Complex Numbers and Quadratic Equations — JEE Main Questions

34 JEE Main practice questions on Complex Numbers and Quadratic Equations, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. If (1 + i√3)^n = 64, where n is a positive integer, find n.

  1. 4
  2. 8
  3. 6
  4. 12

Answer: 6

Express 1+i√3 in polar form: r = √(1+3)=2, θ = tan⁻¹(√3/1)=π/3. By De Moivre's theorem, (1+i√3)^n = 2^n (cos(nπ/3)+i sin(nπ/3)). For this to equal 64 (real positive), we need sin(nπ/3)=0 and 2^n cos(nπ/3)=64. sin(nπ/3)=0 implies nπ/3 = kπ, so n=3k. Then cos(nπ/3)=cos(kπ)=(-1)^k. For n=6 (k=2), 2^6=64 and cos(2π)=1, giving 64. Hence n=6.

2. If arg((z - 1)/(z + 1)) = π/2, then the locus of z is:

  1. the circle |z| = 1
  2. the lower semicircle |z| = 1
  3. the upper semicircle |z| = 1
  4. the line x = 0

Answer: the upper semicircle |z| = 1

arg((z-1)/(z+1)) = π/2 means the angle between vectors from z to 1 and z to -1 is 90°. By Thales theorem, the locus is the circle with diameter from -1 to 1, i.e., |z|=1. Since argument is π/2 (positive), the point lies on the upper semicircle (Im(z)>0).

3. For n ≥ 2, what is the sum of all n-th roots of unity?

  1. -1
  2. 1
  3. n
  4. 0

Answer: 0

The n-th roots of unity are 1, ω, ω²,..., ωⁿ⁻¹ where ω = e^(2πi/n). Their sum is a geometric series: (ωⁿ - 1)/(ω - 1) = (1 - 1)/(ω - 1) = 0. Alternatively, from Vieta's formulas on zⁿ - 1 = 0, the coefficient of zⁿ⁻¹ is zero, so the sum of roots is zero.

4. If |z - 1| = 2|z + 1|, then the locus of z is a circle with centre at:

  1. (-5/3, 0)
  2. (5/3, 0)
  3. (-1/3, 0)
  4. (1/3, 0)

Answer: (-5/3, 0)

Square both sides: |z-1|² = 4|z+1|². Write z = x+iy: (x-1)²+y² = 4[(x+1)²+y²]. Expand: x²-2x+1+y² = 4x²+8x+4+4y². Rearranging: 3x²+10x+3+3y²=0 → x²+y²+(10/3)x+1=0. Complete square: (x+5/3)²+y² = (5/3)²-1 = 25/9-1=16/9. Centre is (-5/3,0).

5. If z is a non-zero complex number satisfying Re(1/z) = 1/2, what is the locus of z?

  1. Circle with centre (0,1) and radius 1, excluding origin
  2. Circle with centre (1,0) and radius 1, excluding origin
  3. Circle with centre (1,0) and radius 2, excluding origin
  4. Circle with centre (0,0) and radius 1, excluding origin

Answer: Circle with centre (1,0) and radius 1, excluding origin

Write z = a + ib. Then 1/z = (a - ib)/(a² + b²). Its real part is a/(a² + b²). Set equal to 1/2: 2a = a² + b². Rearranging gives (a - 1)² + b² = 1, which is a circle with centre (1,0) and radius 1. Since z ≠ 0, the origin is excluded.

6. What is the locus of z if |z - 1| = |z + i|?

  1. x - y = 0
  2. x + y = 1
  3. x - y = 1
  4. x + y = 0

Answer: x + y = 0

The locus is the perpendicular bisector of the segment joining (1,0) and (0,-1). Squaring both sides: |z-1|² = |z+i|². Let z=x+iy. Then (x-1)²+y² = x²+(y+1)². Expand: x²-2x+1+y² = x²+y²+2y+1. Cancel: -2x = 2y => y = -x => x+y=0.

7. If z₁ = 0 and z₂ = 2 are two vertices of an equilateral triangle, find the third vertex z₃.

  1. 1 + i√3 or 1 - i√3
  2. 1 - i√3
  3. 1 + i√3
  4. 2 + i√3

Answer: 1 + i√3 or 1 - i√3

Using rotation, z₃ = z₁ + (z₂ - z₁)e^(±iπ/3) = 2(cos60° ± i sin60°) = 1 ± i√3. The equilateral condition also gives z₃² - 2z₃ + 4 = 0, solving yields z₃ = 1 ± i√3. So both possible third vertices are 1 + i√3 and 1 - i√3.

8. If |z| = 1, then |(z-1)/(z+1)| equals

  1. |cos θ|
  2. |cot(θ/2)| where θ = arg(z)
  3. |sin θ|
  4. |tan(θ/2)| where θ = arg(z)

Answer: |tan(θ/2)| where θ = arg(z)

Let z = cosθ + i sinθ (since |z|=1). Then z-1 = -2 sin²(θ/2) + i 2 sin(θ/2) cos(θ/2) = 2i sin(θ/2) e^{iθ/2}. Similarly, z+1 = 2 cos(θ/2) e^{iθ/2}. Hence (z-1)/(z+1) = i tan(θ/2). Modulus is |tan(θ/2)|.

9. The number of distinct real quadratic factors in the factorisation of x^7 - 1 over reals is

  1. 3
  2. 2
  3. 4
  4. 5

Answer: 3

x^7 - 1 = (x-1) ∏_{k=1}^{3} (x² - 2 cos(2kπ/7) x + 1). The roots are 1 and 6 non-real roots forming 3 conjugate pairs, each giving a real quadratic factor. So there are 3 real quadratic factors.

10. If z₁ = 1 + i and z₂ = 2 - i, what is the conjugate of z₁ z₂?

  1. 1 - 3i
  2. 3 + i
  3. 3 - i
  4. 1 + 3i

Answer: 3 - i

Using the property (z₁ z₂)̄ = z̄₁ z̄₂, we compute z̄₁ = 1 - i and z̄₂ = 2 + i. Their product is (1 - i)(2 + i) = 2 + i - 2i - i² = 2 - i + 1 = 3 - i. Hence the conjugate of z₁ z₂ is 3 - i.

11. If ω is a complex cube root of unity, what is the area of the triangle formed by the points 1, ω, and ω^2 in the Argand plane?

  1. √3/2
  2. √3/4
  3. 3/4
  4. 3√3/4

Answer: 3√3/4

The points 1, ω, ω^2 are vertices of an equilateral triangle inscribed in the unit circle. Side length = √3 (distance between 1 and ω). Area = (√3/4) * (side)^2 = (√3/4)*3 = 3√3/4.

12. The polar form of the complex number -1 + i is:

  1. √2(cos(3π/4) + i sin(3π/4))
  2. √2(cos(-3π/4) + i sin(-3π/4))
  3. √2(cos(π/4) + i sin(π/4))
  4. √2(cos(5π/4) + i sin(5π/4))

Answer: √2(cos(3π/4) + i sin(3π/4))

For z = -1 + i, r = √((-1)² + 1²) = √2. Since a<0, b>0 (Quadrant II), Arg(z) = π - α, where α = tan⁻¹(1/1) = π/4, so Arg(z) = 3π/4. Thus polar form is √2(cos(3π/4) + i sin(3π/4)).

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