Problem
Let $T:\overline{\mathbb C} \to \overline{\mathbb C}$ be a Möbius transformation such that $T(1+2i)=1$, $T(-1+2i)=4$ and $|T(0)|=2$. Show that $|T(bi)|=2$ for all $b \in \mathbb R$.
The exercise asks me to show that the imaginary line is mapped to the circle centered at $0$ of radius $2$. I've tried to solve it in the most basic way, which is to express $T(z)=\dfrac{az+b}{cz+d}$,, solve for $a,b,c,d$ from the information given and try to get to $|T(bi)|=(T(bi)\overline{T(bi)})^{\frac{1}{2}}=2$.
I am still struggling with the operations but I wanted to know: Is there a simpler way or an alternative solution to this one? I would appreciate if someone could suggest another solution using properties of Möbius transformations, or cross-ratio, symmetric points or whatever it is useful in the theory of Möbius transformations to solve the problem.
Möbius transformations map straight lines and circles to straight lines and circles. Also, Möbius transformations preserve symmetry in the sense that if $z,w$ are two points symmetric with respect to a circle or straight line $C$, then $Tz$ and $Tw$ are symmetric with respect to $T(C)$.
The points $1+2i$ and $-1+2i$ are symmetric with respect to the imaginary axis. So the points $1$ and $4$ are symmetric with respect to $C = T(i\mathbb{R})$. And $C$ contains at least one point with modulus $2$.
There is only one straight line in the plane with respect to which $1$ and $4$ are symmetric, the line $\operatorname{Re} z = \frac{5}{2}$. That line contains no point with modulus $2$. So $C$ must be a circle. The midpoint of that circle must lie on the straight line through $1$ and $4$, so on the real axis, and it cannot lie between $1$ and $4$. A circle with respect to which $1$ and $4$ are symmetric and whose midpoint is real $> 4$ lies entirely to the right of the line $\operatorname{Re} z = \frac{5}{2}$, so contains no point with modulus $2$. Thus the midpoint $a$ must lie to the left of $1$. Of two such circles $C_{a,r} = \{z : \lvert z-a\rvert^2 = r^2\}$ with $a < 1$ such that $1$ and $4$ are symmetric with respect to $C_{a,r}$, one is entirely contained in the interior of the other. The two points $1$ and $4$ are symmetric with respect to $C_{0,2}$. Therefore $C = C_{0,2}$ is the only such circle containing any point with modulus $2$.