Question

If z₁ = a + ib and z₂ = c + id are complex numbers such that |z₁| = |z₂| = 1 and Re (z₁${\bar{z}}_{2}$)=0, then the pair of complex numbers a + ic = w₁ and b + id = w₂ satisfies –

• A. |w₁| ¹ 1
• B. |w₂| ¹ 1
• C. Re(w₁${\bar{w}}_{2}$) = 0
• D. None

Answer 1

Answer: (C)

Re(w₁${\bar{w}}_{2}$) = 0

Answer 2

Sol. z₁ = a + ib, z₂ = c + id

|z₁| = |z₂| = 1 Ž a² + b² = c² + d² = 1 w₁ = a + ic, w₂ = b + id

z₁${\bar{z}}_{2}$ = (a + ib) (c – id) = (ac + bd) + i (bc – ad) as Re(z₁${\bar{z}}_{2}$) = 0 Ž ac + bd = 0 Ž ac = – bd

w₁${\bar{w}}_{2}$ = (a + ic) (b – id) = (ab + cd) + i (bc – ad) We have a² + b² = c² + d² Ž a² – c² = d² – b² Ž a² – c² + 2i ac = d² – b² – 2i bd (as ac = – bd)

Ž (a + ic)² = (d – ib)² Ž a + ic = (d – ib) or –d + ib Ž a = d and c = –b or a = –d, b = c Ž c² + d² = b² + d², a² + c² = a² + b² Ž a² + c² = 1, b² + d² = 1 Ž |w₁| = |w₂| = 1 also ab + cd = – cd + cd = 0 Ž Re (w₁${\bar{w}}_{2}$) = 0

Hence (3) is correct answer.