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Let $f$ be an entire function that takes purely real values on some pair of lines intersecting at an angle of $\pi/\sqrt{2}$ radians. Prove that $f$ is a constant function.

I don't really have any clue on how to even start with this. Can anyone help please.

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Hint: prove that if an entire function is real everywhere on the real axis, then it commutes with complex conjugation. Hence, if you mirror one of the "real" lines in the other, you get a new "real" line. Keep mirroring until you know the function is real in a dense subset of the complex plane. –  Henning Makholm Dec 8 '13 at 23:27

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Here is a slightly different approach from what Henning Makholm suggested.

  1. If an entire function $f(z)=\sum c_n z^n$ is real on the real axis, then $ f(z)=\overline{f(\bar z)}$ (both sides are entire, and they agree on the axis). Hence, $c_n$ is real for all $n$.
  2. Apply item 1 to $f(e^{i\theta}z)$ where $\theta=\pi/\sqrt{2}$. Conclude that $c_n e^{in \theta}$ is real for all $n$.
  3. From 1 and 2, conclude that $c_n=0$ for $n\ge 1$.
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Thank you. I understand your solution. However, I have two questions: 1) I was not sure of how to interpret the part of the question where it says "some pair of lines intersecting at an angle of $\pi/\sqrt{2}$". In your solution you assume that one of the lines is the real axis and the other makes an angle of $\pi/\sqrt{2}$ with it. How can I justify that this assumption is WLOG? 2) It seems that the exact angle $\pi/\sqrt{2}$ is quite irrelevant to the problem statement. It would have worked with any other angle. Is that surprising? –  Sourav D Dec 12 '13 at 8:16
    
On second thoughts, I think any angle other than integer multiples of $\pi$ should lead to the same conclusion. –  Sourav D Dec 12 '13 at 8:42
    
@SouravD You can always rotate the plane (by linear change of variable); I did this without mentioning. –  Post No Bulls Dec 12 '13 at 13:19

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