Question

Find the billianear transformation which maps the points z=0,1,∞ on to the points w=0,i,2i.

Answer 1

$z_{1}=\infty, z_{2}=i, z_{3}=0 \& w_{1}=0, w_{2}=i, w_{3}=\infty$

Let the transformation be,

$\begin{gathered} \frac{\left(w-w_{1}\right)\left(w_{2}-w_{3}\right)}{\left(w-w_{3}\right)\left(w_{2}-w_{1}\right)}=\frac{\left(z-z_{1}\right)\left(z_{2}-z_{3}\right)}{\left(z-z_{3}\right)\left(z_{2}-z_{1}\right)} \\ \frac{\left(w-w_{1}\right) w_{3}\left(\frac{w_{2}}{w_{3}}-1\right)}{w_{3}\left(\frac{w}{w_{3}}-1\right)\left(w_{2}-w_{1}\right)}=\frac{z_{1}\left(\frac{z}{z_{1}}-1\right)\left(z_{2}-z_{3}\right)}{\left(z-z_{3}\right) z_{1}\left(\frac{z_{2}}{z_{1}}-1\right)} \\ \frac{\left(w-w_{1}\right)\left(\frac{w_{2}}{w_{3}}-1\right)}{\left(\frac{w}{w_{3}}-1\right)\left(w_{2}-w_{1}\right)}=\frac{\left(\frac{z}{z_{1}}-1\right)\left(z_{2}-z_{3}\right)}{\left(z-z_{3}\right)\left(\frac{z_{2}}{z_{1}}-1\right)} \\ \frac{(w-0)(0-1)}{(0-1)(i-0)}=\frac{(0-1)(i-0)}{(z-0)(0-1)} \\ \frac{w}{i}=\frac{i}{z} \\ w=i^{2} \\ w=\frac{-1}{z} \end{gathered}$