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.

$\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{z}}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{z}}}\mathbf{\text{at}}\mathit{z}\mathbf{=}\mathit{i}\mathit{\pi }$

Short answer

Hence, the residue of the function at $z=i\pi$ is -1 .

Step-by-step solution

Residue Theorem

If $z_0$is an isolated singular point of f(z). Then the integration of the function within any closed curve C is given by:

${\oint }_{c}f\left(z\right)dz=b_1·2\pi i$

Here, $b_1$ is the residue.

Find the Residue

The given function is:$f\left(z\right)=\frac{e^{2z}}{1+e^z}$

As we know, the residue of the function is given by:

$R\left(z=z_0\right)=\frac{g\left(z_0\right)}{h\text{'}\left(z_0\right)}$, for $f\left(z\right)=\frac{g\left(z\right)}{h\left(z\right)}$

According to the question, we have:

$$\begin{gathered} g\left(z\right)=e^{2z} \\ h\left(z\right)=1+e^z\Rightarrow h\text{'}\left(z\right)=e^z \end{gathered}$$

Now, the residue for $z_0=i\pi$ will be:

$\begin{array}{rcl}R\left(i\pi \right)& =& \frac{g\left(i\pi \right)}{h\text{'}\left(i\pi \right)}\\ & =& \frac{e^{2i\pi }}{e^{i\pi }}\\ & =& e^{i\pi }\\ & =& \cos \pi +i\sin \pi ............\left\{e^{i\theta }=\cos \theta +i\sin \theta \right\}\\ & & \end{array}$

Simplication

Simplifying further, we get:

$\begin{array}{rcl}R\left(i\pi \right)& =& \cos \pi +i\sin \pi \\ & =& -1+i·0\\ & =& -1+0\\ & =& -1\\ & & \end{array}$

Hence, the residue of the function at $z=i\pi$is -1.