How can I prove this?
If $f$ is a holomorphic function in a domain $U$ and $f'(z)\neq0$ for all $z\in U$ then every zero of $f$ are simple and positive.
Definition: $q\in U$ is a simple positive zero if $f(q)=0$ and $Det[Df_{q}]>0$, analogous for simple negative zero.
I was thinking in contradiction, we suppose $f$ has a simple negative zero. Then there is a $q_{1}\in U$ such that $f(q_{1})=0$ and $Det[Df_{q_{1}}]<0$. Then I have that $u_{x}v_{y}(q_{1})-v_{x}u_{y}(q_{1})<0$ and I would like to use something like the intermediate value theorem in complex numbers with another point such that $u_{x}v_{y}(p)-v_{x}u_{y}(p)<0$ to have there is one point such that $f'(\zeta)=0$, but I don´t know if I can do that.
Thanks for any help!
Hint: If $f=u+vi$ is holomorphic, it satisfies the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$.