$\displaystyle Let\ the\ bilinear\ transformation\ be\ \ \frac{az+b}{cz+d}\ \ \rightarrow{}\left(i\right) \\[2ex] \displaystyle (\ where\ a,b,c,d\ are\ complex\ constants\ and\ ad-bc\not=0) \\[2ex] $

$ \displaystyle Put\ z=2\ and\ w=1\ in\ (i) \\[2ex] \displaystyle \therefore{}1=\frac{a\left(2\right)+b}{c\left(2\right)+d} \\[2ex] \displaystyle \therefore{}2c+d=2a+b\ \ \ \rightarrow{}\left(ii\right) \\[2ex] \displaystyle Put\ z=i\ and\ w=i\ in\ (i) \\[2ex] \displaystyle \therefore{}i=\frac{a\left(i\right)+b}{c\left(i\right)+d} \\[2ex] \displaystyle \therefore{}ci^2+di=ai+b \\[2ex] \displaystyle \therefore{}-c+di=ai+b\ \ \ \rightarrow{}\left(iii\right) \\[2ex] \displaystyle Put\ z=-2\ and\ w=-1\ in\ (i) \\[2ex] \displaystyle \therefore{}-1=\frac{a\left(-2\right)+b}{c\left(-2\right)+d} \\[2ex] \displaystyle \therefore{}2c-d=-2a+b\ \ \ \rightarrow{}\left(iv\right) \\[2ex] \displaystyle Now\ adding\ (ii)\ and\ (iv) \\[2ex] \displaystyle \therefore{}4c=2b \\[2ex] \displaystyle \therefore{}b=2c\ \ \ \ \rightarrow{}\left(v\right) \\[2ex] \displaystyle Substituting\ (v)\ in\ (ii) \\[2ex] \displaystyle \therefore{}2c+d=2a+2c\ \ \ \\[2ex] \displaystyle \therefore{}d=2a\ \ \ \rightarrow{}\left(vi\right) \\[2ex] \displaystyle Substitute\ (v)\ and\ (vi)\ in\ (iii) \\[2ex] \displaystyle \therefore{}-c+2ai=ai+2c\ \ \ \\[2ex] \displaystyle \therefore{}ai=3c\ \ \ \\[2ex] \displaystyle \therefore{}a=\frac{3c}{i}=-3ci\ \ \ \rightarrow{}\left(vii\right) \\[2ex] \displaystyle From\ (vi)\ and\ (vii) \\[2ex] \displaystyle \therefore{}d=-6ci\ \ \ \ \rightarrow{}\left(viii\right) \\[2ex] \displaystyle Substitute\ (v),\ (vii)\ and\ (viii)\ in\ (i) \\[2ex] \displaystyle \therefore{}w=\frac{-3ci\bullet{}z+2c}{cz-6ci}=\frac{c\left(-3iz+2\right)}{c\left(z-6i\right)} \\[2ex] \displaystyle \therefore{}w=\frac{-3i\bullet{}z+2}{z-6i}\\[2ex] is\ the\ required\ bilinear\ transformation\ \\[2ex]$