Bilinear transformation which maps $z=(\infty, i, 0)$ and $w= (-1, -i, 1)$

Question

I have three equations after simplifying this a bit

$a+c=0$

$ai+b-c=0$

$b-d=0$

How do I proceed further?

If you care to know this is from the chapter Complex Variables

Answer 1

Hint: Use $w(z)=k.\frac{z+a}{z+b}$ to get three equations in three unknowns.

Answer 2

We want $w(z)=\frac{az+b}{cz+d}$, such as $w(\infty)=-1,w(0)=1,w(i)=-i$.

This lead to the system : $$\begin{align}\begin{cases} \frac{a}{c}=-1\\ \frac{b}{d}=1 \\ \frac{ai+b}{ci+d}=-i \end{cases} &\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{ai+b}{-ai+b}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{(ai+b)^2}{b^2+a^2}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ \frac{b^2-a^2+i2ab}{b^2+a^2}=-i \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2-a^2+i2ab=-ib^2-ia^2 \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2-a^2=0 \\2ab=-b^2-a^2 \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2=a^2 \\2ab=-2a^2 \end{cases} \\&\Rightarrow \begin{cases} c=-a\\ b=d \\ b^2=a^2 \\ab=-a^2 \end{cases} \end{align}$$

Consider the two cases $a=b$ and $a=-b$.

If $a=b$ :

$$\begin{cases} c=-a\\ b=d \\ a=b \\a^2=-a^2 \end{cases} \Rightarrow a=b=c=d=0$$

But such a transformation is not defined, so this case is not possible.

If $a=-b$ :

$$\begin{cases} c=-a\\ b=d \\ a=-b \\-a^2=-a^2 \end{cases} \Rightarrow -a=b=c=d$$

So $w(z)$ is of the form $\frac{-az+a}{az+a}=\frac{-z+1}{z+1}$ (if $a \neq 0$).

Let's verify that such an $w$ is solution. $$w(\infty)=-1,w(0)=1,w(i)=\frac{-i+1}{i+1}=-i$$

So $w(z)=\frac{-z+1}{z+1}$ is a solution.