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.

$\frac{\mathbf{cosz}}{\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{sinz}}$at$\mathbf{z}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{6}}$

Short answer

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

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 given function:

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

Rewrite the function as:

$\begin{array}{rcl}f\left(z\right)& =& \frac{p\left(z\right)}{q\left(z\right)}\\ & =& \frac{\cos z}{1-2\sin z}\\ & & \end{array}$ ……. (1)

At $z=\frac{\pi }{6}$

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

$\begin{array}{rcl}R\left(z_0\right)& =& \underset{z\to z_0}{\lim }\left(z-z_0\right)f\left(z\right)\\ & =& \frac{q\left(z_0\right)}{p\text{'}\left(z_0\right)}\end{array}$ ……. (2)

 Determine the residue of the function:

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

$\begin{array}{rcl}R\left(\frac{\pi }{6}\right)& =& \frac{\cos \left(z_0\right)}{-2\sin \left(z_0\right)}\\ & =& -\frac{1}{2}\end{array}$

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