Expand $\frac{1}{e^{2z}+1}$ about $\frac{\pi i}{2}$? How do you expand $\frac{1}{e^{2z}+1}$ about $\frac{\pi i}{2}$? I know this must have some negative powers, but by using binomial theorem, this doesn't work as the function is not small about this point. 
 A: Step 1. Let $z = \frac{\pi i}{2} + \epsilon$, then $\frac{1}{ \mathrm{e}^{2z}+1} = \frac{1}{ \mathrm{e}^{i \pi} \mathrm{e}^{2 \epsilon }+1}$. 
Step 2. Use the series expansion for exponential $\exp(x) \sim 1 + x + \frac{x^2}{2} + \ldots + \frac{1}{k!} x^k + \ldots$.
Step 3. Find expansion of the denominator $\mathrm{e}^{i \pi} \mathrm{e}^{2 \epsilon }+1$.
Step 4. Find the reciprocal, noting that
$$
  \left( a_1 x + a_2 x^2 + \ldots \right)^{-1} = \frac{1}{a_1 x} \left( 1 + \frac{a_2}{a_1} x + \ldots \right)^{-1}
$$
Expansion of $(1+c_1 x + c_2 x^2 + \ldots)^{-1}$ can be found using geometric series:
$$
  \frac{1}{1+c_1 x + c_2 x^2 + \ldots} = 1 - \left(c_1 x + c_2 x^2 + \ldots\right) + \left(c_1 x + c_2 x^2 + \ldots\right)^2 - \left(c_1 x + c_2 x^2 + \ldots\right)^3 + \ldots
$$
A: Write $x:={\pi i\over 2}+t$. Then
$$ {1\over e^{2x}+1} = {1\over 1-e^{2t}}=-{1\over 2t}\ {2t\over e^{2t}-1}\ .$$
The last fraction looks like the generating function of the Bernoulli numbers. A look at the page
http://en.wikipedia.org/wiki/Bernoulli_numbers
shows that in fact we have
$${1\over e^{2x}+1}=-{1\over 2t}\ \sum_{m=0}^\infty B_m{(2t)^m\over m!}\qquad\bigl(t:=x-{\pi i\over 2}\bigr)\ .$$
