Question

Evaluate the following integrals by computing residues at infinity. Check your answers by computing residues at all the finite poles. (It is understood that $\mathbf{\oint }$ means in the positive direction.)

$\mathbf{\oint }\frac{\mathbf{1}\mathbf{-}{\mathbf{z}}^{\mathbf{2}}}{\mathbf{1}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}\frac{\mathbf{d}\mathbf{z}}{\mathbf{z}}$around $\left|z\right|\mathbf{=}\mathbf{2}$

Short answer

Integral, I = -1 by using residue method.

Step-by-step solution

If a function is not analytic at  z = a then f(z)  has a singularity at z = a.

Here integral is given as,$\oint \frac{1-z^2}{1+z^2}·\frac{dz}{z}$.

Around z = 2 as follows:

$\oint f\left(z\right)=2\mathrm{\pi iRes}\left(\mathrm{f}\right(\mathrm{z}),\mathrm{z}=\infty )$

Let's replace z by $\frac{1}{z}$ and compute the residue.

$\oint \frac{1-{\displaystyle \frac{1}{z^2}}}{1+\frac{1}{z^2}}·\frac{dz}{z}$

Residue at infinity

Residue at infinity is as follows:

$\begin{array}{rcl}Res\left(f\right(z),\infty )& =& -Res\left(\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right)\\ & =& -Res\left(\frac{1}{z^2}·\frac{z·\left(1-{\displaystyle \frac{1}{z^2}}\right)}{1+{\displaystyle \frac{1}{z^2}}},0\right)\\ & =& -Res\left(\frac{z^2-1}{z^2(z+1)},0\right)\\ I& =& -Res\\ & =& -\underset{z\to 0}{\lim }\frac{d}{dz}\frac{z^2-1}{z+1}\end{array}$

Simplify further as follows:

$\begin{array}{rcl}I& =& -\underset{z\to 0}{\lim }\frac{\left(z+1\right)\left(2z\right)-\left(z^2-1\right)}{{\left(z+1\right)}^2}\\ & =& -\underset{z\to 0}{\lim }\frac{2z^2+2z-z^2+1}{{\left(z+1\right)}^2}\\ & =& -\underset{z\to 0}{\lim }\frac{2z^2+2z+1}{{\left(z+1\right)}^2}\\ & =& -\underset{z\to 0}{\lim }\frac{{\left(z+1\right)}^2}{{\left(z+1\right)}^2}\end{array}$

So,

$\begin{array}{rcl}I& =& -\frac{{\left(0+1\right)}^2}{{\left(0+1\right)}^2}\\ & =& -1\end{array}$

Hence, the required integral is I = -1 by using residue method.