# 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

## 1 Answer

I am writing, enlarging and enhancing (hopefully...) Mex's answer to his own question (kudos!) and I'll be happy to erase this answer of mine if he decides to write down his.

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.

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Don :) my great pleasure, thank you very much. – La Belle Noiseuse Jun 14 '12 at 14:42
(1)You've very much welcome, @Mex, and (2) thanks for giving the solution to your own question. – DonAntonio Jun 14 '12 at 14:50