# Fundamental Theorem of Calculus in complex analysis?

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.

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.

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 $$y$$.

• "the closed loop property says we can take $\Gamma$ as a straight-line segment" Yes, but $D$ might not be convex. – Arthur Jul 29 '19 at 11:03
• 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. – Daron Jul 29 '19 at 11:11
• Sure, it can be mended. But leaving it as an unqualified statement is objectively false. – Arthur Jul 29 '19 at 11:13
• Objectively false! That's the worst kind of false champ! – Daron Jul 29 '19 at 11:22

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...

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.