Prove that the image of $f: D \rightarrow \mathbb{C}, z\mapsto\frac{i+z}{1+iz}$ is the upper half plane $H=\{z\in\Bbb{C}\mid\text{Im }z>0\}$.

Question

Let $D$ be the open unit disk centered at $0$ in the complex plane and let $f: D \rightarrow \mathbb{C}, z \mapsto \frac{i+z}{1+iz}$. How should I proceed in order to show that $\operatorname{im}(f)$ is the upper half plane $H = \{z \in \mathbb{C} \mid \operatorname{Im}(z) > 0\}$ ? Unfortunately, I have no idea how to start.

Thanks in advance.

Answer 1

Evaluate the imaginary part of the image of the map. With $\;z=a+ib\in D\iff a^2+b^2<1\;$ ,

$$\frac{i+z}{1+iz}=\frac{i+a+ib}{1+ia-b}=\frac{a+i(b+1)}{(1-b)+ia}\cdot\frac{(1-b)-ia}{(1-b)-ia}=$$

$$=\frac{\left[a(1-b)+a(1+b)\right]+i\left[(1+b)(1-b)-a^2\right]}{(1-b)^2+a^2}=$$

and thus the imaginary part of the above is

$$\frac{1-a^2-b^2}{a^2+(b-1)^2}>0\iff a^2-b^2<1$$

and we're done.

Answer 2

It might be easiest to use the circle/line preserving property of Möbius transformations. That is Möbius transformations sends circles/lines to circles/lines. If you can show its sends two points on the circle to the real line, then the boundary of the image must be the real line. You need to only figure out where it sends the interior of the circle, which can be established by looking at the image of one point.

So, for instance, the image of $1$ is $$\frac{i +1}{1 +i} =1$$ and the the image of $-1$ is $$\frac{ -1 + i}{1 - i} = -1$$

Then, the image of $0$ is $\frac{i}{1}=i$, the the interior is the upper half plane.