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Chapter 14: Q27P (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.
$\frac{cosz}{1 - 2 sinz}$ at $z = \frac{蟺}{6}$

Short Answer

Expert verified

The residue of the function at $z = \frac{蟺}{6}$ is $R (\frac{蟺}{6}) = - \frac{1}{2}$

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 given function:

$f (z) = \frac{\cos z}{1 - 2 \sin z}$

Rewrite the function as:

$\begin{array}{rcl} f (z) & = & \frac{p (z)}{q (z)} \\ = & \frac{\cos z}{1 - 2 \sin z} \end{array}$ 鈥︹€�. (1)

At $z = \frac{蟺}{6}$

The function has simple pole at $z = \frac{蟺}{6}$ and the residue of simple pole is given by:

$\begin{array}{rcl} R (z_{0}) & = & \lim_{z 鈫� z_{0}} (z - z_{0}) f (z) \\ = & \frac{q (z_{0})}{p' (z_{0})} \end{array}$ 鈥︹€�. (2)

03

Determine the residue of the function:

Apply equation (2) to the given equation (1) and solve:

$\begin{array}{rcl} R (\frac{蟺}{6}) & = & \frac{\cos (z_{0})}{- 2 \sin (z_{0})} \\ = & - \frac{1}{2} \end{array}$

Therefore, the residue of a function at $z = \frac{蟺}{6}$ is $R (\frac{蟺}{6}) = - \frac{1}{2}$

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