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Is it true that an entire function, whose $Re(s)\gt0$, then it has to be a constant? I think this is true and we could use Louville's theorem to prove this, but not sure.

J.doe
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  • If $f$ is a bounded entire function with $\textrm{Re}(f(z)) > 0$ (I assume this is what you meant?), then apply Liouville's theorem to $g(z) = e^{-f(z)}$. – Ethan Alwaise Dec 14 '16 at 09:12
  • Wait... doesn't this one prove that the statement is wrong? – J.doe Dec 14 '16 at 09:14
  • And yep that's what I meant. Sorry for causing any confusion. – J.doe Dec 14 '16 at 09:14
  • Also: http://math.stackexchange.com/questions/1956954/if-fz-has-positive-real-part-show-that-f-is-a-constant-function, http://math.stackexchange.com/questions/573691/prove-that-a-holomorphic-function-with-postive-real-part-is-constant. – Martin R Dec 14 '16 at 09:34

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I guess you have an entire function $f$ with $Re(f(z))>0$ for all $z$. If yes, then define $g(z)=e^{-f(z)}$.

Show that $|g(z)|<1$ for all $z$.

By Liouville, we get that $g$ is constant. Now its your turn to show that $f$ is constant.

Fred
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