I'm working through a series of lectures in complex analysis, and I've encountered something that doesn't seem to agree with my understanding so far.
Suppose $f$ is a complex-valued function with an isolated singularity at $z_0$. Suppose also that $f$ is analytic in $U=\lbrace r<|z-z_0|<R\rbrace\subseteq\mathbb{C}$, then $f$ has a Laurent series expansion in the annulus $U$ given by
$$f(z)=\sum_{n=-\infty}^\infty a_n(z-z_o)^n$$
Now this part I understand well. The lecturer then proceeds to explain that,
For $r<\rho<R$,
$$\int_{|z-z_0|=\rho}f(z)~dz=\sum_{n=-\infty}^\infty a_n\int_{|z-z_0|=\rho}(z-z_0)^n~dz$$
Now with this part I'm fine as well. I understand that the integral of a sum is equivalent to summing an integral in this case, as well as the fact that $a_n$ is just a constant, and can hence be removed from the integral. What I don't understand, however, is when the lecturer goes about evaluating the integral given that $k\neq-1$. It goes as follows
As $k\neq-1$, then integrand becomes
$$\frac{1}{n+1}(z-z_0)^{n+1}$$
yet, without consider the bounds given by any potential parameterisation of $|z-z_0|=\rho$, by Cauchy's integral theorem, we know that if $f$ is analytic in a simply connected domain, $U\subseteq\mathbb{C}$, then
$$\oint_\gamma f(z)~dz=0$$
for every $\gamma\subseteq U$. Therefore, our original integral, when $k\neq-1$, evaluates to $0$.
This is where I got lost, because if $f$ has an isolated singularity at $z_0$, then the domain is not simply connected, and Cauchy's integral theorem cannot be applied, or am I missing something? I've taken a loot at the question when integrating a Laurent series $f(z)=\sum\limits_{j=-\infty}^{\infty}a_j(z-z_0)^j$, why do all terms for $j\neq-1$ dissappear?, which asks why each of the terms of the laurent series disappear when $k\neq-1$. I can come to grips with why they disappear, given that the integral evaluates to $0$, but I don't understand why we can apply Cauchy's theorem to being with.
Any help is appreciated, thank you.
For $n \ne -1$, the function $(z-z_0)^n$ has the anti-derivative $F_n(z):=\frac{1}{n+1}(z-z_0)^{n+1}$.
If $c(t)=z_0 +\rho e^{it}$ for $t \in [0,2 \pi]$, then we have
$\int_{|z-z_0|=\rho}(z-z_0)^n ~dz= \int_{c}(z-z_0)^n ~dz= F_n(c(2 \pi))-F_n(c(0))=0$.