Q28P - Mathematical Methods in Physical Sciences

Question

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer.

$\frac{\mathbf{z}\mathbf{+}\mathbf{2}}{\left({\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{9}\right)\left({\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\right)}$ at $\mathbf{z}\mathbf{=}\mathbf{3}\mathbf{i}$

Step-by-Step Solution

Answer

The residue of the function at $z=3i$ is $R\left(3i\right)=-\frac{1}{16}+\frac{1}{24}i$

Step 1: Determine the formula

Consider the formula for residue of a function at simple poles is given by:

$\mathbf{R}\left[\mathbf{f}\left(\mathbf{z}\right)\right]\mathbf{=}\underset{\mathbf{z}\mathbf{\to }{\mathbf{z}}_{\mathbf{0}}}{\mathbf{l}\mathbf{i}\mathbf{m}}\left(z-z_0\right)\mathbf{f}\left(z\right)$

Step 2: Determine the residue of the simple pole

Consider the expression for the given function as:

$f\left(z\right)=\frac{z+2}{\left(z^2+9\right)\left(z^2+1\right)}$

At $z = 3i$

The function has simple pole at $z = 3i$ and the residue of simple pole is given by:

$R\left(3i\right)=\underset{z\to 3i}{\lim }\left(z-z_0\right)f\left(z\right)$                                                             ……. (1)

Step 3: Determine the residue of the function:

Substitute the values in equation (1) and solve as:

$\begin{array}{rcl}R\left(3i\right)& =& \underset{z\to z_0}{\lim }\left(z-3i\right)\frac{z+2}{\left(z^2+9\right)\left(z^2+1\right)}\\ & =& \underset{z\to z_0}{\lim }\left(z-3i\right)\frac{z+2}{\left(z+3i\right)\left(z-3i\right)\left(z^2+1\right)}\\ & =& \frac{\left(3i\right)+2}{\left(3i+3i\right)\left({\left(3i\right)}^2+1\right)}\\ & =& \frac{2+3i}{6i\left(-8\right)}\end{array}$

Solve further as:

$\begin{array}{rcl}R\left(3i\right)& =& \frac{2+3i}{-48i}\\ & =& -\frac{1}{64}+\frac{1}{24}i\end{array}$

Therefore, the residue of a function at $\begin{array}{rcl}z & =& 3i\end{array}$ is $R\left(3i\right)=-\frac{1}{16}+\frac{1}{24}$.