This following fact came up in a course of complex analysis I was studying, and I was wondering how to prove it.
Suppose that $f:D \rightarrow \mathbb{C}$ is continuous, and that $\oint f(z) dz=0$. $D$ is a domain, not necessarily simply connected.
Let $\Gamma$ be a curve connecting $z_0,z \in \mathbb{C}$, define $F(z)=\int_{\Gamma} f(w) dw$.
Then, $F$ is an analytic function.
Comments:
It is easy to show that $F$ is well defined, and I was able to do that. I know that there is a real analysis version that is the fundamental theorem of calculus, but to prove analycity I need to show that the C-R eq's hold which is what I am struggling with.
Best Answer
This sort of problem is easier if you don't try to decouple the real and imaginary parts. Just try to compute the complex derivative as follows.
$F(x) -F(y) = \int_{\Gamma_1} f(w) dw - \int_{\Gamma_2} f(w) dw = \int_{\Gamma_1} f(w) dw + \int_{\Gamma_2'} f(w) dw $
where $\Gamma_1$ goes from $z_0$ to $x$ and $\Gamma_2'$ goes from $y$ to $z_0$. Using the closed-loop property we see
$F(x) -F(y) = \int_{\Gamma} f(w) dw $
where $\Gamma$ goes from $y$ to $x$. Again the closed loop property says we can take $\Gamma$ as a straight-line segment.
Now estimate $\frac{F(x) -F(y) }{x-y}$ using the fact that $f(w)$ is very close to $f(z_0)$ for $x$ close to $y$.