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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.

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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 $y$.

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  • $\begingroup$ "the closed loop property says we can take $\Gamma$ as a straight-line segment" Yes, but $D$ might not be convex. $\endgroup$ – Arthur Jul 29 '19 at 11:03
  • $\begingroup$ Is the domain open? In that case just restrict attention to the $y \in D$ in a small convex disc around $x$. Since we're ultimately taking the limit as $y \to x$ this does not cause a problem. $\endgroup$ – Daron Jul 29 '19 at 11:11
  • $\begingroup$ Sure, it can be mended. But leaving it as an unqualified statement is objectively false. $\endgroup$ – Arthur Jul 29 '19 at 11:13
  • $\begingroup$ Objectively false! That's the worst kind of false champ! $\endgroup$ – Daron Jul 29 '19 at 11:22
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The main reason with it holds is that analycity is a local condition, not a global one. Once you know $F$ exists, you can work locally, and assume the interior of $D$ is connected, star-shaped, convex...

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I suppose that when you write that $\oint f(z)\,\mathrm dz=0$, what you mean is that the integral of $f$ over any closed path is equal to $0$.

Let $z\in D$ and consider the quotient$$\frac{F(z+h)-F(z)-hf(z)}h,$$where $h$ is such that $z+h\in D$. Now, let $\Gamma$ be a path in $D$ going from $z_0$ to $z$ and let $\Gamma^\star$ be the same path followed by a path that goes in a straight line from $z$ to $z+h$. Then$$F(z+h)-F(z)=\int_{\Gamma^\star}f(w)\,\mathrm dw-\int_\Gamma f(w)\,\mathrm dw=\int_{\eta}f(w)\,\mathrm dw,$$where $\eta(t)=(1-t)z+t(z+h)$. Therefore,$$F(z+h)-F(z)-hf(z)=\int_\eta f(w)-f(z)\,\mathrm dw\tag1$$Take $\varepsilon>0$. And now take $\delta>0$ such that $\lvert w-z\rvert<\delta\implies\bigl\lvert f(w)-f(z)\bigr\rvert<\varepsilon$. It follows from $(1)$ and from the fact that the length of $\eta$ is $\lvert h\rvert$ that $\left\lvert F(z+h)-F(z)-hf(z)\right\rvert<\lvert h\rvert\varepsilon$. So$$\lvert h\rvert<\delta\implies\left\lvert\frac{F(z+h)-F(z)-hf(z)}h\right\rvert<\frac{\lvert h\rvert\varepsilon}{\lvert h\rvert}=\varepsilon.$$And this proves that$$\lim_{h\to0}\frac{F(z+h)-F(z)-hf(z)}h=0$$which is equivalent to$$\lim_{h\to0}\frac{F(z+h)-F(z)}h=f(z).$$In other words, $F'(z)=f(z)$.

As you can see, the Cauchy-Riemann equations are not needed here.

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