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I posted this question before, and after some clarification I'll try again (since the first post was badly formulated).

If the real part of a holomorphic function is bounded, does that mean that the imaginary part also has to be bounded?

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No, consider $i\log z$ in the the right half plane.

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  • $\begingroup$ Hello! I'm thinking $$ilog(z)=i(Log|z|+iArg(z))=iLog|z|-Arg(z)$$. Am I correct to say that $Re[ilog(z)]=Arg(z)$ is in this case bounded, because of the principle branch that says $$-\pi<Arg(z)<\pi$$, and $Im[ilog(z]=iLog|z|$ is not bounded, since we don't have any restrictions to z? $\endgroup$ – armara Oct 10 '16 at 18:46
  • $\begingroup$ Yes, but Re ilog z = - Arg z. $\endgroup$ – zhw. Oct 10 '16 at 19:03
  • $\begingroup$ Of course, I missed that part. Thank you! $\endgroup$ – armara Oct 10 '16 at 19:25

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