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Let's say we have an entire complex function $f(z)$ such that:

  1. $f(z)$ is real when $z$ is real
  2. $f(z)$ is purely imaginary when $z$ is purely imaginary.

SO basically this entire function maps real/im axis to itself. The question asks me to show this function is an odd function and I have no clue.

qwerty
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  • If you interested in a solution without using the hint see https://math.stackexchange.com/questions/158168/how-to-show-that-f-is-an-odd-function?rq=1 – Kavi Rama Murthy Jan 14 '21 at 09:16
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    The hint is probably about the Schwarz reflection principle. My gut tells me that you should use it, together with the identity theorem, to examine how $f$ acts under two reflections: first along the imaginary axis, than along the real axis (or the other way around, doesn't really matter), since these two reflections correspond to a reflection at the origin, which is really what you have to examine to show that a function is odd (a function $f$ is odd if reflecting at the origin yields $-f$). – Vercassivelaunos Jan 14 '21 at 09:42

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Let $f_1$ be the restriction of $f$ to $\{\Im(z)\ge 0\}$. By Schwarz's reflection, this admits an analytic extension to $\mathbb{C}$ defined in $\Im(z)\le 0$ by $g(z)=\overline{f_1(\bar z)}$. By the identity principle, $g=f$. Since $f$ is purely imaginary on $i\mathbb{R}$, this equation tells us that $f(z)+f(-z)$ is zero on $i\mathbb{R}$. By the identity principle again, $f(z)+f(-z)\equiv 0$, which implies that $f$ is odd.