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.
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.
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.
