Residue of $\frac{z^2 e^{\frac 1z}}{1-z}$

Question

I am not sure of what i do trying to calculate the residue of that function at z=0. I calculate the limit $\lim_{z\to 0} \frac{z^2 e^{\frac 1z}}{1-z}$ and found out it doesn’t exist, so z=0 is an essential singularity. I tryed to write down the its Laurent series to obtain the residue, but I think I was wrong and stop it at $z^2 \sum_{n=0}^\infty z^n \sum_{h=0}^\infty {\frac 1{h!} z^{-h}}$. How can I continue?

Answer 1

Note that$$z^2e^{\frac1z}=z^2+z+\frac12+\frac1{3!z}+\frac1{4!z^2}+\frac1{5!z^3}+\cdots$$and that$$\frac1{1-z}=1+z+z^2+z^3+\cdots$$Therefore, $\operatorname{res}_{z=0}\left(\frac{z^2e^{\frac1z}}{1-z}\right)$, which is the coefficient of $\frac1z$ in$$\left(z^2+z+\frac12+\frac1{3!z}+\frac1{4!z^2}+\frac1{5!z^3}+\cdots\right)\left(1+z+z^2+z^3+\cdots\right),$$is$$\frac1{3!}+\frac1{4!}+\frac1{5!}+\cdots=e-\frac52.$$

Answer 2

Just an informative addition:

Here is a way to calculate the residue without Laurent series using that the sum of all residues is $0$. This includes the residue at infinity:

• Let $f(z)= \frac{z^2 e^{\frac 1z}}{1-z} \Rightarrow \operatorname{Res}_{z=0}f(z) = - \left( \operatorname{Res}_{z=1}f(z) + \operatorname{Res}_{z=\infty}f(z) \right) $

$\operatorname{Res}_{z=1}f(z)$: $$\operatorname{Res}_{z=1}f(z) =\lim_{z\to 1}\frac{(z-1)z^2 e^{\frac 1z}}{1-z} = -e$$

$\operatorname{Res}_{z=\infty}f(z)$:

$$\operatorname{Res}_{z=\infty}f(z) = -\operatorname{Res}_{w=0}\left( \frac{1}{w^2}f\left( \frac{1}{w}\right)\right) =-\operatorname{Res}_{w=0}\frac{1}{w^2}\frac{\frac{1}{w^2} e^w}{1-\frac{1}{w}}$$ $$= -\operatorname{Res}_{w=0}\frac{1}{w^3}\frac{e^w}{w-1} =-\frac{1}{2!}\lim_{w\to 0}\frac{d^2}{dw^2}\left( \frac{w^3}{w^3}\frac{e^w}{w-1}\right)$$ $$ = -\frac{1}{2!}\lim_{w\to 0}\frac{w^2-4w+5}{(w-1)^3}e^w= \frac{5}{2}$$

All together

$$\operatorname{Res}_{z=0}f(z) = - \left( \operatorname{Res}_{z=1}f(z) + \operatorname{Res}_{z=\infty}f(z) \right) = -\left(-e +\frac{5}{2}\right) = e-\frac{5}{2}$$