In Conway's complex analysis while proving proposition 2.10
"If $G$ is an open connected set and $ f : G \to \Bbb{C} $ such that $f'(z)=0 \ \forall z \in G$ then f is constant"
He fixes some $z_{0} \in G$ and defines the set $A=$ {$z \in G: f(z)=f(z_0)$} and shows that this set is both open as well as closed, since it is a non-empty subset of a connected set, we have, $A=G$ which would complete the proof.
I can't understand the proof to show it's open, where he defines $ g(t)=f(zt+(1-t)a) \ , t \in [0,1]$ for some fixed $a \in A $ and $\epsilon >0$ such that $z\in B(a,\epsilon)\subset G$ He wants to show that $B(a,\epsilon) \subset A.$
By proving $g'(t)=0 \ $on $ [0,1]$ he concludes that g is constant.
I thought this result was only true for real valued functions. I am clear about everything else, why the function has been constructed in this way and all. I am just not able to convince myself of the above conclusion.
If $g:[0,1]\to \mathbb C$ is differentiable and $g'=0$ for all $t$ then its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.