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.

$\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{z}}}{\mathbf{4}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\mathbf{z}\mathbf{-}\mathbf{5}}$at$\mathit{z}\mathbf{=}\mathit{l}\mathit{n}\mathbf{2}$

Short answer

The residue of the function at $z=\ln 2$is $R\left(\ln 2\right)=\frac{4}{3}$

Step-by-step solution

Determine the formula 

Simple pole at the residue is given by the function as:

$\mathit{R}\left[f\left(z\right)\right]\mathbf{=}\underset{\mathbf{z}\mathbf{\to }{\mathbf{z}}_{\mathbf{0}}}{\mathbf{lim}}\mathbf{(}\mathit{z}\mathbf{-}{\mathbf{z}}_{\mathbf{0}}\mathbf{)}\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$

Determine the residue of the simple pole

Consider the given function as:

$f\left(z\right)=\frac{e^{2z}}{4\cos hz-5}$

Rewrite the function as:

$\begin{array}{rcl}f\left(z\right)& =& \frac{q\left(z\right)}{p\left(z\right)}\\ & =& \frac{e^{2z}}{4\cos hz-5}\end{array}$ …… (1)

At $\begin{array}{rcl}z& =& \ln 2\end{array}$

The function has simple pole at $\begin{array}{rcl}z& =& \ln 2\end{array}$and the residue of simple pole is given by:

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

 Determine the residue of the function as:

Substitute the values in equation (2) as follows:

$\begin{array}{rcl}R\left(\ln 2\right)& =& \frac{e^{2\ln 2}}{4\sin h\left(\ln 2\right)}\\ & =& \frac{1}{\sin h\left(\ln 2\right)}\\ & =& \frac{4}{3}\\ & & \end{array}$

Therefore, the residue of a function at $\begin{array}{rcl}z& =& \ln 2\\ & & \end{array}$is $\begin{array}{rcl}R\left(\ln 2\right)& =& \frac{4}{3}\\ & & \end{array}$.