Find the coefficients $a^{−1}, a_0, a_1$ in the Laurent expansion ￼$\frac{1}{e^z − 1} = ···+a_{−1}z^{−1} +a_0 +a_1z+...$ Find the coefficients $a^{−1}, a_0, a_1$ in the Laurent expansion ￼$\frac{1}{e^z − 1} = ···+a_{−1}z^{−1} +a_0 +a_1z+...$
on $2π < |z| < 4π.$
I know this should be a very easy problem, but not sure how to start it.  Any help would be great.  It is a past complex qualifying problem.  Thanks.
 A: The coefficients in the Laurent series are given by
$$
a_n = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{(z-c)^{n+1}}\,dz
$$
where $c$ is the center of expansion, and $\gamma$ is a simple closed curve in the annulus of convergence. You can use the residue theorem to compute the relevant integrals:
\begin{align}
a_{-1} &= \frac{1}{2\pi i} \int_\gamma \frac{1}{e^z - 1}\,dz \\
&= \newcommand{\res}{\operatorname{Res}\limits}
\res_{z=0}\Big(\frac{1}{e^z-1}\Big)+
\res_{z=2\pi i}\Big(\frac{1}{e^z-1}\Big)+
\res_{z=-2\pi i}\Big(\frac{1}{e^z-1}\Big) \\
&= 1 + 1 + 1 = 3.
\end{align}
Similarly,
\begin{align}
a_{0} &= \frac{1}{2\pi i} \int_\gamma \frac{1}{z(e^z - 1)}\,dz \\
&= 
\res_{z=0}\Big(\frac{1}{z(e^z-1)}\Big)+
\res_{z=2\pi i}\Big(\frac{1}{z(e^z-1)}\Big)+
\res_{z=-2\pi i}\Big(\frac{1}{z(e^z-1)}\Big) \\
&= -\frac12 - \frac{i}{2\pi} + \frac{i}{2\pi} = -\frac12.
\end{align}
I'll leave the $a_1$ to you.
A: Long division.  Divide $1$ by $z+z^2/2+z^3/6+\cdots$.  
See https://math.stackexchange.com/a/342389/442 for another example worked out.
