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How can I prove Morera's theorem?

Morera's Theorem: If a function $f(z)$ is continuous in a simply connected region and its integration over a closed path gives $0$, then the function is analytic.

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Morera´s theorem states:Let the function $f: \mathbb{C} \longrightarrow \mathbb{C}$. be continuous in a simply connected open set S. If $$\oint_Cf(z)dz=0$$ for a simple closed path $C$ that lies on $S$ then $f$ is an analytic function in $S$

Let $z_0$ in $S$ and define for all $z$ in $S$ $$F(z)=\int_{z_0}^{z} f(z)dz$$ Look that $F(z)$ do not depend on the path of the curve between the integration intervals, so the function $F(z)$ is well defined. Now we need to prove that $F(z)$ is differentiable at any point $z$ in $S$, such that it derivative is $f$, by definition, $F$ and all its derivatives are analytic over $S$. now let $h$ be a complex number such that $z+h$ is in $S$, then $$\dfrac{F(z+h)-F(z)}{h}=\dfrac{1}{h}\int_{z}^{z+h} f(y)dy$$, note now that $f(z)$ is in terms of the variable $y$, so $$|\dfrac{F(z+h)-F(z)}{h}-f(z)|=|\dfrac{1}{h}\int_{z}^{z+h}(f(y)-f(z))dy|$$ The integration variable $y$ lies on the path defined by the last integral, so if $h$ tends to 0, $y$ tends to $z$ by basic complex analysis.Now , how $f$ is continuous at $z$: $f(y)$ tends to $f(z)$ as $y$ tends to $z$, so here you use epsilon, delta definition and show that

$$\lim_{h\to0}{\dfrac{F(z+h)-F(z)}{h}}=f(z)$$

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    $\begingroup$ I like this prof, but where is the assumption of Morera's theorem used? $\endgroup$
    – Conjecture
    Mar 5, 2018 at 19:53
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    $\begingroup$ @PerelMan, In the process of Well-definedness of $F(z)$ the $\oint_C f(z)dz=0$ was used. And in the last part, the continuity of $f$ was used. $\endgroup$
    – phy_math
    Feb 29, 2020 at 13:41
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Let $D$ be a simply connected region. We fix a point $p$ of $D$. Let $z \in D$. Consider $g(z) = \int_{p}^{z} f(\zeta) d\zeta$, where the integral path is any piece-wise smooth curve starting from $p$ and ending at $z$. Then, by the assumption, $g(z)$ is uniquely determined. Since $g'(z) = f(z)$, $g(z)$ is holomorphic on $D$. Hence $f(z)$ is holomorphic on $D$.

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Statement of theorem is incorrect: the line integral of f(z) around C must be 0 for every closed curve c contained in S , otherwise F(z) as defined above might be path dependent and the proof invalid

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    $\begingroup$ Why do the functions $f(z)=(1/z^m)$ where $m$ is an integer and $\leq 2$ and which satisfy $\oint_{\gamma}f(z)dz=0$ contradict the theorem if $\gamma$ contains the origin where $f$ has a pole? $\endgroup$
    – Ian Taylor
    Jul 3, 2017 at 5:53

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