Actual Problem
I feel like by attempting to transfer my problem to a more general form it suffered a lot of important detail. So here i present to you my actual problem. It's part of a proof in an article about Theta functions and this particular aspect is stated without any explanation:
$ k':=\left(\frac{\vartheta_{01}(0;w)}{\vartheta_{00}(0;w)}\right)^2 $ (I use notations like on wikipedia)
Remark: $k'$ is holomorphic on the whole upper half of the complex plane.
The article's author shows that $k'$ maps the red line in below figure to the green lines (positive real and imaginary axis). And then he does the - for me - vague implication that $k'$ maps left side's grey area to right side's grey area.
Could anybody explain why?
PS: I could provide the article but it's in german :\
Old Version of my question
Idea
Does a holomorphic function map the boundary of a set to the boundary of another set? Are all interior points mapped to interior points of a set, which is bounded by the mapped boundary?
Conditions
Let $f$ be a holomorphic function on open set $G$ and $D$ be a connected subset of $G$. Let's say the boundary $\partial D$ of $D$ is mapped to the positiv imaginary axis and the positiv real axis by $f$, i.e. $ f(\partial D)=\left\{ z\in\mathbb{C}: Re(z)=0, > Im(z)\geq0 \right\}\cup \left\{ z\in\mathbb{C}: Im(z)=0,Re(z)\geq0 > \right\} $. So $f(\partial D)$ is a boundary to the first quadrant of the complex plane: $f(\partial D)=\partial \left\{ z\in\mathbb{C}: > Re(z)\geq0, Im(z)\geq0 \right\}$
Question
Is the map of an interior point $z$ of $D$ by $f$ an element of the first quadrant of the complex plane? $ \forall z \in D: > f(z)\in\left\{ z\in\mathbb{C}: Re(z)\geq0, Im(z)\geq0 \right\} $?
Not Necessarily. If $f$ maps $\partial D$ to the two positive axes, then so does $g\colon z\mapsto i f(z)^3$. Note that if $f(z)$ is in the first quadrant, then $g(z)$ is not.
Your problem is that $f(\partial D)$ is the boundary of two distinct (simply) connected regions. (I added "simply" in parentheses because you did not even require $D$ to be simply connected, which leaves a lot of liberty)