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Is there an entire not constant function with bounded real part?

I've seen a solution which involves the exponential and Liouville's theorem.

I wanted to see a more direct proof. If $f$ is an entire function and we have a continuously differentiable function $\gamma:[0,\infty)\rightarrow\mathbb C$ and we know that $\Im (f\circ\gamma)(t)\rightarrow\infty$ (for $t\rightarrow\infty$), can we construct an analogous curve for the real part?

I'm having trouble to get an intuition for how the real part of a holomorphic function looks like in relation to its imaginary part. The only thing I can visualize is that the gradient of the two are orthogonal, but I kind of struggle what that means for the potential of those two gradients, i.e. the real and imaginary part.


marked as duplicate by Davide Giraudo, Nosrati, Claude Leibovici, José Carlos Santos, user91500 Jul 27 '17 at 8:57

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Hint: Postcompose your function with a conformal mapping of a half-plane (containing the image of $f$) to the unit disk. Then use Liouville's theorem.

  • $\begingroup$ Hi, thanks for the hint, but this isn't what I'm looking for. Like I said in the post I know a proof doing just as you say, $f \circ \exp + Liouville$ $\endgroup$ – zkzm1 Jan 10 '17 at 11:27
  • $\begingroup$ @zkzm1 It is not exp, it is linear fractional. $\endgroup$ – Moishe Kohan Jan 10 '17 at 13:10
  • $\begingroup$ I've ment $\exp\circ f$ + Liouville. I've ment it uses the same idea, I am looking for a more "direct" proof, and for some intuition on how the real part behaves in respect to the imaginary part. $\endgroup$ – zkzm1 Jan 10 '17 at 13:14
  • $\begingroup$ The reason I'm looking for a more "direct" proof is the hope that I'd get some more insight on how the real part behaves compared to the imaginary part. $\endgroup$ – zkzm1 Jan 10 '17 at 13:47

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