I've got to show that such transformation $f$ with mapping points $f(0) = -1$, $f(i) = 0$, $f(+\infty) = 1$ has properties $f(D_1) = D_2$ where $D_1 = \{z \in \mathbb{C} \vert \quad \text{Im$(z)$}\geq 0\}$ and $D_2 = \{z \in \mathbb{C}\vert \quad |z|\leq 1\}$
I figured $f = \dfrac{z-i}{z+i}$ includes all the points of the sought transformation. My problem is at the root of showing that property. My idea is to substitute $z = a+i\,b$ and then calculate the absolute value of $f(z)$. Afterwards taking the limit of $|(f(z))|$ where $a \to\infty$ respectively $b\to\infty$ both with maximal value $1$. I have done these calculations before but it seems like a lot of paperwork for such a task, regardless of the fact this is legitimate. That's why I'm asking.