Find a Möbius transformation that maps the unit disk to $\{x+yi \in \mathbb C: x+y \geq 0 \}$ Find a Möbius transformation that maps the unit disk to $\{x+yi \in \mathbb C: x+y \geq 0 \}$.
I was trying to use that the unit disk is $|z| < 1$, and the general form for Möbius transformation is $$f(z)=\frac{az+b}{cz+d}$$
Any help would appreciate!
 A: Let $\Delta$ be the open unit disc and let $C=\partial\Delta$ be the unit circle.
Hint 1. Möbius transformations are homeomorphisms of the Riemann sphere onto itself, and therefore if $m$ is a Möbius transformation then $m(\Delta)$ is almost determined by $m(\partial\Delta)=m(C)$; meanwhile, $C$ is a circle and the boundary of the required image region is a straight line.
Hint 2. If you want to construct a Möbius map that sends three chosen points $\alpha,\beta,\gamma$ to three other chosen points $A,B,C,$ one way to proceed is as follows: write down
$$m(z)=\frac{az+b}{cz+d},$$
use the fact that $\alpha,\beta,\gamma$ map to $A,B,C$ to put constraints on $a,b,c,d,$ and then find a solution to the resulting linear equations.
Answer.

 We choose the points $1,i,-1\in C$ and we want to construct a Möbius map $m$ 
 sending these three points to the points $1-i,0,-1+i$ on the line $x+y=0.$
 After following the advice of Hint 2 above, we might arrive at the candidate
 $$m(z) = \frac{2(z-i)}{(1-i)z+(1+i)}.$$
 So this Möbius map sends the unit circle $C$ to the line $x+y=0.$ This means 
 that $m(\Delta)$ is either $x+y>0$ or $x+y<0.$ To see which, we examine 
 $m(0)$: we have
 $$m(0)=\frac{2(-i)}{1+i} = -1-i,$$
 so we see that $\operatorname{Re}{(m(0))}+\operatorname{Im}{(m(0))}<0.$ This 
 means that the Möbius map we have come up with is not quite what we are 
 looking for, but in this case it is easy to fix this: just take $n(z)=-m(z)$ 
 instead. That is, a Möbius map that fits the required condition is
 $$n(z)=\frac{2(z-i)}{(-1+i)z-(1+i)}.$$

A: The common idea in this and one of your previous recent questions is the upper half plane. The Mobius transfrmations of the upper half plane to itself are those $\frac{az+b}{cz+d}$ with real $a,b,c,d$ and $ad-bc > 0.$ Worth remembering.
Going back and forth between the upper half plane and the standard unit disk are inverse mappings
$$  \frac{z+i}{iz+1} $$ and
$$ \frac{iz+1}{z + i}.  $$
The first one $  \frac{z+i}{iz+1} $ takes $0$ from the interior of the unit disk to $i$ in the interior of the upper half plane. Also, the unit circle goes to the real line, as $1 \rightarrow 1 \; , \; $ $-1 \rightarrow -1 \; , \; $ $-i \rightarrow 0 \; , \; $ $i \rightarrow \infty \; . \; $   
The second one does the reverse. 
We also know that the two transformations are inverse mappings because the product of the associated matrices is a scalar multiple of the identity matrix, therefore the identity mapping ($2iz/(2i) = z$).
$$
\left(
\begin{array}{cc}
1 & i \\
i & 1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
i & 1 \\
1 & i \\
\end{array}
\right) =
\left(
\begin{array}{cc}
2i & 0 \\
0 & 2i \\
\end{array}
\right) = 2 i
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)
$$
