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Complex Numbers question

2002 · Shift 0 · Q98
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Complex Numbers question

2002 · Shift 0 · Q98

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If ∣z−4∣<∣z−2∣\left| {z - 4} \right| \lt \left| {z - 2} \right|∣z−4∣<∣z−2∣, its solution is given by :
  1. A
    Reolimits(z)>0{\mathop{\rm Re} olimits} (z) \gt 0Reolimits(z)>0
  2. B
    Reolimits(z)<0{\mathop{\rm Re} olimits} (z) \lt 0Reolimits(z)<0
  3. C
    Reolimits(z)>3{\mathop{\rm Re} olimits} (z) \gt 3Reolimits(z)>3
  4. D
    Reolimits(z)>2{\mathop{\rm Re} olimits} (z) \gt 2Reolimits(z)>2
View written solutionFree

Correct answer: C

  1. Let z=x+iyz=x+iyz=x+iy, where x=Re(z)x={\rm Re}(z)x=Re(z) and y=Im(z)y={\rm Im}(z)y=Im(z).

  2. Then ∣z−4∣=∣(x−4)+iy∣=(x−4)2+y2|z-4|=|(x-4)+iy|=\sqrt{(x-4)^2+y^2}∣z−4∣=∣(x−4)+iy∣=(x−4)2+y2​ and ∣z−2∣=∣(x−2)+iy∣=(x−2)2+y2.|z-2|=|(x-2)+iy|=\sqrt{(x-2)^2+y^2}.∣z−2∣=∣(x−2)+iy∣=(x−2)2+y2​.

  3. Given ∣z−4∣<∣z−2∣,|z-4|<|z-2|,∣z−4∣<∣z−2∣, squaring both sides (valid since both sides are non-negative), we get (x−4)2+y2<(x−2)2+y2.(x-4)^2+y^2<(x-2)^2+y^2.(x−4)2+y2<(x−2)2+y2.

  4. Cancel y2y^2y2 from both sides: (x−4)2<(x−2)2.(x-4)^2<(x-2)^2.(x−4)2<(x−2)2.

  5. Expand both sides: x2−8x+16<x2−4x+4.x^2-8x+16<x^2-4x+4.x2−8x+16<x2−4x+4. Cancel x2x^2x2: −8x+16<−4x+4.-8x+16<-4x+4.−8x+16<−4x+4.

  6. Rearranging, 12<4x12<4x12<4x so x>3.x>3.x>3.

  7. Since x=Re(z)x={\rm Re}(z)x=Re(z), the solution is Re⁡(z)>3.\operatorname{Re}(z)>3.Re(z)>3.

  8. Checking options:

  • A: Re(z)>0{\rm Re}(z)>0Re(z)>0 — too broad, not exact
  • B: Re(z)<0{\rm Re}(z)<0Re(z)<0 — incorrect
  • C: Re(z)>3{\rm Re}(z)>3Re(z)>3 — correct
  • D: Re(z)>2{\rm Re}(z)>2Re(z)>2 — too broad, not exact

Therefore, the correct option is C.

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