TITLE: Given an entire function which is real on the real axis and imaginary on the imaginary axis, prove that it is an odd function.
QUESTION [8 upvotes]: Given an entire function which is real on the real axis and imaginary on the imaginary axis, prove that it is an odd function.
By a Corollary: If $f$ analytic in a region symmetric with respect to the real axis and if $f$ is real for real $z$, then $f(z) = \overline{f(\bar z)} $.
So that, $f(z) = u(x+iy) + iv(x+iy) = u(x-iy) - iv(x-iy)$
$f(-z) = u(-x-iy) + iv(-x-iy) = u(-x+iy) - iv(-x+iy)$
$-f(-z) = -u(-x+iy) + iv(-x+iy)$
It looks close to the answer but what else can I do by using Schwartz reflection principle??

REPLY [6 votes]: Consider the Taylor series of $f(z)$
$$
f(z)=\sum_{n=0}^{\infty}a_nz^n
$$
Since $f(z)= \overline{f(\bar z)}$, all $a_n$ are real. Since $f$ maps imaginary on the imaginary axis
\begin{align}
f(iy)=\sum_{n=0}^{\infty}a_ni^ny^n&=\sum_{n=0}^{\infty}a_{2n}i^{2n}y^{2n}+\sum_{n=1}^{\infty}a_{2n+1}i^{2n+1}y^{2n+1}
\\
&=\sum_{n=0}^{\infty}(-1)^{n}a_{2n}\:y^{2n}+i\sum_{n=1}^{\infty}(-1)^{n}a_{2n+1}\:y^{2n+1}
\end{align}
is imaginary. So
$$
\sum_{n=0}^{\infty}(-1)^{n}a_{2n}\:y^{2n}=0
$$
which means $a_{2n}=0$ for all $n$. So $f(z)$ is odd.