Prove that $f$ is constant, when $ \operatorname{Re}(f)^m = \operatorname{Im}(f)^n$ Take $f:U\rightarrow \mathbb{C}$ is holomorphic function in $U$ and $U$ connected open subset. If exist $m,n\in \mathbb{N}$ such that $$ [
 \operatorname{Re}(f(z))]^m = [ \operatorname{Im}(f(z))]^n ,$$ $f$ is constant in $U$. 
Note: The exercise requires using the theorem and equations of Cauchy-Riemman
NB: An attempt is in the comments.
Question: In my first attemp, I want to see that is not posible that $det(A_{(x,y)})=0 \ \ \forall (x,y) \in U$
One possibilities is $v^{2n-2} (x,y) =0$. This implies that $f(x,y)=0$. But, is in this point. And others? $f=0$ ever?
 A: Show that $C=\{ z | (\operatorname{re} z)^m = (\operatorname{im} z)^n \}$ contains no open sets.
The open mapping theorem shows that a non constant $f$ maps open sets into open sets.
Alternative:
Note that it is sufficient to show that if $f(z) \neq 0$, then $f'(z) = 0$.
In particular, this implies that $f'(z) = 0$ everywhere in $U$.
To see this, suppose $f'(z) \ne 0$, then we must have $f(z) = 0$, but since
$f'(z) \neq 0$, we have $f(w) \neq 0$ for all $w$ in some neighbourhood of $z$ and hence $f'(w) = 0$ for all $w$ in some neighbourhood of $z$.
Continuity then shows that $f'(z) = 0$, a contradiction.
So, we can suppose that $f(z) \neq 0$. The rest of the argument follows
the OPs comments above.
In particular, we consider $f$ as a map from $\mathbb{R}^2$ to $\mathbb{R}^2$, $(x,y) \mapsto (u(x,y), v(x,y))$, that satisfies
The Cauchy Riemann equations.
Define $\phi(a,b) = a^m-b^n$, then since $\phi(u(x,y),v(x,y)) = 0$, we
have, suppressing the parameters for brevity, ${\partial \phi \over \partial a} {\partial u \over \partial x} +
{\partial \phi \over \partial b} {\partial v \over \partial x} = 0$
and 
${\partial \phi \over \partial a} {\partial u \over \partial y} +
{\partial \phi \over \partial b} {\partial v \over \partial y} = 0$.
Substituting the Cauchy Riemann equalities into the second equation
gives
$-{\partial \phi \over \partial a} {\partial v \over \partial x} +
{\partial \phi \over \partial b} {\partial u \over \partial x} = 0$.
Writing as a matrix gives
$\begin{bmatrix} {\partial \phi \over \partial a} & {\partial \phi \over \partial b}\\
{\partial \phi \over \partial b} & -{\partial \phi \over \partial a}
\end{bmatrix}
\begin{bmatrix} {\partial u \over \partial x} \\ {\partial y \over \partial x}
\end{bmatrix} = 0$ and we see that the left hand side is invertible
iff $({\partial \phi \over \partial a},{\partial \phi \over \partial a}) \neq 0$.
Since we have presumed $f(z) \neq 0$, it is straightforward to check
that $({\partial \phi \over \partial a},{\partial \phi \over \partial a}) \neq 0$ and hence $f'(z) = 0$.
