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Complex Numbers question

2020 · 6 Sep · Shift 1 · Q22
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Complex Numbers question

2020 · 6 Sep · Shift 1 · Q22

JEE MainMathematicsComplex NumbersMCQ+4 / −1
The region represented by {z = x + iy ∈\in∈ C : |z| – Re(z) ≤\le≤ 1} is also given by the inequality : {z = x + iy ∈\in∈ C : |z| – Re(z) ≤\le≤ 1}
  1. A
    y2 ≤\le≤ 2(x+12)2\left( {x + {1 \over 2}} \right)2(x+21​)
  2. B
    y2 ≤x+12\le {x + {1 \over 2}}≤x+21​
  3. C
    y2 ≥\ge≥ 2(x + 1)
  4. D
    y2 ≥\ge≥ x + 1
View written solutionFree

Correct answer: A

  1. Let z=x+iyz=x+iyz=x+iy. Then ∣z∣=x2+y2,Re⁡(z)=x.|z|=\sqrt{x^2+y^2},\qquad \operatorname{Re}(z)=x.∣z∣=x2+y2​,Re(z)=x. The given inequality is ∣z∣−Re⁡(z)≤1|z| - \operatorname{Re}(z) \le 1∣z∣−Re(z)≤1 which becomes x2+y2−x≤1.\sqrt{x^2+y^2}-x\le 1.x2+y2​−x≤1.

  2. Rearrange: x2+y2≤x+1.\sqrt{x^2+y^2}\le x+1.x2+y2​≤x+1. Since the left side is non-negative, we must have x+1≥0.x+1\ge 0.x+1≥0. Now square both sides: x2+y2≤(x+1)2.x^2+y^2\le (x+1)^2.x2+y2≤(x+1)2.

  3. Expand the right-hand side: x2+y2≤x2+2x+1.x^2+y^2\le x^2+2x+1.x2+y2≤x2+2x+1. Cancelling x2x^2x2 from both sides, y2≤2x+1.y^2\le 2x+1.y2≤2x+1. Rewrite this as y2≤2(x+12).y^2\le 2\left(x+\frac12\right).y2≤2(x+21​).

  4. Compare with the options:

  • A: y2≤2(x+12)y^2\le 2\left(x+\frac12\right)y2≤2(x+21​) ✓
  • B: y2≤x+12y^2\le x+\frac12y2≤x+21​ ✗
  • C: y2≥2(x+1)y^2\ge 2(x+1)y2≥2(x+1) ✗
  • D: y2≥x+1y^2\ge x+1y2≥x+1 ✗

Hence the correct option is A.

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