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This question already has an answer here:

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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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.

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  • $\begingroup$ 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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