Q29P - Mathematical Methods in Physical Sciences

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 $\mathit{z}\mathbf{=}\mathit{l}\mathit{n}\mathbf{2}$

Step-by-Step Solution

Answer

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

Step 1: Determine the formula

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

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

Step 2: 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)

Step 3: 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 $\begin{array}{rcl}R\left(\ln 2\right)& =& \frac{4}{3}\\ & & \end{array}$.