If , where and , satisfies the condition that is purely real, then the set of values of is
- A.
- B.and
- C.
- D.None of these
Show answer and solution
Correct option: B
Solution
Let w = α + iβ. Then w̄ = α - iβ.
Compute w - w̄z = (α + iβ) - (α - iβ)z = α(1-z) + iβ(1+z).
So the expression becomes [α(1-z) + iβ(1+z)] / (1-z) = α + iβ(1+z)/(1-z).
For this to be purely real, the imaginary part must be zero: β · Re[(1+z)/(1-z)] ... actually we need Im[β(1+z)/(1-z)] = 0, meaning (1+z)/(1-z) must be real (since β ≠ 0).
Let z = x + iy. Then (1+z)/(1-z) = (1+x+iy)/((1-x)-iy). Multiply numerator and denominator by conjugate of denominator: the imaginary part of the result is [(1+x)(-y) - (iy)(1-x... let me compute directly.
Im[(1+z)/(1-z)] = Im[(1+z)(1-z̄)/|1-z|²] = Im[(1+z)(1-z̄)] / |1-z|².
(1+z)(1-z̄) = 1 - z̄ + z - |z|² = 1 - |z|² + i·2y (since z - z̄ = 2iy... wait: z - z̄ = 2iy, and -z̄ + z = 2iy).
So Im
Compute w - w̄z = (α + iβ) - (α - iβ)z = α(1-z) + iβ(1+z).
So the expression becomes [α(1-z) + iβ(1+z)] / (1-z) = α + iβ(1+z)/(1-z).
For this to be purely real, the imaginary part must be zero: β · Re[(1+z)/(1-z)] ... actually we need Im[β(1+z)/(1-z)] = 0, meaning (1+z)/(1-z) must be real (since β ≠ 0).
Let z = x + iy. Then (1+z)/(1-z) = (1+x+iy)/((1-x)-iy). Multiply numerator and denominator by conjugate of denominator: the imaginary part of the result is [(1+x)(-y) - (iy)(1-x... let me compute directly.
Im[(1+z)/(1-z)] = Im[(1+z)(1-z̄)/|1-z|²] = Im[(1+z)(1-z̄)] / |1-z|².
(1+z)(1-z̄) = 1 - z̄ + z - |z|² = 1 - |z|² + i·2y (since z - z̄ = 2iy... wait: z - z̄ = 2iy, and -z̄ + z = 2iy).
So Im
[AIEEE 2006]