Q24P - 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.

 $\frac{\mathbf{1}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{2}\mathbf{z}}{{\mathbf{z}}^{\mathbf{3}}}\mathbf{\text{ at }}\mathit{z}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

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

Step 1: Given information

The given function is: $f\left(z\right)=\frac{1-\cos 2z}{z^3}$.

Step 2: 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.

Step 3:Find the Residue

Since,  $z=0$ is a pole of higher order of 3.

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: $z_0=0\text{ and }n=3$

Now, the residue at $z_0=0$ will be:

 $\begin{array}{rcl}\underset{z=0}{Res}f\left(z\right)& =& \frac{1}{{\displaystyle \left(3-1\right)}!}\underset{z\to 0}{\lim }\left\{\frac{d^{3-1}}{d{z}^{3-1}}\left(\left(\frac{1-\cos 2z}{z^3}\right)·{\left(z-0\right)}^3\right)\right\}\\ & =& \frac{1}{\left(2\right)!}\underset{z\to 0}{\lim }\left\{\frac{d^2}{d{z}^2}\left(1-\cos 2z\right)\right\}\\ & =& \frac{{\displaystyle 1}}{{\displaystyle \left(2\right)!}}\underset{z\to 0}{\lim }\left\{4\cos 2z\right\}\\ & =& \frac{1}{2}·4=2\end{array}$

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