# How to show that $f$ is an odd function?

An entire function $f$ takes real $z$ to real and purely imaginary to purely imaginary. We need to show that $f$ is an odd function. well, $f=\sum_{n=0}^{\infty}a_nz^n$ what I can say is $f(\mathbb{R})\subseteq\mathbb{R}$ and $f(\mathbb{iR})\subseteq\mathbb{iR}$

How to proceed, please give me hint.

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ahhh! I just got it since $f(\mathbb{R})\subseteq\mathbb{R}$ so each $a_n$ is real again since $f(\mathbb{iR})\subseteq\mathbb{iR}$ each $a_n=0$ for even $n$ am I right? –  La Belle Noiseuse Jun 14 '12 at 9:08
You may be expected to show that the two conditions respectively force (i) the $a_n$ to be real and (ii) the even coefficients to be $0$. –  André Nicolas Jun 14 '12 at 9:28
@Mex that is correct. –  TenaliRaman Jun 14 '12 at 9:30
You may also use Schwarz Reflection Principle and show $f(-z) = -f(z)$. –  TenaliRaman Jun 14 '12 at 9:32
@Mex: I encourage you to answer your own question and accept it, if you think you have the correct answer. –  mixedmath Jun 14 '12 at 10:38

We have that $$f(z)=\sum_{n=0}^\infty a_nz^n$$ because $\,f\,$ is entire, and by the given conditions we have$$(1)\,\,f(r)=\sum_{n=0}^\infty a_nr^n\in\mathbb{R}\,,\,\,\,r\in\mathbb{R}$$$$(2)\,\,f(ir)=\sum_{n=0}^\infty a_n(ir)^n\in i\mathbb{R}\,\,,\,\,r\in\mathbb{R}$$but we have that $$\sum_{n=0}^\infty a_n(ir)^n=\sum_{n=0}^\infty i^n (a_nr^n)=\sum_{n=0}^\infty (-1)^na_{2n}r^{2n}+i\sum_{n=0}^\infty (-1)^na_{2n+1}r^{2n+1}$$and as the above is purely imaginary we get that $\,a_{2n}=0\,,\,\forall n\in\mathbb{N}\,$ , so the power series of the function has zero coefficients for the even powers of $\,z\,$ and is thus a sum of odd powers and trivially then an odd function.