Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 6M

Year: Dec 2015

Let bilinear transformation be $w=\frac{az+b}{cz+d} →(1)$ (where a, b, c, d are complex constants & ad – bc ≠ 0)

Put z=1 and w=I in (1)

$∴ i = \frac {a(1)+b}{c(1)+d} \\ ∴ ic+ id = a + b → (2)$

Put z=I & w=0 in (1)

$∴ 0 = {a(i)+b}{c(i)+d}\\ ∴ 0 = ai+b\\ ∴ b = -ai → (3)$

Put z=-1 & w=-I in (1)

$∴ -i = \frac{a(-1)+b}{c(-1)+d}\\ ∴ ic - id =-a + b → (4)$

Adding (2) & (4), we get, 2 ic =2b

$∴ c = \frac{b}{i} = \frac{-ai}{i} = -a → (5)$

Substitute (3) , (5) & (6) in (1)

$ ∴ w = \frac{a2 - ai}{-a2 - ai}\\ ∴ w = \frac{-a(-2+i)}{-a(2+i)}\\ ∴ w = \frac{i-2}{z+i} \text{is the bilinear transformation.}$