Q26P - 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.

 $\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{\pi iz}}}{\mathbf{1}\mathbf{-}{\mathbf{z}}^{\mathbf{3}}}$at  $\mathbf{z}\mathbf{=}{\mathbf{e}}^{\frac{\mathbf{2}\mathbf{\pi i}}{\mathbf{3}}}$

Step-by-Step Solution

Answer

The residue of the function at $z=e^{\frac{2\pi i}{3}}$ is $R\left(z_0\right)=-e^{-\sqrt{3}\pi }\left(\frac{1}{6}-i\frac{\sqrt{3}}{6}\right)$

Step 1: Determine the formula:

Residue of a function at simple poles is given by:

$\mathbf{R}\left[\mathbf{f}\left(\mathbf{z}\right)\right]\mathbf{=}\underset{\mathbf{z}\mathbf{\to }{\mathbf{z}}_{\mathbf{0}}}{\mathbf{l}\mathbf{i}\mathbf{m}}\left(z-z_0\right)\mathbf{f}\left(z\right)$

Step 2: Determine the residue of simple pole

Consider the function is written as:                                                         

$\begin{array}{rcl}f\left(z\right)& =& \frac{q\left(z\right)}{p\left(z\right)}\\ & =& \frac{e^{2\pi iz}}{1-z^3}\\ & & \end{array}$                                                     …….. (1)

At $z_0=e^{2\pi i/3}$

The function has simple pole at $z_0=e^{\frac{2\pi i}{3}}$ 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)

Step 3: Determine the residue of the function:

From equation (2), solve as:

$\begin{array}{rcl}R\left(z_0\right)& =& {\left[\frac{e^{2\pi iz}}{3z^2}\right]}_{z=e^{\frac{2\pi i}{3}}}\\ & =& \frac{\exp \left[2\pi i\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right]}{3{\left[\exp \left(\frac{2\pi i}{3}\right)\right]}^2}\\ & =& -\frac{e^{-\sqrt{3}\pi }{e}^{-\pi i}}{3e^{\frac{4\pi i}{3}}}\\ & =& -\frac{e^{-\sqrt{3}\pi }{e}^{-\frac{7\pi i}{3}}}{3}\end{array}$

Solve further to obtain,

 $\begin{array}{rcl}R\left(z_0\right)& =& -\frac{e^{-\sqrt{3}\pi }}{3}\left[\cos \left(\frac{7\pi i}{3}\right)-i\sin \left(\frac{7\pi i}{3}\right)\right]\\ & =& -e^{-\sqrt{3}\pi }\left(\frac{1}{6}-i\frac{\sqrt{3}}{6}\right)\end{array}$

 

Therefore, the residue of a function at $z=e^{\frac{2\pi i}{3}}$ is $R\left(z_0\right)=-e^{-\sqrt{3}\pi }\left(\frac{1}{6}-i\frac{\sqrt{3}}{6}\right)$.