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I’m trying to understand a step on this answer:

So $f\colon \Omega\to{\mathbb C}$ is $C^1$ and satisfies the CR equations; therefore it is an analytic function of $z=x+iy\in\Omega$.

Assume that $\Omega$ is simply connected and chose a point $z_0\in\Omega$. Then by a standard theorem of complex analysis the function $$F(z)\ :=\ h(z_0)+\int_\gamma f(z)\ dz\ , \qquad \hbox{$\gamma\ $ a path from $z_0$ to $z$}\ ,$$ is an analytic primitive of $f$ in $\Omega$.

More concretely, I’m trying to understand why the function $F$ is well-defined and why is it a primitive of $f$.

My complex analysis is a bit rusty and I don’t immediately see what standard theorem is being referred to here. Is it Cauchy’s integral theorem (also called Cauchy-Goursat’s theorem)? I thought so because I found the same proof (with a bit more detail) on these notes (Theorem 5.3) and Cauchy’s theorem seems to be mentioned there as a justification for that step, but I don’t understand yet how it can be applied. Any help would be appreciated.

dahemar
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  • just parameterize the path $\gamma $ and observe that the value of $F$ doesn't depend on the path, just on the end points. To do that you can represent $f$ by it Taylor series – Masacroso Mar 24 '23 at 11:03

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This is essentially (part of the proof of) the theorem of Morera.On the Wikipedia page, it also explains why this expression is well-defined.

Or, from another point of view, you can see it as a generalization of the real result that $$F(x) = F(0)+\int\limits_0^x f(x) dx$$ is a primitive for a continuous function $f$.

Nuke_Gunray
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    Thank you, I will look into that! Isn’t this theorem a generalization of Cauchy’s, though? I’m just trying to understand the proof when $\Omega$ is simply connected, we could even assume it’s a disk. – dahemar Mar 24 '23 at 11:35
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    @gisame It's not a generalization, but rather a corollary of it: Because of Cauchy and the simply connectedness of $\Omega$, the expression $$\int\limits_\gamma f(z)dz$$ only depends on the endpoints of $\gamma$, and so $$F(z)= F(z_0)+\int\limits_\gamma f(z)dz$$ is well-defined for any path $\gamma$ that goes from $z_0$ to $z$. – Nuke_Gunray Mar 24 '23 at 11:38
  • I see. Just a final question if you don’t mind, it says on the Wikipedia article that “using the continuity of $f$ to estimate difference quotients, we get that $F′(z) = f(z)$”. What does it mean exactly by estimating difference quotients? – dahemar Mar 24 '23 at 12:36
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    @gisame Well, it's more or less the same as in the real proof: Let $F(z):= F(z_0)\int\limits_\gamma f(z)d z$ for any path $\gamma$ that connects $z_0$ with $z$. Then, we want to see that $$F'(z_1)=\lim\limits_{h\rightarrow 0} \frac{F(z_1+h)-F(z_1)}{h}, ()$$ so we have to somehow estimate the numerator. So, let $\gamma$ be a path from $z_0$ to $z_1$, and $\gamma_h$ a path from $z_1$ to $z_1+h$. Then $$F(z_1+h)=\int\limits_{\gamma+\gamma_h} f(z)dz = \int\limits_\gamma f(z)dz+\int\limits_{\gamma_h}f(z)dz,$$ so the right side of $()$ just becomes $$\frac{1}{h}\int\limits_{\gamma_h} f(z)d z.$$ – Nuke_Gunray Mar 24 '23 at 23:48
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    Then, $\gamma_h$ can be chosen to be just the connecting line between $z$ and $z+h$, i.e. $\gamma(t) = z_1+th$ for $t\in[0,1]$, so $$\frac{1}{h}\int\limits_{\gamma_h} f(z)dz=\frac{1}{h}\int\limits_0^1 f(z_1+th)\cdot h dt = \int\limits_0^1 f(z_1+th) dt.$$ And now, using continuity of $f$, one can show that for $\lim\limits_{h\rightarrow0}$, this expression converges towards $f(z_1)$. – Nuke_Gunray Mar 24 '23 at 23:53
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    Oh, I just saw that the equation $(*)$ is missing something very important: We want to show that $$F'(z_1) := \lim\limits_{h\rightarrow 0}\frac{F(z_1+h)-F(z_1)}{h} = f(z_1).$$ – Nuke_Gunray Mar 24 '23 at 23:55