For two non-zero complex numbers and , if and , then which of the following are possible?
A. and
B. and
C. and
D. and
Choose the correct answer from the options given below:
A. and
B. and
C. and
D. and
Choose the correct answer from the options given below:
- A.A and C
- B.B and D
- C.B and C
- D.A and B
Show answer and solution
Correct option: C
Solution
Let z₁ = ai and z₂ = bi (since Re(z₁ + z₂) = 0 forces Re(z₁) = -Re(z₂), but Re(z₁ + z₂) = 0 doesn't force both real parts to be zero individually).
Let z₁ = p + ai and z₂ = -p + bi (so Re(z₁ + z₂) = 0).
Now Re(z₁z₂) = Re((p + ai)(-p + bi)) = Re(-p² + pbi - api + ai·bi) = Re(-p² + (pb - pa)i - ab) = -p² - ab = 0.
So p² + ab = 0, meaning ab = -p².
Since p² ≥ 0, we need ab ≤ 0. If p ≠ 0, then ab = -p² < 0, so a and b have opposite signs — one positive, one negative. This corresponds to option C (Im(z₁) > 0, Im(z₂) < 0) or its mirror, which is option B (Im(z₁) < 0, Im(z₂) > 0).
If p = 0, then ab = 0, but z₁ and z₂ are non-zero, so at least one of a, b is zero, making one of them zero — contradiction since both must be non-zero. Thus p ≠ 0 and ab < 0 strictly, confirming only B and C are possible (opposite signs for imaginary parts), ruling out A and D (same signs).
Let z₁ = p + ai and z₂ = -p + bi (so Re(z₁ + z₂) = 0).
Now Re(z₁z₂) = Re((p + ai)(-p + bi)) = Re(-p² + pbi - api + ai·bi) = Re(-p² + (pb - pa)i - ab) = -p² - ab = 0.
So p² + ab = 0, meaning ab = -p².
Since p² ≥ 0, we need ab ≤ 0. If p ≠ 0, then ab = -p² < 0, so a and b have opposite signs — one positive, one negative. This corresponds to option C (Im(z₁) > 0, Im(z₂) < 0) or its mirror, which is option B (Im(z₁) < 0, Im(z₂) > 0).
If p = 0, then ab = 0, but z₁ and z₂ are non-zero, so at least one of a, b is zero, making one of them zero — contradiction since both must be non-zero. Thus p ≠ 0 and ab < 0 strictly, confirming only B and C are possible (opposite signs for imaginary parts), ruling out A and D (same signs).
[JEE Main 2023 · 29 Jan · Shift 1]