Find the bilinear transformation which maps the points z1=2, z2=2i , and z3=−2 into the points ω1=1 , ω2=i, and ω3=−1 .

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\frac{z}{2}

\frac{z}{3}

\frac{z}{5}

answer is A.

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Detailed Solution

Let us assume

ω = f (z) = \frac{az + b}{cz + d}

………(i) be the required bilinear transformation, then
Putting z = 2 and ω=1 in above equation (1), we have

f (2) = 1

\Rightarrow \frac{a (2) + b}{c (2) + d} = 1

2 a + b = 2 c + d

……………..(ii)
Again, putting

z = 2 i and ω = i

in equation (i), we have

f (2 i) = i

\Rightarrow \frac{a (2 i) + b}{c (2 i) + d} = i 2 ia + b = i (2 ic + d)

\Rightarrow 2 ia + b = 2 i^{2} c + id

\Rightarrow (b - 2 i^{2} c) + i (2 a - d) = 0

\Rightarrow (b + 2 c) + i (2 a - d) = 0

Now equating real parts and imaginary parts of this equation with 0, we have
(b + 2c) = 0 …………(iii) and (2a - d) = 0 ……..(iv)
Again, putting z = -2 and ω=−1 in equation (i), we have
f(−2)=−1

\Rightarrow \frac{a (- 2) + b}{c (- 2) + d} = - 1

-2a+b=-1(-2c+d)

\Rightarrow b - 2 a = 2 c - d . . . . . . . . . (v)

From equation (iii), and (iv), we get
b= −2c and d= 2a……..(vi)
Now substituting d = 2a in equation (i), we get
b= 2c but b= −2c from (iii)
∴ b = −b
⇒b+b = 0
⇒2b = 0
⇒b = 0
∵ b = 2c and b = 0
⇒ c = 0
Now,