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Chapter 14: Q28P (page 687)

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

Short Answer

Expert verified

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

Step by step solution

01

Determine the formula

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

$R [f (z)] = \underset{z 鈫� z_{0}}{l i m} (z - z_{0}) f (z)$

02

Determine the residue of the simple pole

Consider the expression for the given function as:

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

At $z = 3 i$

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

$R (3 i) = \lim_{z 鈫� 3 i} (z - z_{0}) f (z)$ 鈥︹€�. (1)

03

Determine the residue of the function:

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

$\begin{array}{rcl} R (3 i) & = & \lim_{z 鈫� z_{0}} (z - 3 i) \frac{z + 2}{(z^{2} + 9) (z^{2} + 1)} \\ = & \lim_{z 鈫� z_{0}} (z - 3 i) \frac{z + 2}{(z + 3 i) (z - 3 i) (z^{2} + 1)} \\ = & \frac{(3 i) + 2}{(3 i + 3 i) ({(3 i)}^{2} + 1)} \\ = & \frac{2 + 3 i}{6 i (- 8)} \end{array}$

Solve further as:

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

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

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