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.

$\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{z}}\mathbf{-}\mathbf{1}}{{\mathbf{z}}^{\mathbf{2}}}\mathbf{\text{at}}\mathit{z}\mathbf{=}\mathbf{0}$

Short answer

Hence, the residue of the function at $z_0=0$ is 2.

Step-by-step solution

Given information

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

Residue Theorem

If $z_0$ is an isolated singular point of $f\left(z\right)$. 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.

Step 3:Find the Residue

Since, $z=0$ is a pole of higher order of 2.

So, the residue of this type of function is given by:

$\underset{z=z_0}{Res}f\left(z\right)=\frac{1}{\left(n-1\right)!}\underset{z\to z_0}{\lim }\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left(f\left(z\right)·{\left(z-z_0\right)}^n\right)\right\}$

According to the question, we have:$z_0=0\text{and}n=2$

Now, the residue at$z_0=0$will be:

$\begin{array}{rcl}\underset{z=0}{Res}f\left(z\right)& =& \frac{1}{\left(2-1\right)!}\underset{z\to 0}{\lim }\left\{\frac{d^{2-1}}{d{z}^{2-1}}\left(\left(\frac{e^{2z}-1}{z^2}\right)·{\left(z-0\right)}^2\right)\right\}\\ & =& \frac{1}{\left(1\right)!}\underset{z\to 0}{\lim }\left\{\frac{d}{dz}\left(e^{2z}-1\right)\right\}\\ & =& \underset{z\to 0}{\lim }\left\{2e^{2z}\right\}\\ & =& 2\end{array}$

Hence, the residue of the function at $z_0=0$ is 2.