Complex function with real image Let $f:C \subset \mathbb{C} \rightarrow \mathbb{C}$ be an analytic function and $C$ a connected subset. I want to prove that if $f(z)$ is real for all $z \in C$, then $f$ is constant.
Write $f$ as a power series $f(z)=\sum_{i=0}^n a_n (z-a)^n$ with $a_n \in \mathbb{R}$ and $a \in C$. 
Since $f(z)$ is real, the imaginary part of the power series is null. I want to know if I can conclude that $a_n=0$ for all $n \geq 1$, for, if $a_1=0$, then $f'(a)=0$ for all $a$ in $C$ and by connectedness, $f$ is constant. (Or, $f(z)=a_0=f(a)\Rightarrow f $ is constant)
 A: If you are not allowed to use the CR equations for the proof the use of power series is even more tabu. But it is sufficient to work with the definition of complex differentiability.
Consider a point $z\in{\rm dom}(f)$ and assume that $f'(z)=:A\ne0$. There is an $\epsilon>0$ such that
$$\left|{f(z+h)-f(z)\over h}-A\right|<{|A|\over2}\qquad\bigl(|h|<\epsilon\bigr)\ .$$
It follows that 
$$|f(z+h)-f(z)-Ah|<{|A h|\over2}\qquad\bigl(0<|h|<\epsilon\bigr)$$
or
$$f(z+h)-f(z)=Ah\left(1+{\Theta_h\over2}\right)\qquad\bigl(0<|h|<\epsilon\bigr)\tag{1}$$
for some complex $\Theta_h$ of absolute value $\leq1$. Now choose $$h:=t i\bar A$$ with $t>0$ small enough. Then the left hand side of $(1)$ is real by assumption, but the right hand side is definitively not real.
From this we conclude that in fact $f'(z)=0$ for all $z\in{\rm dom}(f)$. This immediately implies that all partial derivatives of $u$, $v$ are $\equiv0$; and as a consequence  $f$ is constant.
A: One approach (that doesn't require a power series representation) is to mimic the proof you'd use to deduce the Cauchy-Riemann equations. Namely, write $f(z) = u(z) + iv(z)$ with $u$ and $v$ real-valued. By hypothesis, $v \equiv 0$, and the complex limit
$$
\lim_{h \to 0} \frac{f(z + h) - f(z)}{h}
$$
exists. Let $h$ approach $0$ along the real axis, then along the imaginary axis, expressing the results in terms of partials of $u$. The fact that these limits are equal tells you something about $u$....
