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Let $G$ be a non-empty connected open subset of $\mathbb{C}$ which is symmetric with respect to the real axis. Let $f$ be a holomorphic function on $G$ such that $f$ is real valued on $G\cap \mathbb{R}$. Then prove that $f(\overline{z})=\overline{f(z)}$ for all $z\in G$.

Can we use principle of analytic continuation to show this. I am not sure about how to use it. Please help!!

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Since $G$ is open, the set $G\cap\mathbb{R}$ is open in the real line; it cannot be empty, or the set $G$ would not be connected. Hence $G\cap\mathbb{R}$ contains a proper interval and therefore it has an accumulation point.

The function $g(z)=f(z)-\overline{f(\bar{z})}$ is holomorphic on $G$ and is zero on $G\cap\mathbb{R}$. A holomorphic function which vanishes on a set with an accumulation point is globally zero.

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Reduction 1
It suffices to prove that $\overline {f(\overline{z})}=f(z)$
Reduction 2
For that it suffices to prove that $\overline {f(\overline{z})}$ is holomorphic, since then the two holomorphic functions $\overline {f(\overline{z})}$ and $f(z)$ will be equal because they coincide on any real interval $(r,s)\subset G\cap \mathbb R$ .
The proof
The magic of Wirtinger's calculus immediately proves that $\overline {f(\overline{z})}$ is holomorphic:$$\frac { \partial} {\partial \overline z } (\overline {f(\overline{z})})= \overline{ \frac { \partial} {\partial z } ( {f(\overline{z})}) }=0 $$ The last equality because $f(\bar z)$ is antiholomorphic (or by using the formula for $\frac { \partial} {\partial z } $ of the composition of two smooth functions, applied here to $f$ and complex conjugation)

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