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:

  1. What is the difference if $\Omega$ is convex or not? Does this change the theorem at all?
  2. 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.
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    $\begingroup$ "What happens if b is on the boundary" -- Try it with the unit disk, $f(z) = 1$, and, e.g., $b = 1$. $\endgroup$ – amsmath Mar 28 at 1:58
  • $\begingroup$ @amsmath If I use Cauchy's integral formula, it simply gives us $1$, but actually attempting to evaluate the integral shows that it is not convergent, so it does not necessarily hold for the boundary points? $\endgroup$ – Jake Mar 28 at 2:07
  • $\begingroup$ Exactly. The integral does not converge. About $\Omega$: The convexity is indeed superfluous. $\endgroup$ – amsmath Mar 28 at 2:11
  • $\begingroup$ @amsmath, thank you! Greatly appreciate your help. $\endgroup$ – Jake Mar 28 at 2:14

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