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Thread: F-Theory, Identity Theorem

  1. #1
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    F-Theory, Identity Theorem

    Hi,

    I'm a little stuck with this exercise.

    Let $\displaystyle f:\mathbb{C}\to \mathbb{C}$ be a holomorphic function which satisfies $\displaystyle f(z)\in\mathbb{R}\quad\forall z\in\mathbb{R}$. I want to show that
    $\displaystyle f(\overline{z}) = \overline{f(z)}$.

    We do have a non-discrete set $\displaystyle \mathbb{R}$ on which this identity holds, so if both functions were holomorphic, the proof was complete using the identity theorem.
    But why are the functions $\displaystyle z\mapsto f(\overline{z})$ and $\displaystyle z\mapsto \overline{f(z)}$ holomorphic?

    Thank you!
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  2. #2
    MHF Contributor chisigma's Avatar
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    If $\displaystyle f(z)$ il holomorphic $\displaystyle \forall z\in\mathbb{C}$, it will be true also for $\displaystyle z=0$, so that $\displaystyle f(z)$ can be written as…

    $\displaystyle f(z)=\sum_{n=0}^{\infty} a_{n}\cdot z^{n}$ (1)

    But $\displaystyle f(z) \in\mathbb{R}$ $\displaystyle \forall z\in\mathbb{R}$, so that all the $\displaystyle a_{n}$ in (1) are real. Since $\displaystyle \overline {z^n}= \overline {z}^{n}$ for (1) is $\displaystyle f(\overline {z})= \overline {f(z)}$…

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$



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  3. #3
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    Oh, thank you! So I was on the wrong train with wanting to use the theorem.
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