Find a Mobius Transformation that maps the points $\{0,1,\infty\}$ to $\{-1,-i,1\}$

Question

Subbing in $0$ for $z$, I got that $b = -d$, where the Mobius transformation coefficients are $a,b,c,d$.

Subbing in $\infty$ for $z$, I found that $a = c$.

I cannot find out how to solve for $a$ and $b$ though. I know the mobius transformation is now $\frac{az+b}{az-b}$, and the inverse transformation is $\frac{-bz-b}{-az+a}$ but can't seem to solve it.

Answer 1

What you want is $\frac{a+b}{a-b}=i$, as I’m sure you saw. Your next step should have been to recognize that the left-hand side is homogeneous in the unknowns, so you might as well assume that $a=1$, giving you \begin{align} \frac{1+b}{1-b}&=-i\\ 1+b&=(-1+b)i=-i+bi\\ 1+i&=(-1+i)b\\ b&=\frac{1+i}{-1+i}=\frac12(1+i)(-1-i)=-i\\ T(z)&=\frac{z-i}{z+i} \end{align}

Answer 2

$T(z)=\frac {az+b}{cz+d}$. $T(\infty)=\frac ac=1$. Put $c=1$. Now $T(1)=\frac{1+b}{1-b}=-i \implies 1+b=-i+bi \implies b=\frac{i+1}{i-1}=-i$.

Thus $T(z)=\frac{z-i}{z+i}$.