Find the bilinear transformation under which 1,i,-1 from the z- plane are mapped onto 0,1,∞ of w plane . Also shows that under this transformation the unit circle in w- plane is mapped on to a straight line in the z- plane. Write the name of this line .

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

Marks: 6M

Year: May 2014

PART I

Let the bilinear transformation be $w= \frac{az+b}{cz+d}$ ---- (1)

Let z= -1 & w= ∞ in (1)

$:. \frac{1}{w} = \frac{cz+d}{az+b}$

$0 = \frac{-c+d}{-a+b}$

:. c = d -----(2)

Let z=1 and w=0

$:. 0 = \frac{a+b}{c+d}$

:. a= -b -----(3)

$1 = \frac{ai+b}{ci+d}$

:. ci+d = ai + b

di+ d =-bi+b

d(1+i) = b(1-i)

$:. d= \frac{(1-i)}{(1+i)} b= -ib$ -----(4)

:. d= c

c= -ib ------(5)

Now replacing a,c &d in terms of –b in eq (1) we get

$w = \frac{-bz+b}{-ibz-ib}$

$w = \frac{-b(z-1)}{-ib(z+1)}$

$:. w={(z-1)}{i(z+1)}$

This is the required transformation

Part II

|w| = 1 : is the unit circle in w plane

$= \frac{z-1}{i(z+1)} = 1$

:. (z-1) = |i| (z+1)

:. |x+iy-1|= |i| (x+iy+1)

:. |(x-1)+iy| = |(x+1)i-y|

Taking magnitude we get

$\sqrt{(x - 1)^2 + y^2} = \sqrt{(x+1)^2 + (-y)^2}$

:. Squaring both the sides we get

$=(x - 1)^2 +y^2 = (x + 1)^2 + y^2$

$X^2 – 2x + 1 = x^2 + 2x +1$

:. 4x = 0

:. x= 0 is a straight line in z-plane

:. Line x=0 is the y-axis.