JEE Main 2020 · 6 September · morning · Single correct · +4 / −1

The region represented by
z = x + iy ∈ C : |z| – Re(z) ≤ 1 is also given by the
inequality : z = x + iy ∈ C : |z| – Re(z) ≤ 1

  1. (a) y² ≤ 2( x + 1/2 )
  2. (b) y² ≤ x + 1/2
  3. (c) y² ≥ 2(x + 1)
  4. (d) y² ≥ x + 1
Answer and solution

Answer (a) y² ≤ 2( x + 1/2 )

Solution

Maths sir · 4 steps

Move Re(z) across, then square both sides. The x² cancels and a parabola is left.

  1. Rewrite the condition. With z = x + iy, the condition is √(x² + y²) - x ≤ 1, that is √(x² + y²) ≤ 1 + x. The left side is never negative, so the right side 1 + x cannot be negative either.
  2. Square both sides. Both sides are non-negative, so squaring keeps the inequality the same way. This gives x² + y² ≤ 1 + 2x + x², and the x² cancels.

    y² ≤ 2x + 1 = 2(x + 1/2)

  3. Check the condition. Since y² ≥ 0, this already forces x ≥ -1/2. Then 1 + x ≥ 1/2 > 0, so the condition from the first step holds automatically and no extra restriction is needed.
  4. So. The region is y² ≤ 2(x + 1/2): the inside of a parabola with vertex at (-1/2, 0) that opens to the right. That is option (a).

The idea to keep. For |z| compared with something, isolate the modulus first, make sure both sides are non-negative, then square. Always check whether that sign condition adds any extra restriction.

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 2020.