Cauchy's Integral Theorem: Let $f:G\to \mathbb{C}$ be a holomorphic function in a simply connected domain $G\subseteq \mathbb{C}$ and let $\gamma$ be a closed path in $G$. Then $$ \int_\gamma f(z)\,\mathrm{d}z=0. $$
One example I struggle to apply this theorem mathematically. I have to determine the value of $$ \int_{\partial B(0,1)}\frac{1}{(z-a)(z-b)}\,\mathrm{d}z $$ for $\vert a\rvert, \vert b\rvert>1$. Here, I notice that $a$ and $b$ doesn't lie in $B(0,1)$. Here's what I did cronologically:
One of my friends once said that it's okay to use this theorem, just because that the integrand is holomorphic on the simply connected domain $B(0,1)$. I didn't think about it until I read this theorem carefully [when I discovered that $\partial B(0,1)$ isn't contained in $B(0,1)$]. So, I tried to fix the argument somehow.
I define $f(z):=\frac{1}{(z-a)(z-b)}$ for $z\in \mathbb{C}\setminus \{a,b\}$. Clearly, $f$ is holomorphic. But then I can't use the theorem, because $\mathbb{C}\setminus \{a,b\}$ is not simply connected. So, I wonder, what if I make the domain "smaller" that doesn't contain "holes" such as $a$ and $b$?
- I choose $R$ satisfying $1<R<\min\{\vert a\rvert,\vert b\rvert\}$. Then, define $g(z):=f(z)$ for $z\in K(0,R)$. Since $B(0,R)$ is simply connected, and $\partial B(0,1)$ is a closed path that is contained in $B(0,R)$, the theorem says that $\int_{\partial B(0,1)}g(z)\,\mathrm{d}z=0$. I do not know if this is a valid proof.
If there's a proper way to prove, please let me know.