1. ## residue

Find the residue of the following.

(a) $\text{Res}(f, z_0=0)$, where $f(z)=\frac{e^z}{\sin(2z^2)}$

(b) $\text{Res}(f, z_0=0)$, where $f(z) = \frac{\cos(\frac{1}{z})}{e^z}
$

We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.

2. Originally Posted by pascal4542
Find the residue of the following.

(a) $\text{Res}(f, z_0=0)$, where $f(z)=\frac{e^z}{\sin(2z^2)}$

(b) $\text{Res}(f, z_0=0)$, where $f(z) = \frac{\cos(\frac{1}{z})}{e^z}
$

We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.
If $z_0$ is a pole of order $m$, then $\text{Res}\!\left(f;z_0\right)=\frac{1}{m-1}\lim_{z\to z_0}\frac{\,d^{m-1}}{\,dz^{m-1}}\left[(z-z_0)^mf\!\left(z\right)\right]$.

In the case its a simple pole, we have $\text{Res}\!\left(f;z_0\right)=\lim_{z\to z_0}(z-z_0)f\!\left(z\right)$.

Are you allowed to use these?

3. Originally Posted by Chris L T521
If $z_0$ is a pole of order $m$, then $\text{Res}\!\left(f;z_0\right)=\frac{1}{m-1}\lim_{z\to z_0}\frac{\,d^{m-1}}{\,dz^{m-1}}\left[(z-z_0)^mf\!\left(z\right)\right]$.

In the case its a simple pole, we have $\text{Res}\!\left(f;z_0\right)=\lim_{z\to z_0}(z-z_0)f\!\left(z\right)$.

Are you allowed to use these?
Yes I am. I will try that. Thank you.

4. Originally Posted by pascal4542
Yes I am. I will try that. Thank you.

I was wondering if you can check my work.

For (a) $\text{Res}(f, z_0=0)$, where $f(z)=\frac{e^z}{\sin(2z^2)}$ we have

$\frac{1+z+z^2/2!+z^3/3!+\cdots}{2z^2 -(2z^2)^3/3!+(2z^2)^5/5!- \cdots}$

$= \frac{1}{2z^2}+\frac{1}{2z}+1/4+z/12$

So the residue is $1/2$.

For (b) $\text{Res}(f, z_0=0)$, where $f(z) = \frac{\cos(\frac{1}{z})}{e^z}$ we have

$\cos(\frac{1}{z}) \cdot (1-z+z^2/2-z^3/6+z^4/24+\cdots).$

I am stuck. I don't know how to get the residue. I know that $z_0$ is an essential singularity. I need help on part (b).

5. You have a Taylor series multiplied by a non-terminating singular series so the residue is an infinite sum due to the term-wise multiplication of $z^n$ by $z^{-n}$ so then:

\begin{aligned}\frac{\cos(1/z)}{e^z}&=\sum_{n=0}^{\infty}\frac{(-1)^n}{1}\frac{1}{(2n)! z^{2n}} \sum_{n=0}^{\infty}\frac{(-1)^n z^n}{n!}\\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{(-1)^k}{(2k)!z^{2k}}\frac{(-1)^{n-k}z^{n-k}}{(n-k)!}
\end{aligned}

and note the coefficients on the $1/z$ terms are when the exponent of z is $n=3k-1$. Now, can you extract out all those terms and come up with the expression:

$\mathop\text{Res}\limits_{z=0} \frac{\cos(1/z)}{e^z}=\sum_{k=0}^{\infty} \frac{(-1)^{3k-1}}{(2k)!(2k-1)!}$