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.

$\frac{{\mathbf{e}}^{\mathbf{i}\mathbf{z}}}{\mathbf{9}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{4}}\mathbf{\text{at}}\mathit{z}\mathbf{=}\frac{\mathbf{2}\mathbf{i}}{\mathbf{3}}$

Short answer

Hence, the residue of the function at $z_0=\frac{2i}{3}$ is $-0.042785i$.

Step-by-step solution

Given information

The given function is: $f\left(z\right)=\frac{e^{iz}}{9z^2+4}$.

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

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:

$$\begin{gathered} g\left(z\right)=e^{iz} \\ h\left(z\right)=9z^2+4\Rightarrow h\text{'}\left(z\right)=18z \end{gathered}$$

Now, the residue at$z_0=\frac{2i}{3}$ will be:

$\begin{array}{rcl}R\left(\frac{2i}{3}\right)& =& \frac{g\left(\frac{2i}{3}\right)}{h\text{'}\left(\frac{2i}{3}\right)}\\ & =& \frac{e^{i·\frac{2i}{3}}}{18\left(\frac{2i}{3}\right)}\\ & =& \frac{e^{-\frac{2}{3}}}{12i}=-i\left(\frac{e^{-\frac{2}{3}}}{12}\right)\\ & =& -0.042785i\end{array}$

Hence, the residue of the function at $z_0=\frac{2i}{3}$ is $-0.042785i$.