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Complex Numbers question

2024 · 27 Jan · Shift 1 · Q32
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  5. /2024 · 27 Jan · Shift 1 · Q32

Complex Numbers question

2024 · 27 Jan · Shift 1 · Q32

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}S=\{z \in C:|z-i|=|z+i|=|z-1|\}S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}, then, n(S)n(S)n(S) is :
  1. A
    1
  2. B
    2
  3. C
    3
  4. D
    0
View written solutionFree

Correct answer: A

  1. We need the set S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}.S=\{z\in \mathbb{C}: |z-i|=|z+i|=|z-1|\}.S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}. Let z=x+iy,z=x+iy,z=x+iy, where x,y∈Rx,y\in \mathbb{R}x,y∈R.

  2. Interpret the moduli geometrically:

  • ∣z−i∣|z-i|∣z−i∣ is the distance of point (x,y)(x,y)(x,y) from (0,1)(0,1)(0,1)
  • ∣z+i∣|z+i|∣z+i∣ is the distance from (0,−1)(0,-1)(0,−1)
  • ∣z−1∣|z-1|∣z−1∣ is the distance from (1,0)(1,0)(1,0)

So we need points in the plane equidistant from all three points: (0,1), (0,−1), (1,0).(0,1),\ (0,-1),\ (1,0).(0,1), (0,−1), (1,0).

  1. First use ∣z−i∣=∣z+i∣.|z-i|=|z+i|.∣z−i∣=∣z+i∣. Squaring both sides: x2+(y−1)2=x2+(y+1)2.x^2+(y-1)^2=x^2+(y+1)^2.x2+(y−1)2=x2+(y+1)2. This gives (y−1)2=(y+1)2.(y-1)^2=(y+1)^2.(y−1)2=(y+1)2. Expanding: y2−2y+1=y2+2y+1y^2-2y+1=y^2+2y+1y2−2y+1=y2+2y+1 −2y=2y-2y=2y−2y=2y 4y=04y=04y=0 y=0.y=0.y=0.

Thus any such point must lie on the real axis.

  1. Now use ∣z−i∣=∣z−1∣|z-i|=|z-1|∣z−i∣=∣z−1∣ with y=0y=0y=0. Then ∣z−i∣2=x2+(0−1)2=x2+1,|z-i|^2=x^2+(0-1)^2=x^2+1,∣z−i∣2=x2+(0−1)2=x2+1, and ∣z−1∣2=(x−1)2+02=(x−1)2.|z-1|^2=(x-1)^2+0^2=(x-1)^2.∣z−1∣2=(x−1)2+02=(x−1)2. So x2+1=(x−1)2=x2−2x+1.x^2+1=(x-1)^2=x^2-2x+1.x2+1=(x−1)2=x2−2x+1. Hence x2+1=x2−2x+1x^2+1=x^2-2x+1x2+1=x2−2x+1 −2x=0-2x=0−2x=0 x=0.x=0.x=0.

Therefore z=0.z=0.z=0.

  1. Check: ∣0−i∣=1,∣0+i∣=1,∣0−1∣=1,|0-i|=1,\quad |0+i|=1,\quad |0-1|=1,∣0−i∣=1,∣0+i∣=1,∣0−1∣=1, so it satisfies all conditions.

Hence S={0}.S=\{0\}.S={0}. Therefore, n(S)=1.n(S)=1.n(S)=1.

  1. Comparing with the stored correct answer: Stored answer = A = 1, which matches our result.
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