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 $\mathbf{z}\mathbf{=}{\mathbf{e}}^{\frac{\mathbf{2}\mathbf{\pi i}}{\mathbf{3}}}$

Short 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-by-step solution

Determine the formula: 

Residue of a function at simple poles is given by:

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

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)

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 $R\left(z_0\right)=-e^{-\sqrt{3}\pi }\left(\frac{1}{6}-i\frac{\sqrt{3}}{6}\right)$.