Question about Holomorphic functions I try to show:
Let $f: \mathbb{C} \longrightarrow \mathbb{C}  $ be holomorphic with $$\Re(f)+\Im(f)=1 $$  then $ f $  is constant.
($\Re$ = Real Part, $\Im$ = Imaginary Part)
I have certain ideas one is to  use Liouvilles theorem about bounded functions but I am not sure if the given information already implies that $f$ is bounded. Consider for example $f(x)=1 + (1 - i) \Re(x)$ then $f$ is not bounded (here it fails because $f$ is not holomorphic). But I don't know how I could use the fact that $f$ is holomorphic to show that its bounded.

This is a special case of a very interesting Liouville-type theorem: 
Theorem Let $\Omega$ be an open subset of the complex plane and $f\colon \Omega \to \mathbb{C}$ be holomorphic. If $f(\Omega)$ is contained in a 1-dimensional manifold (i.e. a smooth curve) then $f$ is constant. 
The geometric intuition behind this is explained very well in Needham's Visual complex analysis. Locally $f$ acts like a rotation composed with a dilation (amplitwist) so it must be full rank or rank 0, it cannot be rank 1. If the range of $f$ lies in a curve then the rank of $f$ cannot exceed 1 and so it must be 0 everywhere. This means that $f$ must be constant.
