JEE Main 2023 · 10 April · evening · Single correct · +4 / −1

Let S = { z = x + iy:(2z - 3i)/(4z + 2i) is a real number }. Then which of the following is NOT correct?

  1. (a) y + x² + y² ≠ - 1/4
  2. (b) (x,y) = ( 0, - 1/2 )
  3. (c) x = 0
  4. (d) y ∈ ( - ∞ , - 1/2 ) ∪ ( - 1/2,∞ )
Answer and solution

Answer (b) (x,y) = ( 0, - 1/2 )

Solution

Maths sir · 3 steps

Make the imaginary part zero, then throw out the one point where the denominator vanishes.

  1. Split into parts. Put z = x + iy. The numerator is 2x + i(2y-3) and the denominator is 4x + i(4y+2). Multiply both by the conjugate 4x - i(4y+2), so the denominator becomes real and positive.

    Im(numerator) = 4x(2y-3) - 2x(4y+2) = -16x

  2. Read off S. The fraction is real only when -16x = 0, that is x = 0. The denominator must also be non-zero, so (x, y) = (0, -1/2) is removed. So S is the imaginary axis without the point -i/2.
    OptionCheck with x = 0, y ≠ -1/2Correct?
    (a) y + x² + y² ≠ -1/4Becomes (y + 1/2)² ≠ 0, which holds since y ≠ -1/2Yes
    (b) (x, y) = (0, -1/2)This point makes the denominator 0, so it is not in SNo
    (c) x = 0Exactly the condition we foundYes
    (d) y ≠ -1/2Every other y is allowedYes
  3. So. The statement that is NOT correct is (b): the point (0, -1/2) makes the expression undefined, so it does not belong to S.

The idea to keep. For a fraction to be real, set its imaginary part to zero. Then always remove the points where the denominator becomes zero, because the expression is undefined there.

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

Study Complex Numbers

Asked about 160 times in JEE Main since 2020. This question is from 2023.