# How do I compute the residue at this pole?

Let $$f(z) = \frac{e^{az}}{1 + e^z} \qquad (0 < a < 1)$$ I know this function has poles at $z = (2k + 1) i \pi$ with $k \in \mathbb{Z}$. Let's say I need to find the residue at the pole $z = \pi i$. How would I compute this? I tried using the formula $$Res(f(z), z= \pi i) = \lim_{z \to \pi i} (z - z_0) f(z)$$ but that doesn't seem to work.

• You can use the series expansion of $f(z)$ about $z=i\pi$ and take the coefficient of the $(z-i\pi)^{-1}$ term. – user170231 Jan 12 '16 at 20:45

Let $\zeta=z-i \pi$. Then the function of $\zeta$ is
$$\frac{e^{i a (\zeta+i \pi)}}{1+e^{(\zeta+ i \pi)}} = e^{i \pi a} \frac{e^{i a \zeta}}{1-e^{\zeta}}$$
Now take the Laurent expansion of this about $\zeta=0$ and see that the coefficient of $\zeta^{-1}$ is clearly $-e^{i \pi a}$.
• I tried taking the Laurent expansion, but still I don't see that the coefficient is $-e^{i \pi a}$. I expand the denominator as a geometric series right? And the numerator is just complex exponential expansion? – Kamil Jan 12 '16 at 21:04
• @Kamil: That's right. The $1$'s cancel in the denominator, leaving a $\zeta$ as the leading term. The rest should be simple. – Ron Gordon Jan 12 '16 at 21:05