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Find the Bi-linear Transformation which maps the points 1,i,-1 of z plane onto i,0,-i of w-plane
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  • Since it is given Bilinear transformation hence we have

$w=\dfrac{az+b}{cz+d}$............(i)

  • Given z=1,i,-1 maps onto w=i,0,-i plane.
  • Substituting the different values of z into eq (i) we get,

$i=\dfrac{a+b}{c+d}$,    $0=\dfrac{ai+b}{ci+d}$,    $-i=\dfrac{-a+b}{-c+d}$

  • From the first we get a+b=i(c+d)
  • From the last i(c+d)=-a+b
  • From the second we get ai+b=0
  • Substituting value of b,d, we get,

ai-a=ci+c       $\therefore c=a \dfrac{(i-1)}{(i+1)}$

$\therefore$ c = a.$\dfrac{(i-1).(i-1)}{(i+1).(i-1)}=a.\dfrac{(i^2-2i+1)}{i^2-1}=ai$

$\therefore $ c=d=ai

  $\therefore$ w = $\dfrac{az-a}{aiz+ai}=\dfrac{z-1}{i(z+1)}=-i.\dfrac{(z-1)}{(z+1)}$

 

  • The unit circle in w plane is given by |w|=1.Hence we need to show under this transformation it is mapped onto a straight line
  • Now when,$|w|=1$

$|-i.\dfrac{(z-1)}{(z+1)}|=1$

$\therefore$ $|z-1|=|z+1|$.........$(|i|=1)$

$\therefore$ $|(x-1)-iy|=|(x+1)+iy|$

$\therefore$ (x-1)2 + y2 = (x+1)2 + y2

     2x = -2x    $\therefore$ 4x = 0      $\therefore$ x = 0

  • Hence the map is the y-axis
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