Laurent series representation of given function

Question

I wanted to find Laurent series representation of function $1/(e^{z} -1)$. So I took minus common and apply the series formula of $1/(1-z)$ and then I use series formula for each $e^{zn}$. But I am getting very different answer what book have provided. What's wrong in my attempt.

Answer 1

Assuming you mean around $\;z=0\;$ , observe we need to worry only for small values of $\;|z|\;$ , so we can assume $\;|z|<1\;$ and thus:

$$e^z-1=z+\frac{z^2}2+\mathcal O(z^3)\implies$$

$$\frac1{e^z-1}=\frac1{z\left(1+\frac z2+\mathcal O(z^2)\right)}=\frac1z\cdot\left(1-\frac z2+\frac{z^2}4-\ldots\right)=\frac1z-\frac12+\frac z4-\ldots$$

If you need more addends just add them...