# Question about Domain and the Open Disk in Cauchy's Integral Formula

In my class, this is our version of Cauchy's integral formula:

Let $$\Omega\subseteq\mathbb{C}$$ be an open convex set, $$D\left(a,r\right)=\left\{z\in\mathbb{C}:\left|z-a\right| be the open disk centered at $$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.
• "What happens if b is on the boundary" -- Try it with the unit disk, $f(z) = 1$, and, e.g., $b = 1$. – amsmath Mar 28 at 1:58
• @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? – Jake Mar 28 at 2:07
• Exactly. The integral does not converge. About $\Omega$: The convexity is indeed superfluous. – amsmath Mar 28 at 2:11
• @amsmath, thank you! Greatly appreciate your help. – Jake Mar 28 at 2:14