(\mathrm{z}& =& \infty )=2\mathrm{\pi i}\left(\frac{-1}{2}\right)=\mathrm{\pi i}\end{array}$