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Given that $w=f(z)$ is a regular function of $z$ such that $f'(z)\neq0$. Is $\overline{f(z)}=f(\overline z)$ always?

This suggests that it is true for basic operations but is it always true when $f$ is an elementary or non elementary function? I am unable to come up with a counterexample.

I encountered this step in class while proving $\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)\log|f'(z)|=0$

José Carlos Santos
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ZSMJ
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  • https://math.stackexchange.com/questions/186991/complex-conjugate-of-complex-function – Ahmed Hossam Aug 09 '19 at 14:18
  • $f'$ for a complex function doesn't make sense unless $f$ is holomorphic / analytic / entire. Did you just means "continuous, nonconstant" or something? – Arthur Aug 09 '19 at 14:19
  • What does regular mean in this context? Consider any function $f$ for which there is an $a \in \Bbb R$ such that $f(a) \not\in \Bbb R$. Then, $f(\bar a) = f(a) \neq \overline {f(a)}$. – Travis Willse Aug 09 '19 at 14:19
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    $f(z) = iz$ is a counterexample. – Martin R Aug 09 '19 at 14:21
  • @MartinR Thanks, I wonder why the professor allowed this step then. – ZSMJ Aug 09 '19 at 14:26
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    @Travis We're doing intro to analytic functions so http://mathworld.wolfram.com/RegularFunction.html would be the definition. – ZSMJ Aug 09 '19 at 14:27
  • @Arthur It just says regular as in here http://mathworld.wolfram.com/RegularFunction.html – ZSMJ Aug 09 '19 at 14:30
  • I wonder where you needed that in order to prove $(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\log|f'(z)|=0$, it is not necessary. See for example https://math.stackexchange.com/questions/308683/how-do-you-prove-that-lnfz-is-harmonic – Martin R Aug 09 '19 at 14:59

2 Answers2

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If $\omega\in\mathbb C\setminus\mathbb R$ and if $f(z)=\omega z$, then we don't have$$(\forall z\in\mathbb C):\overline{f(z)}=f\left(\overline z\right),$$since, for instance, $\overline{f(1)}=\overline{\omega}$, whereas $f\left(\overline 1\right)=f(1)=\omega\neq\overline\omega$.

José Carlos Santos
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This is not true in general. If we choose any regular function $f$ for which there is an $a \in \Bbb R$ such that $f(a) \not\in \Bbb R$, then $$f(\bar a) = f(a) \neq \overline{f(a)} .$$ There are many such functions: If a chosen regular function $f$ whose domain intersects $\Bbb R$ admits no such $a$ then $i f$ does.

The statement is true, however, in an important case: The Schwarz Reflection Principle (sometimes called the Riemann-Schwarz Principle) implies that the identity holds when an analytic function $f$ (with connected domain) maps real values to real values, that is, when $f(\Bbb R \cap \operatorname{dom} f) \subseteq \Bbb R$.

Travis Willse
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