For two non-zero complex numbers z₁ and z₂, if Re(z₁ z₂)=0 and Re(z₁+z₂)=0, then which of the following are possible?
A. Im(z₁)>0 and Im(z₂) > 0
B. Im(z₁) < 0 and Im(z₂) > 0
C. Im(z₁) > 0 and Im(z₂) < 0
D. Im(z₁) < 0 and Im(z₂) < 0
Choose the correct answer from the options given below :

Solution

Maths sir · 4 steps

The two conditions force bd = -a² < 0, so the imaginary parts must have opposite signs.

1. Use the sum. Let z₁ = a + ib and z₂ = c + id. Then Re(z₁ + z₂) = a + c = 0, so c = -a.
2. Use the product. Re(z₁ z₂) = ac - bd. Put c = -a and set it equal to zero.

   -a² - bd = 0 ⇒ bd = -a²

3. Non-zero check. If a = 0, then c = 0 and bd = 0, so z₁ = ib or z₂ = id would be zero. Both numbers are non-zero, so a ≠ 0 and bd = -a² < 0.
4. So. Im(z₁) = b and Im(z₂) = d must have opposite signs, so only B and C are possible. For example, z₁ = 1 + i and z₂ = -1 - i give z₁ z₂ = -2i, which is case C. Swapping them gives case B. The answer is (d).

The idea to keep. Turn conditions on Re and Im into equations in a, b, c, d. When a product like bd equals minus a square, the signs are fixed. Always use the non-zero condition to rule out the edge case.

Answer from the NTA final answer key. Solution by PrepWiser.