How to construct Möbius map to tranform so 2D region to another?

Question

I want to know what is the general step-by-step method to find a Möbius map $w(x) = \dfrac{ax+b}{cx+d}$ that transforms a region in a complex plane to another region (2D).

I know it takes circle and lines to circle and lines, so for example if I wanted to transform the unit circle in z-plane to the real line in w-plane, I would need $w(-1)=-1$, $w(i)=0$ and $w(1)=1$. These fix 3 of $a,b,c,d$ and the remaining just factor out.

Now I need to find a map that maps the unit upper semi-disc to the upper right quadrant. I let $w(-1)=0$, $w(i)=i$ and $w(1)=2i$ and found ${w(z)=\frac{(z+1)(1+i)}{-iz+1}}$. But this is different from the given solution ${w(z)=\frac{1+z}{1-z}}$. It gives a hint that I should consider the angle-preserving property of conformal map at the point $-1,1$, but I am not so sure about that.

Answer 1

The boundary of the upper semi-disk is made of a line segment and and arc of a circle, meeting at $\pm1$. The boundary of the upper right quadrant is two rays, meeting at $0$ and $\infty$. So map $-1$ to $0$ and $1$ to $\infty$. To map $-1$ to $0$, you need $z+1$ as a factor of the numerator. To map $1$ to $\infty$ you need $z-1$ as a factor of the denominator. That explains the "given" solution.

Your mapping $1$ to $2i$ I find questionable, to say the least. The complex number $1$ is at the intersection of the two parts of the boundary of the upper half disk, so it should map to one of the two points of intersection of the two parts of the boundary of the upper right quadrant.

PS:

• We need to map the two intersection points $\pm1$ of the unit circle and the real axis to the two intersection points $0$ and $\infty$ of the real and imaginary axes. If we want $-1\mapsto0$ and $1\mapsto\infty$ (and we could do it the other way around) then we need $1+z$ to be a factor of the numerator and $1-z$ to be a factor of the denominator.

• But then we're not done. Look at the point $-1$ in the domain, and you see the two circles (one of which is a straight line) crossing each other and divinding the vicinity of $-1$ into four regions, which let us label $A,B,C,D$ in counterclockwise order starting with $A$ in the upper half-disk, $B$ to its left, $C$ below that, and $D$ to the right of $C$ and below $A$. We need $A$ to be the one that maps to the upper right quadrant.

• Only one (complex) degree of freedom remains to be chosen once we've decided that our linear fractional transformation is $z\mapsto a\dfrac{1+z}{1-z}$, and that is choosing the complex number $a$. If we choose it so that the positive part of the real axis maps to any particular ray going out from $0$, then the upper half of the circle will map to a ray $90^\circ$ counterclockwise from that one, since the mapping is conformal. That means we need $0\mapsto\text{some positive number}$. But we have $0\mapsto a$. So we just need $a$ to be a positive number. If we want $0\mapsto1$ then we need $a=1$.