What is the Residue of this function?

Question

What is $$\operatorname*{Res}_{z=i \pi}{\frac{e^{(1-a)z}}{\left (1+e^{z} \right )^n}}$$

where $0<a<1$ and $n$ is an integer??

Answer 1

Let $z=\zeta + i \pi$, where $\zeta=z-i \pi$ is taken to be a small quantity. The residue at $z=i \pi$ will be equal to the coefficient of $\zeta^{-1}$ in the Laurent expansion of the above function about $\zeta=0$.

So this function may be written as

$$f(z) = e^{i \pi (1-a)} \frac{e^{(1-a) \zeta}}{\left (1-e^{\zeta}\right )^n} $$

Noting that

$$1-e^{\zeta} = -\left ( \zeta + \frac1{2!}\zeta^2 + \frac1{3!} \zeta^3+\cdots \right )$$

So that the Laurent expansion about $\zeta=0$ is

$$e^{i \pi (1-a)} (-1)^n \zeta^{-n} \left (1+ (1-a) \zeta + \frac1{2!} (1-a)^2 \zeta^2 + \cdots \right ) \times \\ \left [1 - \binom{n}{n-1}\left ( \frac1{2!} \zeta + \frac1{3!} \zeta^2 + \cdots \right ) + \binom{n+1}{n-1}\left ( \frac1{2!} \zeta + \frac1{3!} \zeta^2 + \cdots \right )^2 + \cdots \right ] $$

It is extremely difficult to find the coefficient of $\zeta^{-1}$ for arbitrary $n$. Your best bet is to find that coefficient for particular values of $n$ as they come. I will illustrate for $n=3$. In that case, the Laurent expansion is

$$e^{i \pi (1-a)} (-1)^3 \zeta^{-3} \left (1+ (1-a) \zeta + \frac1{2!} (1-a)^2 \zeta^2 + \cdots \right ) \times \\ \left [1 - \binom{3}{2}\left ( \frac1{2!} \zeta + \frac1{3!} \zeta^2 + \cdots \right ) + \binom{4}{2}\left ( \frac1{2!} \zeta + \frac1{3!} \zeta^2 + \cdots \right )^2 + \cdots \right ] $$

Note that, for the expansions, we only need to look at the coefficients of $\zeta^2$ because we have a $\zeta^3$ in the denominator. So, the terms are, to terms in $\zeta^2$ and below,

$$e^{i \pi (1-a)} (-1)^3 \zeta^{-3} \left (1+ (1-a) \zeta + \frac1{2!} (1-a)^2 \zeta^2 + \cdots \right ) \left [1 - \frac32 \zeta + \zeta^2 + \cdots \right ] \\ = e^{-i \pi a} \zeta^{-3} \left [\cdots + \left (1-\frac32 (1-a) + \frac12 (1-a)^2 + \right ) \zeta^2 + \cdots\right ] $$

Thus, the residue of the original function at $z=i \pi$ is, for $n=3$,

$$\left [1-\frac32 (1-a) + \frac12 (1-a)^2 \right ] e^{-i \pi a} = \frac12 a(1+a) e^{-i \pi a}$$