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If f is an entire function satisfying f(z+1)=f(z) and f(z+i)=f(z) then f must be constant throughout the complex plane. I would appreciate if someone helps me know how to prove the above statement.

  • Restrict to a fundamental domain, use compactness, then Liouville's. –  Aug 31 '15 at 01:06

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This function is bounded and entire, since is is both 1 and i periodic and continuous. It is therefore constant.

ncmathsadist
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  • How did you conclude that f is bounded? –  Aug 31 '15 at 01:21
  • This function is 1 and i periodic. Its range is just its image of the square cornered by the $0$, $1$, $i$ and $1 + i$. That square is compact, which gives you a bounded range. Now apply Liouville. – ncmathsadist Aug 31 '15 at 01:39