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$\mathbf{z}\mathbf{=}\mathbf{3}\mathbf{i}$

Short answer

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

Step-by-step solution

Determine the formula 

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

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

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)

 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 $R\left(3i\right)=-\frac{1}{16}+\frac{1}{24}$.