# How should I expand $\frac{1}{z(e^z-1)}$ to find the residue and order at the pole z = 0?

$\frac{1}{z(e^z-1)}$ is the function I want to expand. I tried using the expansion for $e^z$ and got $$\frac{1}{z^2+z^3/2!+z^4/3!+...}$$ Can I put this fraction into the $b_n/(z-z_0)^n$ form, or did I take the wrong approach?

$$\frac1{z(e^z-1)}=\frac1z\cdot\frac1{z\left(1+\frac z2+\mathcal O(z^2)\right)}=\frac1{z^2}\left(1-\frac z2+\frac{z^2}4-\ldots\right)=\frac1{z^2}-\frac1{2z}+\ldots$$
So the pole is double and its residue is $\;-\dfrac12\;$ .
• @user224740 Since we can assume $\;|z|\;$ is very small, and in fact $\;|z|<2\;$ , then $\;\frac1{1+z}=1-z+z^2-z^3+\ldots\;$ Observe we don't need to worry about elements with high exponents as you're only interested in the pole's order and in the function's residue at it. – Timbuc May 6 '15 at 14:51
$z=0$ is a simple zero both for $z$ and $e^z-1$, hence $z=0$ is a double pole for $f(z)=\frac{1}{z(e^z-1)}$ and: $$\operatorname{Res}(f(z),z=0) = \left.\frac{d}{dz}(z^2 f(z))\right|_{z=0}=\left.\frac{d}{dz}\frac{z}{e^z-1}\right|_{z=0}=B_1=\color{red}{-\frac{1}{2}}$$ by the definition of Bernoulli numbers.