Question

 Find the mobius transformation which maps the 

circle |𝑧| < 1 conformally onto |𝑤 − 1| < 1 and 

make the points z=0,1 correspond to w=1/2,0 

respectively. Is the transformation uniquely 

determined

Answer 1

The Mobius transformation is generally given by

$f(z) = \frac{az+b}{cz+d}$

As $f(0)\neq \infty$, we conclude that $d \neq 0$ and thus we can choose $d=2$ (as in Mobius transformation multiplying $a,b,c,d$ by a same non-zero constant gives the same transformation). Knowing the images if $z=0, z=1$ we find:

• $\frac{b}{d}=\frac{1}{2}, b=1$
• $a+b=0, a=-1$

Now suppose $z\in \mathbb{C}, |z|=1$, then this tranformation should carry it onto $|w-1|=1$, thus

$|f(z)-1|=|\frac{-z+1-(cz+2)}{cz+2}|=|\frac{-(c+1)z-1}{cz+2}|$

$|f(z)-1|=\frac{(1+(c+1)\cos \theta)^2+(c+1)^2\sin^2\theta }{(c\cos\theta +2)^2+c^2 \sin^2\theta}=1$

$1+(c+1)^2\cos^2\theta+2(c+1)\cos\theta +(c+1)^2\sin^2\theta =c^2\cos^2\theta+4c\cos\theta +4+c^2\sin^2\theta$

$1+(1+c)^2+2(c+1)\cos \theta = c^2 +4c\cos\theta+4$

$(1+2c)+2(1-c)\cos \theta =3$

$2(1-c)\cos \theta =2(1-c)$

As this should be satisfied for all $\theta \in [0;2\pi]$, we have $c=1$ and thus the Mobius transformation is

$f(z)=\frac{-z+1}{z+2}$ and the couple $[-1:1:1:2]$ is unique up to a multiplication by a scalar.