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.

$f(z) = u(z) + iv(z), \quad u(z) = 1-v(z)$:
$$\frac{\partial u}{\partial x} = -\frac{\partial v}{\partial x} = \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x}$$
$$\implies \quad 0 = \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$$
q.e.d.
