# Real and imaginary part of holomorphic function

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?

No, consider $i\log z$ in the the right half plane.
• 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? – armara Oct 10 '16 at 18:46