I am trying to prove the claim that:
For any positive integer $n \in \mathbb{N}$, \begin{align} \text{Res} \Big ( \frac{1}{(1-e^{-z})^n} ;0 \Big ) = 1. \end{align}
Here's my working:
Let $f(z) = \frac{1}{(1-e^{-z})^n}$. The zeros of $1-e^{-z}$ are the points $z = 2\pi k i, k \in \mathbb{Z}$. Therefore, if we consider $C$ to be the unit circle, it contains only one of the poles, that at $0$. By the Residue Theorem,
\begin{align} \int_{C} \frac{dz}{(1-e^{-z})^n} = 2 \pi i \; \text{Res} \Big ( \frac{1}{(1-e^{-z})^n} ;0 \Big ). \end{align}
We compute the integral by means of a the substitution $w = 1 - e^{-z}$. Since $dw = e^{-z} dz$, a simple substitution shows that,
\begin{align} \int_{C} \frac{dz}{(1-e^{-z})^n} = \int_{C^*} \frac{dw}{w^n(1 - w)}, \end{align}
where $C^*$ is the image $C$ under the non-constant analytic map $w(z)$. The function $\frac{1}{w^n(1 - w)}$ has poles at $w = 0, 1$. We show that $0$ is in the region bounded by $C^{*}$, and $1$ is not in the region bounded by $C^{*}$.
Consider the unit disk, $D(0, 1)$. Since $D$ is compact, its image, $D^*$ is compact as well. Now $z = 1 \in D^*$ if and only if $e^{-z} = 0$ for some $z \in D(0,1)$, which is impossible. Therefore, $1 \notin D^*$. Since $C^{*} \subset D^{*}$, we have that $C^*$ doesn't enclose $1$. For $z = 0$ in the unit disk, $D(0,1)$, we have that $w(0) = 1 - 1 = 0 \in D^{*}$. It remains to show that $C^*$ encloses $0$.
Assuming it encloses $0$, by the Residue Theorem, we have:
\begin{align} \int_{C^*} \frac{dw}{w^n(1 - w)} = 2 \pi i \; {n}(C^{*}, 0) \; \text{Res} \Big ( \frac{1}{w^n(1 - w)} ;0 \Big ), \end{align}
where ${n}(C^{*}, 0)$ is the winding number of $C^{*}$ around $0$. Assuming the winding number is $1$, since $\text{Res} \Big ( \frac{1}{w^n(1 - w)} ;0 \Big ) = 1$, we have:
\begin{align} 2 \pi i \text{Res} \Big ( \frac{1}{(1-e^{-z})^n} ;0 \Big ) = \int_{C} \frac{dz}{(1-e^{-z})^n} = \int_{C^*} \frac{dw}{w^n(1 - w)} = 2 \pi i, \end{align}
implying that $\text{Res} \Big ( \frac{1}{(1-e^{-z})^n} ;0 \Big ) = 1$.
Questions:
Can I argue that $C^{*}$ does in fact enclose $0$?
Is it true that ${n}(C^{*}, 0)$ must be 1? I know it isn't $0$ if I can prove the claim in 1, but I'm not sure if it must be 1? I'm reading a book that doesn't discuss winding numbers in a lot of detail, since for the most part the closed curves are chosen to be circles. More generally, is the image of a regular closed curve a regular closed curve? The textbook defines a regular curve to be one which gives a winding number of $0$ or $1$.