Conformal mapping, Residue

Question

I have solved the second question till $z=0$ is a simple pole... But it's residue is mentioned as $2$. How come the laurent series expansion helps in finding $a_{-1}$, is there any other way to obtain the series expansion.

Answer 1

For a simple pole at $z=z_0$ an easy way to find the residue is $${\rm Res}_{z=z_0}\Bigl(\frac{p(z)}{q(z)}\Bigr)=\frac{p(z_0)}{q'(z_0)}\ :$$ this is true provided that $p$ and $q$ are analytic at $z_0$ and $q(z_0)=0$ and $q'(z_0)\ne0$. In this case we have $$\frac{p(z)}{q(z)}=\frac{e^z+1}{e^z-1}$$ and the residue is $$\left.\frac{e^z+1}{e^z}\right|_{z=0}=2\ .$$

Comment. In the above formula suppose that $p(z_0)=0$. Then $z_0$ is not in fact a simple pole but a removable singularity, and the formula gives residue $0$, as it should.