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Let $f,g : \mathbb{C} \to \mathbb{C}$ two holomorphic functions with $(f \circ g)(z)=0 \, \forall z \in \mathbb{C}$. Show that $f$ or $g$ is constant.

What I've tried: If $f$ and $g$ are holomorphic, then $\frac{\partial f}{\partial \overline{z}} = \frac{\partial g}{\partial \overline{z}} = 0$.

$\frac{\partial (f \circ g)}{\partial z} = \frac{\partial f}{\partial z} \frac{\partial g}{\partial z} + \frac{\partial f}{\partial \overline{z}} \frac{\partial \overline{g}}{\partial z} = \frac{\partial f}{\partial z} \frac{\partial g}{\partial z} = 0 \implies \frac{\partial f}{\partial z} = 0 $ or $\frac{\partial g}{\partial z} = 0 \implies f = $ constant or $ g = $ constant.

Is my proof ok? I've seen another proofs with connected spaces but I don't understood them.
MathLearner
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  • But you've proved everything (if you are allowed to use that $f' = 0$ implies $f$ being constant). – Invincible Mar 28 '23 at 22:36
  • @Vladislav can $f' = 0 \notimplies f$ is constant? – MathLearner Mar 28 '23 at 22:37
  • https://math.stackexchange.com/questions/751403/how-do-i-prove-that-fz-0-implies-f-is-constant – Invincible Mar 28 '23 at 22:39
  • @Vladislav I was thinking that if I write $\frac{df}{dz} = \frac{1}{2}(\frac{df}{dx} + \frac{1}{i} \frac{df}{dy}) = 0 \implies \frac{df}{dx} = 0$ and $\frac{df}{dy} = 0 \implies \frac{du}{dx} =\frac{du}{dy} = \frac{dv}{dx} = \frac{dv}{dy} = 0$. Is good? – MathLearner Mar 28 '23 at 22:42
  • Your proof works for a holomorphic function on a path connected open set. $\mathbb{C}$ is path connected so everything is OK. – Invincible Mar 28 '23 at 22:48
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    Note that $f(x) g(x) = 0 $ does not mean that $f(x) = 0 $ or $g(x) = 0$ on all of the domain. It means that (at least) one of them is 0 for each $x$. You can't guarantee (yet) that $\forall x,$ (say) $f(x) = 0 $. So we don't have the first implication sign (assuming you meant that it = 0 on all of the domain, as opposed to just point wise). – Calvin Lin Mar 28 '23 at 22:50
  • @CalvinLin oh. Ok. I think I missed that. Can you help me please to continue? – MathLearner Mar 28 '23 at 22:54

1 Answers1

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Suppose $g$ is not a constant. By Open Mapping Theorem, the range of $g$ is an open set. Since $f$ vanishes on the range of $g$ it follows, by the Identity Theorem, that $f\equiv 0$.

geetha290krm
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