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Prove that if $f(z)$ is a complex holomorphic function in the domain $D$ and $f'(z)=0$ so $f(z)$ is constant


My thoughts was to say that since $f'(z)=0$ so it is obvious $f(z)$ is constant I don't understand what the "holomorphic" is giving me here, any hints? or spoilers?

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Since $f$ is holomorphic, $f$ is analytic then :

$$f(z)=\sum_{k=0}^{\infty}a_kz^k$$ $$f'(z)=\sum_{k=1}^{\infty}ka_{k}z^{k-1}=0$$ So $$\forall k\in \mathbb{N}, a_{k+1}=0$$ and $$f(z)=a_0$$

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  • $\begingroup$ Why $f(z)=\sum\limits_{k=0}^{\infty}a_zz^k$? $\endgroup$ – 3SAT Jan 28 '16 at 21:58
  • $\begingroup$ The function can be represented by a power series. $\endgroup$ – imranfat Jan 28 '16 at 22:00
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    $\begingroup$ $f$ can be represented by a power series in each circle contained in $D$, but not necessarily by a single power series. $\endgroup$ – Martin R Jan 28 '16 at 22:26
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It suffices that $f$ be complex differentiable, for then it counts as an antiderivative of $f'$, in which case, for any two points $a$ and $b$ connected by a path $\gamma$ in the domain $D$, we have $$f(b)-f(a) = \int_\gamma f'(z) dz = \int_\gamma 0 dz = 0,$$ so $f(a) = f(b)$. Since $a$ and $b$ are arbitrary, we see that $f$ assumes only one value on $D$.

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  • $\begingroup$ The fundamental theorem of calculus does not work this way on the complex plane; your first equality is incorrect. There are obvious counter-examples if $f$ has a pole (not on $\gamma$), so you need a more global statement about $f$. $\endgroup$ – T.J. Gaffney Jan 28 '16 at 22:31
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    $\begingroup$ @Gaffney: I think you are mistaken. The first equality holds for any path from $a$ to $b$ in $D$. – $\int_\gamma f(z) dz$ may depend on the path $\gamma$, but $\int_\gamma f'(z) dz$ can always be computed like this. $\endgroup$ – Martin R Jan 28 '16 at 22:34
  • $\begingroup$ @Gaffney: Moreover, $f$ has no poles, since $f$ is complex differentiable in $D$. $\endgroup$ – Unit Jan 29 '16 at 2:02

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