JEE Mathematics Complex Numbers MCQs Set 28

Answer: (b) P-1, Q-2, R-3, S-4

Explanation:
1. P: z_j = z_{10−k} gives z_kz_j = 1, so True.
2. Q: z_1·z = z_k always has the solution z = z_k/z_1, so the statement is False.
3. R: ∏|1 − z_k| = 10 (the product of 1 − z_k over the non-trivial 10th roots), so the value is 1. S: ∑cos(2kπ/10) = −1, so the value is 2.

Teacher's Note:
1. Use xⁿ − 1 = (x − 1)∏(x − z_k) at x = 1.
2. The sum of all non-trivial nth roots of unity is −1.

Answer: (c) \( -\frac12 \)

Explanation:
1. α is a primitive 11th root of unity, so α + α² + … + α^{10} = −1.
2. α^{11−k} is the conjugate of α^k, so the sum of the first five terms has real part equal to half of −1.
3. Real part = −1/2.

Teacher's Note:
1. Pair α^k with α^{11 − k}.
2. The sum of all non-trivial 11th roots is −1.

Answer: (d) Re(z) > 3

Explanation:
1. z is closer to 4 than to 2, so it lies to the right of the perpendicular bisector x = 3.
2. Hence Re(z) > 3.

Teacher's Note:
1. Interpret |z − a| as a distance.
2. Find the perpendicular bisector of 2 and 4.

Answer: (a) A, C and D only

Explanation:
1. z − z_1 = t(z_2 − z_1) and z − z_2 = (1 − t)(z_1 − z_2): z lies on the segment between z_1 and z_2.
2. (A) holds because z is between them, (C) is the collinearity condition, and (D) holds since t > 0.
3. (B) is false: z − z_1 and z − z_2 point in opposite directions, so their arguments differ by π.

Teacher's Note:
1. Write z − z_1 and z − z_2 in terms of z_2 − z_1.
2. A point of the segment is collinear with both ends.

Answer: (b) 2

Explanation:
1. The expression is 3|z − 1| + 2|z − 2|, a weighted sum of distances from 1 and 2.
2. On the real segment between 1 and 2 it equals 3t + 2(1 − t) = 2 + t for z = 1 + t, which is smallest at z = 1.
3. The minimum is 2 (attained at z = 1).

Teacher's Note:
1. Factor out the coefficients of z.
2. The weighted sum is smallest at the point with the larger weight.

Answer: (c) 5

Explanation:
1. |2z − 6 + 5i| = 2|z − 3 + 5i/2|, twice the distance from z to the point 3 − 5i/2.
2. The centre 3 + 2i is at distance 9/2 from that point, and the disc has radius 2, so the least distance is 9/2 − 2 = 5/2.
3. Minimum = 2 × 5/2 = 5.

Teacher's Note:
1. Factor out 2.
2. The nearest point of a disc is along the line to its centre.

Answer: (d) B and D only

Explanation:
1. z_2 = (1 − i)z_1/2, so z_1z_2 = (1 − i)z_1²/2 with argument −π/4 + 2arg z_1 = π/2 (mod 2π), i.e. arg z_1 = 3π/8 (mod π).
2. arg z_2 = arg z_1 − π/4 = π/8 (mod π).
3. The principal values in (−π, π] are π/8 and π/8 − π = −7π/8.

Teacher's Note:
1. Express z_2 in terms of z_1.
2. Arguments are determined only up to a multiple of π here because z_1 appears squared.

Answer: (a) \( p^2+q^2+r^2=2(pq+qr+rp) \)

Explanation:
1. q = ω²p and r = ωp.
2. p + q + r = p(1 + ω² + ω) = 0, so (A) fails (a + b + c ≠ 0 in general).
3. p² + q² + r² = p²(1 + ω⁴ + ω²) = 0 and pq + qr + rp = p²(ω² + 1 + ω) = 0, so p² + q² + r² = 2(pq + qr + rp).
4. (B) would need a² + b² + c² = 0, which is false for real a, b, c ≠ 0.

Teacher's Note:
1. Express q and r as multiples of p.
2. Use 1 + ω + ω² = 0.

Answer: (b) C and D only

Explanation:
1. (x − 1)³ = −8, so α − 1, β − 1, γ − 1 are −2, −2ω, −2ω² in some order.
2. If they are in the order 1 : ω : ω², each ratio is ω² and the sum is 3ω²; in the other cyclic order each ratio is ω and the sum is 3ω.

Teacher's Note:
1. Substitute y = x − 1.
2. Ratios of cube roots of −8 are powers of ω.

Answer: (c) circle of radius 2

Explanation:
1. |z_1 − 2z_2| = |2 − z_1z̄_2| gives |z_1|² + 4|z_2|² = 4 + |z_1|²|z_2|² (cross terms cancel).
2. So (|z_1|² − 4)(1 − |z_2|²) = 0, and since |z_2| ≠ 1, |z_1| = 2.

Teacher's Note:
1. Square both sides and expand.
2. Factorise the result.

Answer: (d) 1

Explanation:
1. The first term a and the ratio r are both (√3 + i)/2, whose modulus is √(3 + 1)/2 = 1.
2. |a_n| = |a||r|^{n−1} = 1.

Teacher's Note:
1. Find |a| and |r| separately.
2. The modulus of a product is the product of the moduli.

Answer: (a) \( x^2+y^2 \)

Explanation:
1. Take the modulus squared of both sides: ∏(4 + k²) = |x + iy|².
2. So 5·8·13···(4 + n²) = x² + y².

Teacher's Note:
1. Each factor satisfies |2 + ki|² = 4 + k².
2. The modulus of a product is the product of the moduli.

Answer: (b) A, B, C and D (all of them)

Explanation:
1. (z⁵ − 32)(z⁵ + 31) = 0, so z⁵ = 32 or z⁵ = −31.
2. z⁵ = 32: arguments 0°, 72°, 144°, 216°, 288°. z⁵ = −31: arguments 36°, 108°, 180°, 252°, 324°.
3. Real part positive: 0, 72, 288, 36, 324 (5); negative: 144, 216, 108, 180, 252 (5). Imaginary part ≥ 0: 0, 72, 144, 36, 108, 180 (6); negative: 216, 288, 252, 324 (4).

Teacher's Note:
1. Factorise as a quadratic in z⁵.
2. List the arguments of the roots.

Answer: (c) \( e^{-\pi/4+2n\pi}\cos\left(\frac12\log_e2\right) \)

Explanation:
1. (1 − i)^{−i} = exp(−i log_e(1 − i)) and log_e(1 − i) = (1/2)log_e 2 + i(2nπ − π/4).
2. −i times this is −(i/2)log_e 2 + (2nπ − π/4), so the number equals e^{2nπ − π/4}[cos((1/2)log_e 2) − i sin((1/2)log_e 2)].
3. Real part: e^{−π/4 + 2nπ}cos((1/2)log_e 2).

Teacher's Note:
1. Use a^b = e^{b log a}.
2. Separate the real part of the exponent.

Answer: (d) 7

Explanation:
1. Divide numerator and denominator by z: the ratio is (u + 1)/(u − 1) with u = z + 1/z.
2. The ratio is real exactly when u is real. For non-real z, z + 1/z is real only if |z| = 1.
3. So 7|z| = 7.

Teacher's Note:
1. Write the ratio in terms of z + 1/z.
2. A non-real z has z + 1/z real only when it lies on the unit circle.

Answer: (a) {1}

Explanation:
1. The sum of the roots is −i(a − 1). For a conjugate pair the sum is real, so a − 1 = 0.
2. For a = 1 the equation is x² + 5 = 0 with roots ±i√5, a conjugate imaginary pair.

Teacher's Note:
1. The sum of a pair of conjugates is real.
2. Check that the roots for the value found really are non-real.

Answer: (b) 7

Explanation:
1. With circumradius R, A_1A_k = 2R sin((k − 1)π/n).
2. The condition becomes 1/sin(π/n) = 1/sin(2π/n) + 1/sin(3π/n); it holds for n = 7 (the heptagon identity).
3. For n = 5, 6, 9 it fails (checked numerically).

Teacher's Note:
1. Write each side as 2R sin(kπ/n).
2. Test the values in the options.

Answer: (c) \( -\frac\pi2 \)

Explanation:
1. 1/|z − 1| is largest when |z − 1| is smallest. The point 1 lies inside the circle |z − (2 − i)| = √5, so the nearest point of S to 1 is on the circle, on the line from the centre 2 − i through 1: z_0 = (2 − i) + √5(−1 + i)/√2.
2. 4 − z_0 − z̄_0 = 4 − 2Re z_0 = √10 (real, positive); z_0 − z̄_0 + 2i = 2i(Im z_0 + 1) = i√10.
3. The quotient is √10/(i√10) = −i, with principal argument −π/2.

Teacher's Note:
1. Find the point of S closest to 1.
2. z_0 − z̄_0 = 2i Im z_0.

Answer: (d) \( \theta \)

Explanation:
1. z = e^{iθ}: (1 + e^{iθ})/(1 + e^{−iθ}) = e^{iθ}(1 + e^{−iθ})/(1 + e^{−iθ}) = e^{iθ}.
2. The argument is θ.

Teacher's Note:
1. Factor e^{iθ} out of the numerator.
2. 1 + z̄ = 1 + 1/z.

Answer: (a) 8

Explanation:
1. With z = e^{iθ}: z/z̄ + z̄/z = 2cos 2θ, so |2cos 2θ| = 1.
2. cos 2θ = ±1/2. For 2θ in [0, 4π) each sign gives 4 values of 2θ.
3. That is 8 values of θ in [0, 2π).

Teacher's Note:
1. Write z = e^{iθ}.
2. Count solutions of cos 2θ = ±1/2 in [0, 4π).

Answer: (b) the imaginary axis

Explanation:
1. Square both sides: |z²|² − 2Re(z²) + 1 = |z|⁴ + 2|z|² + 1.
2. This gives −2(x² − y²) = 2(x² + y²), so x = 0.
3. z lies on the imaginary axis.

Teacher's Note:
1. Use |z² − 1|² = |z|⁴ − 2Re(z²) + 1.
2. Replace Re(z²) = x² − y².

Answer: (c) \( \frac{4m\pi}{n(n+1)} \)

Explanation:
1. The product equals cos(θ + 2θ + … + nθ) + i sin(…), and θ(1 + 2 + … + n) = n(n + 1)θ/2.
2. This must equal 2mπ, so θ = 4mπ/(n(n + 1)).

Teacher's Note:
1. Add the arguments of the factors.
2. The total argument must be a multiple of 2π.

Answer: (d) A, B and D only

Explanation:
1. (A) is false: −1 − i lies in the third quadrant, so its principal argument is −3π/4.
2. (B) is false: as t → 0⁺, arg(−1 + it) → π, but for t < 0 it is close to −π, so f is discontinuous at t = 0.
3. (C) is true: the difference of arguments is a multiple of 2π. (D) is false in general: the cross-ratio condition describes an arc of a circle through z_1, z_2, z_3 (a line only if the points are collinear).

Teacher's Note:
1. Check the quadrant of −1 − i.
2. The principal argument jumps by 2π across the negative real axis.

Answer: (a) A, B, C and D (all of them)

Explanation:
1. x = e^{±iθ} and y = e^{±iφ}.
2. With x = e^{iθ}, y = e^{iφ} we get xⁿ + 1/xⁿ = 2cos nθ, xy + 1/(xy) = 2cos(θ + φ) and x^m y^n + 1/(x^m y^n) = 2cos(mθ + nφ); with x = e^{iθ}, y = e^{−iφ} we get x/y + y/x = 2cos(θ − φ).
3. So each statement is true for a suitable choice of the roots.

Teacher's Note:
1. Solve x + 1/x = 2cos θ.
2. Each choice of sign gives one of the formulas.

Answer: (b) \( a^3+b^3+c^3-3abc \)

Explanation:
1. With {α, β} = {ω, ω²}, y = c + aω + bω² and z = c + aω² + bω.
2. (c + a + b)(c + aω + bω²)(c + aω² + bω) = a³ + b³ + c³ − 3abc, the standard factorisation.

Teacher's Note:
1. Reorder the terms so that the factorisation of a³ + b³ + c³ − 3abc applies.
2. The three factors use 1, ω, ω² cyclically.