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.
 A: 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.
A: $f(z) = u(z) + iv(z), \quad u(z) = 1-v(z)$:
$$\frac{\partial u}{\partial x} = -\frac{\partial v}{\partial x} = \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x}$$
$$\implies \quad 0 = \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}$$
q.e.d.
A: Hint: use Cauchy-Riemann equations.
A: The image of the plane under this map must line inside of a line.  Apply the open mapping theorem:  nonconstant analytic functions are open maps.  Since the image of any open set cannot be open, this function must be constant.
A: The condition is equivalent to $g(z)=(1-i)f(z)$ is real for all complex $z$. Note that if $f=k+ik$, then $(1-i)k(1+i)$ is real. The imaginary part of $g(z)$ vanishes - it's an equivalent way to phrase your condition.
The derivatives of $g(z)$ with respect to $z$, which exists because $g(z)$ is holomorphic just like $f(z)$, can be calculated as 
$$\frac{dg(z)}{dz} = \frac{dg(z)}{dz_1} = \frac{\partial g_1(z)}{\partial z_1} + i \frac{\partial g_2(z)}{\partial z_1} =
\frac{\partial g_2(z)}{\partial z_2} + i\frac{\partial g_2(z)}{\partial z_1}$$
by the Cauchy-Riemann equations. I used one of the Cauchy-Riemann equation to rewrite the partial derivative of the real part (the first term) to a derivative of the imaginary part. Note that the first step was to define the derivative with respect to $z$ as the derivative with respect to the real part of $z$. It's really the same thing for holomorphic functions.
If the first or second term has a wrong sign, or if the $i$ should be in the first term and not the second, it doesn't make a difference. But because the imaginary part of $g$, $g_2$, was assumed to be zero, $dg_2(z)/dz$ is zero as well, and therefore $g$ is constant, and therefore $f$ is constant, too.
A more general message: Pretty much any similar constraint unnaturally depending both on the real and imaginary part - instead of the whole $z$ itself - will force any holomorphic function to be constant. Holomorphic functions are really meant to make the operations "real part" and "imaginary part" inappropriate. That's because the "real part" and "imaginary part" are not holomorphic functions themselves.
A: We can use Picard's Theorem in this one too. Our function is entire and $0,2$ are not in the image of $f$. This forces $f$ to be constant. 
