In my class, this is our version of Cauchy's integral formula:
Let $\Omega\subseteq\mathbb{C}$ be an open convex set, $D\left(a,r\right)=\left\{z\in\mathbb{C}:\left|z-a\right|<r\right\}\subseteq\Omega$ be the open disk centered at $a\in\mathbb{C}$ with radius $r>0$, the closure of $D\left(a,r\right)$ be contained in $\Omega$, $b\in D\left(a,r\right)$, and $f\left(z\right)$ be holomorphic on $\Omega$. Then, $$f\left(b\right)=\frac{1}{2\pi i}\int_{\gamma}\frac{f\left(z\right)}{z-b},$$ where $\gamma=\left\{a+re^{i\theta}\in\mathbb{C}:\theta\in\left[0,2\pi\right]\right\}$.
I've seen some other versions (like on Wikipedia) of this where $\Omega$ is only an open set in $\mathbb{C}$. My questions are:
- What is the difference if $\Omega$ is convex or not? Does this change the theorem at all?
- What happens if $b$ is on the boundary of $D\left(a,r\right)$? Does Cauchy's integral formula still hold? I am actually solving the integral $\int_{\left\lvert z\right\rvert=1}\frac{\cos z}{\left(z-i\right)\left(z+4\right)}\text{d}z$. I have set $f\left(z\right)=\frac{\cos z}{z+4}$ and $\Omega=\mathbb{C}\setminus\left\{-4\right\}$, but I realized that $z=i$ lies on the boundary of $D\left(0,1\right)$ and this made me think about what happens at the boundary.