Find the Bi linear Transformation which maps the points 1, i, -1 of z plane onto 0, 1, ∞ of w-plane.

$\text{Let the bilinear transformation be } \ w = \dfrac{az+b}{cz+d}...(A)$ $ \$\text{where a,b,c,d are complex constants and ad-bc} \ne 0) \ Put \: z=1 \: and \: w= 0 \:in \: (A)$ $\therefore, 0=\dfrac{a+b}{c+d} \implies a=-b...(B) $ $\text{Put z=i,w=1 in (A)}, \ 1 = \dfrac{ai+b}{ci+d} $ $\ \therefore, ai+b = ci +d ...(C) \ \text{Put z=-1 ,w=∞ in (A)} $ $\ \infty = \dfrac{-a+b}{-c+d} \ -c+d =0 $ $\ \therefore, c= d...(D) \ \text{From (B),(C)and (D),} $ $\ a(-1+i)=c(i+1) \ c=d=\dfrac{a(-1+i)}{1+i} $ $\ w = \dfrac{az+b}{cz+d} \ w = \dfrac{(az-a)}{a\dfrac{(-1+i)}{(1+i)}z} +a\dfrac{(-1+i)}{(1+i)}$ $ \ w= \dfrac{(z-1)(1+i)}{(z+1)(-1+i)}$ $ \ Now, \dfrac{1+i}{-1+i} = \dfrac{-(1+i)^2}{(-1+i)(-1-i)} = \dfrac{-1-2i-i^2}{1-i^2 } =\dfrac{-2i}{2} = -i$ $\bbox[3pt, yellow]{\therefore, w =-i \dfrac{z-1}{z+1}}$