Suppose a complex valued function $f$ is entire, maps $\mathbb{R}$ to $\mathbb{R}$, and maps the imaginary axis into the imaginary axis.

I see that $f(x)=\overline{f(\bar{x})}$ on the whole real axis, and thus the identity theorem implies that $f(z)=\overline{f(\bar{z})}$ for all $z\in\mathbb{C}$. Then for $ai$ on the imaginary axis, it follows that $$ f(ai)=\overline{f(\overline{ai})}=\overline{f(-ai)}=-f(-ai). $$

Does this relation somehow extend to arbitrary $z$ so that $f$ is odd on the whole complex plane?


Yes: Analytic functions have isolated zeroes or else they are identically zero (see this answer for an argument showing this). Consider the function $F(z)=f(zi)+f(-zi)$. This function has all reals $z=a$ as zeroes, so its zeroes are not isolated.

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    $\begingroup$ So is it correct to think I can basically use the identity theorem again? $F(z)$ is identically $0$, since it agrees with the $0$ function on the real line, which clearly has limit points in $\mathbb{C}$, so $F$ is identically $0$ on $\mathbb{C}$. $\endgroup$ – Ben Nevis May 7 '12 at 22:35
  • $\begingroup$ Yes, that's exactly the point. $\endgroup$ – Andrés E. Caicedo May 7 '12 at 22:39
  • $\begingroup$ Appreciate it, thanks! $\endgroup$ – Ben Nevis May 7 '12 at 22:41

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