JEE Mathematics Complex Numbers MCQs Set 30

Answer: (d) lies in the interval (1, 2)

Explanation:
1. |z + 1/2| ≥ |z| − 1/2 ≥ 2 − 1/2 = 3/2, with equality at z = −2.
2. So the minimum is 3/2, which lies in (1, 2).

Teacher's Note:
1. Use |a + b| ≥ |a| − |b|.
2. Check that equality is attained.

Answer: (a) C and D only

Explanation:
1. w = e^{iπ/6}, so P consists of the 12 points at angles 0°, 30°, …, 330° on the unit circle.
2. H1 ∩ P: angles −30°, 0°, 30°. H2 ∩ P: angles 150°, 180°, 210°.
3. The angles between a point of the first set and a point of the second are 120°, 150°, 180°, i.e. 2π/3, 5π/6 and π; so (C) and (D) are possible.

Teacher's Note:
1. List the points of P by their angles.
2. Re z > 1/2 means the angle is within 60° of the positive real axis.

Answer: (b) \( \frac{\pi^2}{3} \)

Explanation:
1. (k − nω)(k − nω²) = k² + nk + n², so each term is n/(k² + kn + n²) = (1/n)/((k/n)² + k/n + 1).
2. L = ∫₀¹ dx/(x² + x + 1) = (2/√3)[tan⁻¹((2x + 1)/√3)]₀¹ = (2/√3)(π/3 − π/6) = π/(3√3).
3. 9L² = 9π²/27 = π²/3.

Teacher's Note:
1. Use (k − nω)(k − nω²) = k² + kn + n² (ω + ω² = −1).
2. Recognise a Riemann sum.

Answer: (c) 1

Explanation:
1. Add all rows to the first row: every entry becomes z (using 1 + ω + ω² = 0), then the determinant is z times a determinant that reduces to z².
2. The determinant equals z³ (checked with sympy), so z = 0 is the only root (a triple root).

Teacher's Note:
1. Use R1 → R1 + R2 + R3.
2. The determinant reduces to a power of z.

Answer: (d) B and D only

Explanation:
1. (1 + i)² = 2i, so (1 + i)^{2n}/2ⁿ = iⁿ and 2ⁿ/(1 + i)^{2n} = i^{−n}.
2. (1 − i)² = −2i, so (1 − i)^{2n}/2ⁿ = (−i)ⁿ = i^{−n} and 2ⁿ/(1 − i)^{2n} = iⁿ.
3. (B) = iⁿ + i^{−n} and (D) = i^{−n} + iⁿ, while (A) and (C) both give 2iⁿ.

Teacher's Note:
1. Evaluate (1 ± i)² first.
2. Compare each option with iⁿ and i^{−n}.

Answer: (a) (1, 1)

Explanation:
1. 1 + ω = −ω², so (1 + ω)⁷ = −ω¹⁴ = −ω² (as ω¹⁴ = ω²).
2. −ω² = 1 + ω, so A = 1 and B = 1.

Teacher's Note:
1. Use 1 + ω = −ω².
2. Reduce the power of ω modulo 3.

Answer: (b) 3364

Explanation:
1. Real part: 3p − 7q = 0, so p = 7k and q = 3k for an integer k ≠ 0.
2. |z|² = |3 + 7i|²(p² + q²) = 58 × 58k².
3. Least value at k = ±1: 3364.

Teacher's Note:
1. Set the real part of the product to zero.
2. Minimise over the integer multiple k.

Answer: (c) \( x^2-102x+101=0 \)

Explanation:
1. α = ω, so b = ∑ 1 = 101.
2. ∑_{k=0}^{100} ω^{2k}: the first 99 terms give 0 (33 complete cycles), leaving k = 99, 100: 1 + ω² . So a = (1 + ω)(1 + ω²) = 1.
3. The roots 1 and 101 give x² − 102x + 101 = 0.

Teacher's Note:
1. α³ = 1.
2. Group the terms of the sum in threes.

Answer: (d) \( 12\pi \)

Explanation:
1. z³ = e^{±2πi/3}, so the roots have arguments (in degrees) 40°, 80°, 160°, 200°, 280°, 320°.
2. In (180°, 540°] these become 400°, 440°, 520°, 200°, 280°, 320°.
3. Sum = 2160° = 12π.

Teacher's Note:
1. Solve z³ = ω and z³ = ω² first.
2. Add 360° to those arguments that are at most 180°.

Answer: (a) 1

Explanation:
1. α, β = e^{±iπ/3}. 101 ≡ 5 and 107 ≡ 5 (mod 6).
2. α^{101} + β^{107} = e^{5iπ/3} + e^{−5iπ/3} = 2cos(5π/3) = 1.

Teacher's Note:
1. The roots have order 6.
2. Reduce exponents modulo 6.

Answer: (b) the exterior of the circle with centre 1 + 0i and radius 10

Explanation:
1. Let t = |z − 1|. For base 1/2 the inequality means 0 < (t + 4)/(3t − 2) < 1/2.
2. Positivity needs 3t − 2 > 0 and then 2(t + 4) < 3t − 2, i.e. t > 10.
3. So |z − 1| > 10: the exterior of that circle.

Teacher's Note:
1. A logarithm to a base less than 1 reverses the inequality.
2. Check that the argument is positive.

Answer: (c) 4

Explanation:
1. |2z_1| = 2 and |3z_2| = 6.
2. By the triangle inequality, |2z_1 + 3z_2| ≥ 6 − 2 = 4, attained when the two vectors point in opposite directions.

Teacher's Note:
1. Find the lengths of 2z_1 and 3z_2.
2. The minimum occurs for opposite directions.

Answer: (d) 0

Explanation:
1. ω = (z − 1)(z̄ + 1)/|z + 1|² = (|z|² − 1 + z − z̄)/|z + 1|².
2. With |z| = 1 this is (z − z̄)/|z + 1|² = 2i Im(z)/|z + 1|², purely imaginary.
3. So Re(ω) = 0.

Teacher's Note:
1. Multiply numerator and denominator by the conjugate of z + 1.
2. Use z z̄ = |z|² = 1.

Answer: (a) A, B and D only

Explanation:
1. z = w makes both sides 0, so (A) is possible.
2. If z w̄ = 1 then z = 1/w̄, |z|² = 1/|w|², and the left side equals w/|w|² − w while the right side is 1/w̄ − w = w/|w|² − w: equal, so (B) holds. (D) is the conjugate form of the same relation.
3. With |z| = |w| = 2 the condition gives 4w − 4z = z − w, i.e. z = w, contradicting z ≠ w: (C) is impossible.

Teacher's Note:
1. Test each option by substitution.
2. z w̄ = 1 and z̄ w = 1 are conjugates of each other.

Answer: (b) an arc of a circle

Explanation:
1. arg((z − z_1)/(z − z_2)) = θ with 0 < θ < π is the locus of points seeing the segment z_1z_2 at a fixed angle.
2. That locus is an arc of a circle through 2 and 4.

Teacher's Note:
1. The argument of the quotient is the angle subtended at z by the segment.
2. Recall the angle in a segment theorem.

Answer: (c) B and C only

Explanation:
1. For z = kω with k ≥ 0: z² = k²ω², z|z| = k²ω, |z²| = k², and k²(ω² + ω + 1) = 0.
2. For k < 0, |z| = −k and z|z| = −k²ω, which fails; so k must be non-negative (and positive to be non-zero).
3. The same holds for kω². So (B) and (C) are correct; (A) and (D) allow negative k.

Teacher's Note:
1. Check the sign of k carefully since |z| = |k|.
2. Use 1 + ω + ω² = 0.

Answer: (d) A, B and C only

Explanation:
1. The roots are 2, i, −i and a fourth real root r with product of roots 3/2 (constant 3 over leading 2): 2·1·r = 3/2, so r = 3/4.
2. P(z) = 2(z − 2)(z² + 1)(z − 3/4) = 2z⁴ − (11/2)z³ + 5z² − (11/2)z + 3.
3. So a = −11/2, b = 5, c = −11/2; a = −11 is false.

Teacher's Note:
1. Real coefficients force −i to be a root.
2. Use the product of the roots to find the fourth root.

Answer: (a) 0

Explanation:
1. Row 2 equals ω^n times row 1: (ω^n, ω^{2n}, ω^{3n} = 1).
2. Two proportional rows make the determinant 0.

Teacher's Note:
1. Compare the first two rows.
2. Remember ω³ = 1.

Answer: (b) 4

Explanation:
1. Re(z²) = x² − y² = 0 gives y = ±x.
2. x² + y² = 4 gives x = ±√2 for each sign of y.
3. The solutions are (±√2, ±√2): 4 solutions.

Teacher's Note:
1. Write z = x + iy.
2. Solve x² = y² together with x² + y² = 4.

Answer: (c) 0

Explanation:
1. (1 + i tan α)/(1 − i tan α) = (cos α + i sin α)/(cos α − i sin α) = e^{2iα}.
2. So the first term is e^{2inα}, which equals the second term (1 + i tan nα)/(1 − i tan nα).
3. The difference is 0.

Teacher's Note:
1. Write the tangent in terms of sine and cosine.
2. Apply De Moivre's theorem.

Answer: (d) \( \frac{20\pi}{3} \)

Explanation:
1. 1 − √3 i = 2e^{−iπ/3}, so the Im condition says arg(z − (1 − √3 i)) lies in (−π/3, 2π/3). The point 1 − √3 i lies on the line through the origin at angle −60°, so S2 is the half-plane above the line y = −√3 x.
2. S3 is the right half-plane, S1 the disc of radius 4.
3. S is a sector of radius 4 from angle −60° to 90°, i.e. 5π/6: area = (1/2)(16)(5π/6) = 20π/3.

Teacher's Note:
1. Rewrite the imaginary-part condition as an argument condition.
2. The three conditions bound a sector of the disc.

Answer: (a) 13

Explanation:
1. x = 9^{(1/3)/(1 − 1/3)} = 9^{1/2} = 3 and y = 4^{(1/3)/(1 + 1/3)} = 4^{1/4} = √2.
2. z = (1/(1 + i))/(1 − 1/(1 + i)) = 1/i = −i.
3. P = 3 − i√2, whose principal argument is −tan⁻¹(√2/3), so a = 2 and b = 3.
4. a² + b² = 13.

Teacher's Note:
1. Sum each geometric progression in the exponent.
2. Write P in the form 3 − i√2.

Answer: (b) \( t(1-i),\ t\in\mathbb R \)

Explanation:
1. Put z = x + iy: (1 + i)(x + iy) = (x − y) + i(x + y) and (1 − i)(x − iy) = (x − y) − i(x + y).
2. Equating imaginary parts: x + y = 0, so y = −x.
3. z = x(1 − i) with x real.

Teacher's Note:
1. Expand both sides and compare real and imaginary parts.
2. Check the answer by substituting z = t(1 − i).

Answer: (c) 0

Explanation:
1. Let a = e^{iα}, b = e^{iβ}, c = e^{iγ}. Then |a + b + c|² = 3 + 2∑cos(α − β) = 0, so a + b + c = 0.
2. Also ab + bc + ca = abc(1/a + 1/b + 1/c) = abc·conj(a + b + c) = 0, so ∑[cos(α + β) + i sin(α + β)] = 0.
3. Hence both sums are 0.

Teacher's Note:
1. Use the unit complex numbers e^{iα}, e^{iβ}, e^{iγ}.
2. Their sum is zero.

Answer: (d) \( \frac{9\pi}{10} \)

Explanation:
1. With θ = 6π/5: sin θ = 2 sin(θ/2) cos(θ/2) and 1 + cos θ = 2cos²(θ/2).
2. Z = 2cos(3π/5)[sin(3π/5) + i cos(3π/5)], and sin a + i cos a = e^{i(π/2 − a)}, which has argument π/2 − 3π/5 = −π/10.
3. Since cos(3π/5) < 0, the real factor is negative, which adds π: argument = −π/10 + π = 9π/10.

Teacher's Note:
1. Use half-angle formulae.
2. Take care of the sign of cos(3π/5).