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Complex Numbers question

2004 · Shift 0 · Q99
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Complex Numbers question

2004 · Shift 0 · Q99

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If  ∣z2−1∣=∣z∣2+1\,\left| {{z^2} - 1} \right| = {\left| z \right|^2} + 1​z2−1​=∣z∣2+1, then z lies on :
  1. A
    an ellipse
  2. B
    the imaginary axis
  3. C
    a circle
  4. D
    the real axis
View written solutionFree

Correct answer: B

  1. Let z=x+iyz=x+iyz=x+iy so that ∣z∣2=x2+y2.|z|^2=x^2+y^2.∣z∣2=x2+y2.

  2. Compute z2−1z^2-1z2−1: z2=(x+iy)2=x2−y2+2ixy,z^2=(x+iy)^2=x^2-y^2+2ixy,z2=(x+iy)2=x2−y2+2ixy, so z2−1=(x2−y2−1)+2ixy.z^2-1=(x^2-y^2-1)+2ixy.z2−1=(x2−y2−1)+2ixy. Hence, ∣z2−1∣2=(x2−y2−1)2+(2xy)2.|z^2-1|^2=(x^2-y^2-1)^2+(2xy)^2.∣z2−1∣2=(x2−y2−1)2+(2xy)2.

  3. Given ∣z2−1∣=∣z∣2+1=x2+y2+1.|z^2-1|=|z|^2+1=x^2+y^2+1.∣z2−1∣=∣z∣2+1=x2+y2+1. Since both sides are nonnegative, square both sides: ∣z2−1∣2=(x2+y2+1)2.|z^2-1|^2=(x^2+y^2+1)^2.∣z2−1∣2=(x2+y2+1)2. Thus, (x2−y2−1)2+4x2y2=(x2+y2+1)2.(x^2-y^2-1)^2+4x^2y^2=(x^2+y^2+1)^2.(x2−y2−1)2+4x2y2=(x2+y2+1)2.

  4. Simplify the left side: Using (a−b)2+4x2y2=(x2−y2−1)2+4x2y2,(a-b)^2+4x^2y^2=(x^2-y^2-1)^2+4x^2y^2,(a−b)2+4x2y2=(x2−y2−1)2+4x2y2, expand: (x2−y2−1)2+4x2y2=x4+y4+1−2x2y2−2x2+2y2+4x2y2(x^2-y^2-1)^2+4x^2y^2=x^4+y^4+1-2x^2y^2-2x^2+2y^2+4x^2y^2(x2−y2−1)2+4x2y2=x4+y4+1−2x2y2−2x2+2y2+4x2y2 =x4+y4+1+2x2y2−2x2+2y2=x^4+y^4+1+2x^2y^2-2x^2+2y^2=x4+y4+1+2x2y2−2x2+2y2 =(x2+y2)2+1−2x2+2y2.=(x^2+y^2)^2+1-2x^2+2y^2.=(x2+y2)2+1−2x2+2y2.

The right side is (x2+y2+1)2=(x2+y2)2+1+2x2+2y2.(x^2+y^2+1)^2=(x^2+y^2)^2+1+2x^2+2y^2.(x2+y2+1)2=(x2+y2)2+1+2x2+2y2.

  1. Equate both sides: (x2+y2)2+1−2x2+2y2=(x2+y2)2+1+2x2+2y2.(x^2+y^2)^2+1-2x^2+2y^2=(x^2+y^2)^2+1+2x^2+2y^2.(x2+y2)2+1−2x2+2y2=(x2+y2)2+1+2x2+2y2. Cancel common terms: −2x2=2x2-2x^2=2x^2−2x2=2x2 4x2=04x^2=04x2=0 x=0.x=0.x=0.

  2. Therefore, z=iy,z=iy,z=iy, which means zzz lies on the imaginary axis.

  3. Check: If z=iyz=iyz=iy, then z2=−y2,z^2=-y^2,z2=−y2, so ∣z2−1∣=∣−y2−1∣=y2+1,|z^2-1|=|-y^2-1|=y^2+1,∣z2−1∣=∣−y2−1∣=y2+1, and ∣z∣2+1=y2+1.|z|^2+1=y^2+1.∣z∣2+1=y2+1. So every point on the imaginary axis satisfies the condition.

Therefore, the locus is the imaginary axis.

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