Laurent series representation of given function

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.

• How can we tell unless you include your attempt? The actual Laurent series involves Bernoulli numbers. – Lord Shark the Unknown Apr 20 at 11:32
• You cannot use the series for $\frac 1 {1-z}$ and then change $z$ to $e^{z}$. This is because you would require $|e^{z}| <1$ or $\Re z <0$ for this to be valid. – Kavi Rama Murthy Apr 20 at 11:53
• @Kavi Rama Murthy.. thanks sir, got my mistake. – Believer Apr 20 at 12:11

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:
$$e^z-1=z+\frac{z^2}2+\mathcal O(z^3)\implies$$
$$\frac1{e^z-1}=\frac1{z\left(1+\frac z2+\mathcal O(z^2)\right)}=\frac1z\cdot\left(1-\frac z2+\frac{z^2}4-\ldots\right)=\frac1z-\frac12+\frac z4-\ldots$$
• I don't get $+\frac{z}4$. – Lord Shark the Unknown Apr 20 at 11:58
• @LordSharktheUnknown I think that what you don't get is $\;+\frac12\;$ but $\;-\frac12\;$, and then $\;+\frac z4\;$ . Anyway, your comment made me see the mistake. Thanks. – DonAntonio Apr 20 at 12:06
• I still don't get $+\frac z4$. I get $+\frac z{12}$. – Lord Shark the Unknown Apr 20 at 12:08