How to find the Laurent series of $g(z)=\frac{1}{e^{2z}-1}$, for $0<|z|<\pi/2$?

Question

How to find the Laurent series representation of $g(z)=\frac{1}{e^{2z}-1}$, for $0<|z|<\pi/2$? Use geometric series?

Answer 1

For the sake of simplification just let $z:=2z$ and you can re-produce the final results for the $2z$ case afterwards.

Now, since $\frac{1}{e^z-1}$ has a simple pole at $0$, its Laurent Series must not have a coefficient on $z^k$ for $k<-1$.

Let the coefficient of $z^k$ be $a_k$. Then, it should be :

$$\bigg(\frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ \mathcal O\big(z^4\big)\bigg)\bigg(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}+\mathcal O\big(z^6\big)\bigg)=1$$

Now, if you carry out the multiplications, you can equate the coefficients and find relations for the.