Provided that $f(u)$ is holomorphic, prove that $u$ is constant Let $u: \mathbb{C} \rightarrow \mathbb {R}$ be a real valued function, so that $f(z)=\cos \left( u(z) \right) +i \cdot \sin \left( u(z) \right)$ is holomorphic in $\mathbb{C}$. Show that $u$ is constant.
I tried using Cauchy-Reimann equations and the chain rule, but I'm not sure that I can say that $u$ has a derivative. In order to be more formal, I tried to address $u$ as a real vector valued function with a constant value in the second coordinate $u \left( \matrix {x\\y} \right) = \left( \matrix {u_1(x,y)\\0} \right)$ and then $f(u) = f(u_1)$. It's obvious that if $u$ is differentiable (in the complex sense) then it is constant. But I couldn't show it.
I also using Euler's formula for the complex exponent $f(z) = e^{i \cdot u(z)}$ and adding it's conjugate to it in order to get a real valued harmonic function $h(z) = f(z) + \overline{f(z)} = 2 \cdot\cos(u(z))$, since $h$ is both real valued and holomophfic - it is constant, hence $u$ is constant. 
BUT:


*

*I'm not sure why $u$ can't be something like $u(x,y) = 2 \pi \lceil x\rceil$ ?

*What are the general conditions for both $f$ and $\overline{f}$ to be holomorfic?



Anyway I'm pretty sure I'm off the right track so new ideas will be gladly welcomed.

EDIT
My instructor approved that $u$ has to be continuous so now the problem is solved pretty easily. Thanks for all your help! 

Hint: Assume that $u$ is at least continuous, by the little Picard theorem, a non constant holomorphic function avoids at most one point. Your function $f$  is bounded so it is constant. This implies that $u$ is constant.
