JEE Mathematics Complex Numbers MCQs Set 27

Answer: (a) 2i

Explanation:
1. Let z_1 = 2r e^{iα} and z_2 = r e^{iβ}, with α + β = 3π/2.
2. z̄_1/z_2 = 2 e^{−i(α + β)} = 2 e^{−3πi/2} = 2i.

Teacher's Note:
1. Write z_1 and z_2 in polar form.
2. The argument of z̄_1/z_2 is −(α + β).

Answer: (b) 4

Explanation:
1. Consecutive points α_k, α_{k+1} on the unit circle subtend π/7 at the centre, so |α_{k+1} − α_k| = 2sin(π/14) for every k.
2. The numerator has 12 such terms and the denominator 3, each also a consecutive pair (4k − 1 and 4k − 2 differ by 1).
3. Ratio = 12/3 = 4.

Teacher's Note:
1. All the chords have the same length.
2. Count the terms.

Answer: (c) B, C and D only

Explanation:
1. |z_1 + z_2|² = |z_1|² + |z_2|² + z_1z̄_2 + z̄_1z_2, so the condition is z_1z̄_2 + z̄_1z_2 = 0 (C).
2. Dividing by |z_2|², Re(z_1/z_2) = 0, so z_1/z_2 is purely imaginary (B) and its argument is ±π/2 (D).
3. (A) is false: a real non-zero ratio would give Re ≠ 0.

Teacher's Note:
1. Expand |z_1 + z_2|².
2. The real part of z_1z̄_2 is the real part of z_1/z_2 times |z_2|².

Answer: (d) \( z_1+z_3=z_2+z_4 \)

Explanation:
1. In a parallelogram the diagonals bisect each other.
2. The midpoint of z_1z_3 equals the midpoint of z_2z_4, so z_1 + z_3 = z_2 + z_4.

Teacher's Note:
1. Use the midpoint of each diagonal.
2. Vertices are taken in order, so z_1z_3 and z_2z_4 are diagonals.

Answer: (a) 1

Explanation:
1. z is equidistant from 1, −1 and i, so it is the circumcentre of the triangle with these vertices.
2. The circumcentre is z = 0 (all three distances equal 1), and it is unique.

Teacher's Note:
1. The three points are not collinear.
2. A triangle has exactly one circumcentre.

Answer: (b) A, C and D only

Explanation:
1. The common modulus makes the circumcentre the origin, and for an equilateral triangle it is also the centroid, so ∑z = 0 (A).
2. z_2 = z_1ω and z_3 = z_1ω² (or swapped): z_2³ = z_3³ = z_1³ (D) and ∑z_iz_j = z_1²(ω + ω² + ω³) = 0 (C).
3. z_1z_2z_3 = z_1³, which equals 1 only if z_1³ = 1, so (B) is not always true.

Teacher's Note:
1. The centroid and circumcentre of an equilateral triangle coincide.
2. Write z_2 = ωz_1 and z_3 = ω²z_1.

Answer: (c) 3

Explanation:
1. Since a + b + c = 0, ∑cos(θ + α) = Re(e^{iθ}(a + b + c)) = 0.
2. With x + y + z = 0 we have x³ + y³ + z³ = 3xyz, so ∑cos³(θ + α) = 3∏cos(θ + α).
3. The ratio is 3.

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

Answer: (d) \( \sin^{-1}\left(\frac1{\sqrt3}\right) \)

Explanation:
1. Multiply by the conjugate: (2 + 3is)(1 + 2is) = (2 − 6s²) + i(7s), where s = sin θ.
2. Purely imaginary means 2 − 6s² = 0, i.e. sin²θ = 1/3.
3. θ = sin⁻¹(1/√3).

Teacher's Note:
1. Rationalise the denominator.
2. The real part of the numerator must vanish.

Answer: (a) \( z_2=-2,\ z_3=1-i\sqrt3 \)

Explanation:
1. z_1 = 2e^{iπ/3}. The other vertices are z_1 rotated by ±120°: 2e^{iπ} = −2 and 2e^{−iπ/3} = 1 − i√3.
2. So {z_2, z_3} = {−2, 1 − i√3}, and the stated option lists them in this order.

Teacher's Note:
1. Rotate z_1 by 2π/3 and by −2π/3 about the origin.
2. All three vertices must have modulus 2.

Answer: (b) z = 5 + 5i

Explanation:
1. arg z_1 = tan⁻¹(√3/3) = 30° and arg z_2 = tan⁻¹(6/(2√3)) = 60°.
2. The bisector of the angle between them is the ray at 45°, which contains 5 + 5i.
3. The third option has argument 30°, and −1 − i points the opposite way (225°).

Teacher's Note:
1. Find the argument of each vector.
2. The bisector makes the average angle.

Answer: (c) 0

Explanation:
1. If |z − 1| ≥ |z + 1| (Re z ≤ 0), the condition is |z| = |z − 1|, which gives Re z = 1/2, a contradiction.
2. If |z + 1| ≥ |z − 1| (Re z ≥ 0), the condition is |z| = |z + 1|, which gives Re z = −1/2, again a contradiction.
3. So no z satisfies it.

Teacher's Note:
1. Split into the cases according to which distance is larger.
2. Check each solution against the case assumption.

Answer: (d) \( \frac13\left[2^n+2\cos\frac{n\pi}{3}\right] \)

Explanation:
1. Adding (1 + 1)ⁿ, (1 + ω)ⁿ and (1 + ω²)ⁿ keeps exactly the terms with suffix divisible by 3 (each multiplied by 3).
2. 1 + ω = −ω² = e^{iπ/3} and 1 + ω² = e^{−iπ/3}, so (1 + ω)ⁿ + (1 + ω²)ⁿ = 2cos(nπ/3).
3. The sum is (1/3)[2ⁿ + 2cos(nπ/3)] (checked: n = 3 gives C_0 + C_3 = 2).

Teacher's Note:
1. Use the roots of unity filter.
2. 1 + ω = e^{iπ/3}.

Answer: (a) 1

Explanation:
1. ∑ sin(2πk/7) = 0 and ∑ cos(2πk/7) = −1 for k = 1, …, 6.
2. α = 0 − i(−1) = i, so α⁴ = i⁴ = 1.

Teacher's Note:
1. Sum of the non-trivial 7th roots of unity is −1.
2. Separate real and imaginary parts.

Answer: (b) \( \beta\in(1,\infty) \)

Explanation:
1. The roots are non-real (a real double root would not be distinct, and distinct real roots cannot both have real part 1) and form a conjugate pair 1 ± ik with k ≠ 0.
2. β = product of roots = 1 + k² > 1.

Teacher's Note:
1. Real coefficients give conjugate roots.
2. The product of conjugates is |z|².

Answer: (c) −256

Explanation:
1. The roots are −1 ± i = √2e^{±3πi/4}.
2. α^{15} + β^{15} = 2(√2)^{15}cos(45π/4) = 2·2^{7.5}·cos(5π/4).
3. cos(5π/4) = −√2/2, so the value is −2^{8} = −256.

Teacher's Note:
1. Write the roots in polar form.
2. Reduce 45π/4 modulo 2π.

Answer: (d) g(1) ≠ f(1)

Explanation:
1. h(ω) = ωf(1) + ω²g(1) = 0; real coefficients give also h(ω²) = ω²f(1) + ωg(1) = 0.
2. The determinant ω·ω − ω²·ω² = ω² − ω ≠ 0, so f(1) = g(1) = 0.
3. Then h(1) = 0. The statement g(1) ≠ f(1) is false since both are 0.

Teacher's Note:
1. Use the conjugate root ω².
2. Solve the two linear equations for f(1) and g(1).

Answer: (a) \( -\frac{5\pi}{12} \)

Explanation:
1. Numerator argument: 5(π/4) + 2(π/3) = 5π/4 + 2π/3 = 23π/12.
2. −2i(−√3 + i) = 2√3 i + 2 has argument π/3 (it equals 2 + 2√3 i).
3. Argument of Z = 23π/12 − π/3 = 19π/12 ≡ −5π/12 in (−π, π].

Teacher's Note:
1. Add the arguments of the numerator factors and subtract that of the denominator.
2. Reduce the angle to (−π, π].

Answer: (b) \( \frac34 \)

Explanation:
1. With z = x + iy, Im(z² + z + 1) = y(2x + 1) = 0, so x = −1/2 (y ≠ 0).
2. Then a = x² − y² + x + 1 = 3/4 − y² < 3/4.
3. So a can be any real number below 3/4, but not 3/4.

Teacher's Note:
1. Set the imaginary part of z² + z + 1 to zero.
2. y² must be strictly positive.

Answer: (c) \( \frac{3-\sqrt3}{2} \)

Explanation:
1. S is the sector of radius 4 between the ray at −60° and the positive imaginary axis; the point 1 − 3i lies just below the ray at −60°.
2. The distance from (1, −3) to the line √3x + y = 0 is |√3 − 3|/2 = (3 − √3)/2.
3. The foot of the perpendicular lies on the ray at distance about 3.1 from the origin, inside the radius 4, so this is the minimum.

Teacher's Note:
1. Find the nearest boundary of S.
2. Check that the foot of the perpendicular lies on the boundary segment.

Answer: (d) \( \frac3{11} \)

Explanation:
1. S1 and S2 touch internally, so |z_1 − z_2| = 7 − 4 = 3.
2. If S has centre C and radius r: |C − z_1| = 7 − r and |C − z_2| = 4 + r, so the sum is 11.
3. The locus is an ellipse with foci z_1, z_2: 2a = 11, 2c = 3, so e = 3/11.

Teacher's Note:
1. Write the tangency conditions in terms of the radius r.
2. Eliminate r by adding.

Answer: (a) i

Explanation:
1. zω̄ has modulus |z||ω| = 1 and argument arg z − arg ω = π/2.
2. So zω̄ = i.

Teacher's Note:
1. Take the modulus and argument of zω̄.
2. arg(ω̄) = −arg ω.

Answer: (b) \( \pi \)

Explanation:
1. sin x sin(60° − x) sin(60° + x) = (1/4) sin 3x, so Z = −(1/4) sin 3x.
2. For x in (0, π/3), 3x lies in (0, π) and sin 3x > 0.
3. Z is a negative real number, so its principal argument is π.

Teacher's Note:
1. Use the triple-angle product identity.
2. Note the order sin(x − 60°) = −sin(60° − x).

Answer: (c) \( 3iz-3i\bar z+2=0 \)

Explanation:
1. The circles are (x − 2)² + y² = 9 and (x − 2)² + (y − 3)² = 16. Subtracting: 6y − 9 = −7, so y = 1/3.
2. With z − z̄ = 2iy, 3i(z − z̄) = −6y, so y = 1/3 reads −6y + 2 = 0, i.e. 3iz − 3i z̄ + 2 = 0.

Teacher's Note:
1. Subtract the two equations in x and y.
2. Convert y using z − z̄ = 2iy.

Answer: (d) 5

Explanation:
1. |z_1| = 1 and |2z_2| = 4.
2. The maximum of |z_1 − 2z_2| is 1 + 4 = 5, attained when z_1 and 2z_2 point in opposite directions.

Teacher's Note:
1. Use |a − b| ≤ |a| + |b|.
2. Equality needs opposite directions.

Answer: (a) \( \left(-\frac{21}{10},-\frac56\right) \)

Explanation:
1. The circle has centre (0, a) and radius a + 4 (so a > −4); the disc has centre (2, 0) and radius 1.
2. They have a common point when |d − (a + 4)| < 1 with d = √(a² + 4).
3. Solving, d < a + 5 gives a > −21/10 and d > a + 3 gives a < −5/6 (checked numerically).

Teacher's Note:
1. The circle must meet the open disc: compare centre distance with the radii.
2. Square the inequalities carefully.