I want to find the Möbius transformation from the disk $\{|z-1|\leq2\}$ to the half-plane $\{Re(z)\geq3\}$ that moves the point $0$ to $4+4i$.
I know that by specifying the values at 3 points, the Möbius transformation is determined.
If I impose that $$T(0)=4+4i, T(3)=3+i, T(-1)=\infty$$
I get the relations $c=d$, $b=(4+4i)c$, $c=\frac{3}{8}a$ and therefore my transformation is $$T(z)=\frac{az+\frac{3(4+4i)}{8}a}{3az/8+3/8}=\frac{8z+12+12i}{3z+3}$$
This transformation does fulfill my 3 initial conditions, however the are points outside the disk $\{|z-1|\leq2\}$ that, with this transformation, also move to the half-plane $\{Re(z)\geq3\}$. Does this mean I am wrong? If I am correct why can I ensure that all the points inside my disk go to the half-plane $\{Re(z)\geq3\}$?
Best Answer
Regarding your guess about the images under $T$ of the three points $0$, $3$, and $-1$, since those three points all lie on a line, with $3$ and $-1$ on the boundary of the disc, you may conclude that their three images $T(0)=4+4i$, $T(3)$, and $T(-1)$ must all lie on a line (which includes $\infty$) or a circle, with $T(3)$ and $T(-1)$ all on the boundary of the region $Re(z)\ge 3$. You've chosen $T(-1)=\infty$, and so the choice of $T(3)$ is still open. But there are still infinitely many points to choose for $T(3)$ on the boundary line $Re(z)=3$.
You chose $T(3) = 3+3i$.
But that cannot be correct, because the $0$, $3$, and $-1$ lie on a diameter that hits the boundary circle at a right angle at the point $3$, whereas the line through $\infty$, $4+4i$, and $3+3i$ definitely does not hit the boundary line $Re(z)=3$ at a right angle, which would violate conformality of $T$.
On the other hand, the line through $\infty$, $4+4i$, and $3+4i$ does hit the boundary line $Re(z)=3$ at a right angle.....