MHB What Are the Singularity and Residue of \( \frac{e^{z^2}}{(z-i)^3} \)?

[aruwin]

Hello.
Can you check this for me, please?
Find the singularity of $\frac{e^{z^2}}{(1-z)^3}$ and find the residue for each singularity.

My solution:
There is a triple pole at z=i, therefore

$$Res_{|z=i|}f(z)=\frac{1}{2}\lim_{{z}\to{i}}\frac{d^2}{dz^2}(z-i)^3\frac{e^{z^2}}{(1-z)^3}=\lim_{{z}\to{i}}\frac{d^2}{dz^2}e^{z^2}=\lim_{{z}\to{i}}4z^2e^{z^2}$$$$=\frac{1}{2}\frac{-4}{e}=\frac{-2}{e}$$

 

[Euge]

Hello.
Can you check this for me, please?
Find the singularity of $\frac{e^{z^2}}{(1-z)^3}$ and find the residue for each singularity.

My solution:
There is a triple pole at z=i, therefore

$$Res_{|z=i|}f(z)=\frac{1}{2}\lim_{{z}\to{i}}\frac{d^2}{dz^2}(z-i)^3\frac{e^{z^2}}{(1-z)^3}=\lim_{{z}\to{i}}\frac{d^2}{dz^2}e^{z^2}=\lim_{{z}\to{i}}4z^2e^{z^2}$$$$=\frac{1}{2}\frac{-4}{e}=\frac{-2}{e}$$

The triple pole is at $z = 1$, not $z = i$.

 

[Opalg]

Also, the second derivative of $e^{z^2}$ is not $4z^2e^{z^2}$.

 

[alyafey22]

Hint :

Expand $e^{z^2}$ around $z=1$.

 

[aruwin]

Now I realize that I wrote the function wrong. It's actually
$$f(z)=\frac{e^{z^2}}{(z-i)^3}$$

Now after correcting a few things, I got
$$\frac{-1}{e}$$ as the answer.