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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??

Richard
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1 Answers1

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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.

Eugene Zhang
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  • There is an error with the indices, it should be in the first line $f(iy)= \sum_{n=0}^\infty a_{2n}i^{2n}y^{2n}+ \sum_{n=0}^\infty a_{2n+1}i^{2n+1}y^{2n+1}$ and then $\sum_{n=0}^\infty (-1)^na_{2n}y^{2n}+ i\sum_{n=0}^\infty (-1)^na_{2n+1}y^{2n+1}$. Other choices are possible as well, however like it is written it has the starting index problem and the $1/2$ once too much problem. – w_w Apr 30 '20 at 01:47
  • Thanks a lot. I have fixed it – Eugene Zhang Apr 30 '20 at 03:14