# Is there an entire function with bounded real part [duplicate]

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, user91500Jul 27 '17 at 8:57

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.
• 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, $f \circ \exp + Liouville$ – zkzm1 Jan 10 '17 at 11:27
• 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. – zkzm1 Jan 10 '17 at 13:14