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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?2012-06-14
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    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 $0$.2012-06-14
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    @Mex that is correct.2012-06-14
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    You may also use [Schwarz Reflection Principle](http://mathworld.wolfram.com/SchwarzReflectionPrinciple.html) and show $f(-z) = -f(z)$.2012-06-14
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    thank you Raman and Nicolas2012-06-14
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    @Mex: I encourage you to answer your own question and accept it, if you think you have the correct answer.2012-06-14
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    +1 For a nice question and for providing the answer to it, @Mex. Is there any relation to Mexico in your nick?2012-06-14
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    No! :) nothing related to Mexico :)2012-06-14

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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.2012-06-14
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    (1)You've very much welcome, @Mex, and (2) thanks for giving the solution to your own question.2012-06-14