JEE Mathematics Complex Numbers MCQs Set 31

Answer: (a) \( \bar a\bar z+az=0 \)

Explanation:
1. With z = x + iy the line is 2(px + qy) + b = 0, with intercepts −b/(2p) and −b/(2q).
2. The midpoint is (−b/(4p), −b/(4q)), which lies on the line px − qy = 0 as b varies.
3. px − qy = Re(az) = (az + āz̄)/2, so the locus is āz̄ + az = 0.

Teacher's Note:
1. Find the intercepts in terms of p and q.
2. Express px − qy as the real part of az.

Answer: (b) [1, 9]

Explanation:
1. The region is the filled ellipse with foci 1 and −3, and 2a = 8: centre −1, semi-major axis 4, so it meets the real axis at −5 and 3.
2. The point 4 lies on the major axis. The nearest point of the region is 3 (distance 1) and the farthest is −5 (distance 9).
3. |z − 4| ranges over [1, 9].

Teacher's Note:
1. Identify the ellipse and its vertices.
2. The extreme distances from a point on the major axis occur at the vertices.

Answer: (c) \( \frac i2z+\frac12+i \)

Explanation:
1. Let the remainder be az + b. Then f(i) = ai + b = i and f(−i) = −ai + b = 1 + i.
2. Adding: 2b = 1 + 2i, so b = 1/2 + i. Subtracting: 2ai = −1, so a = i/2.
3. Remainder: (i/2)z + 1/2 + i.

Teacher's Note:
1. The remainder has degree at most 1.
2. Use f(i) and f(−i).

Answer: (d) \( x=-2,\ y=\pm2 \)

Explanation:
1. z̄_2 = 8y² + 20i, which must equal z_1 = 9y² − 4 − 10ix.
2. Real parts: 9y² − 4 = 8y², so y² = 4 and y = ±2. Imaginary parts: −10x = 20, so x = −2.

Teacher's Note:
1. Write the conjugate of z_2.
2. Equate real and imaginary parts.

Answer: (a) \( x^2+x+2=0 \)

Explanation:
1. α + β = a + a² + … + a⁶ = −1.
2. αβ = (a + a² + a⁴)(a³ + a⁵ + a⁶) = a⁴ + a⁶ + a⁷ + a⁵ + a⁷ + a⁸ + a⁷ + a⁹ + a¹⁰ = 3 + (a + a² + a³ + a⁴ + a⁵ + a⁶) = 3 − 1 = 2.
3. The equation is x² + x + 2 = 0.

Teacher's Note:
1. Find the sum and the product of the roots.
2. a⁷ = 1.

Answer: (b) 7

Explanation:
1. The disc has centre −3 and radius 3; the distance from 1 to the centre is 4.
2. The greatest distance from 1 to a point of the disc is 4 + 3 = 7.

Teacher's Note:
1. The farthest point lies on the line through the centre.
2. Add the radius to the centre distance.

Answer: (c) 20

Explanation:
1. OPQR is a square with side |z| (OQ is OP rotated through 90°).
2. Triangle PQR is half of the square: |z|²/2 = 200, so |z| = 20.

Teacher's Note:
1. Notice that iz is z turned through a right angle.
2. The diagonal PQ splits the square into two equal triangles.

Answer: (d) −m

Explanation:
1. The roots are the non-trivial (2m + 1)th roots of unity. Since |Z_r| = 1, 1/(Z_r − 1) + 1/(Z̄_r − 1) = −1 (the real part of 1/(Z − 1) is −1/2).
2. The roots form m conjugate pairs, so the sum is −m.

Teacher's Note:
1. Use the pairing of conjugate roots.
2. For |Z| = 1, Re(1/(Z − 1)) = −1/2.

Answer: (a) \( 2+i\text{ and }1-3i \)

Explanation:
1. Rewrite as z² − (3 − 2i)z + (5 − 5i) = 0. Discriminant = (3 − 2i)² − 4(5 − 5i) = 5 − 12i − 20 + 20i = −15 + 8i = (1 + 4i)².
2. z = [(3 − 2i) ± (1 + 4i)]/2 = 2 + i or 1 − 3i.

Teacher's Note:
1. Compute the discriminant carefully.
2. Find the square root of −15 + 8i.

Answer: (b) \( \pm(4+3i) \)

Explanation:
1. (4 + 3i)² = 16 − 9 + 24i = 7 + 24i.
2. So the square roots are ±(4 + 3i).

Teacher's Note:
1. Let (a + ib)² = 7 + 24i: a² − b² = 7 and ab = 12.
2. Try the integer pairs with ab = 12.

Answer: (c) \( \frac{25}{4} \)

Explanation:
1. |z_2 + iz_1| = |iz_1| + |z_2| means z_2 and iz_1 have the same direction, so z_2 = (4/3)i z_1. Take z_1 = 3: A = 3, B = 4i, C = (4i − 3i)/(1 − i) = i/(1 − i) = (−1 + i)/2.
2. Area = (1/2)|(B − A) × (C − A)| = (1/2)|(−3)(1/2) − (4)(−7/2)| = (1/2)(12.5) = 25/4.

Teacher's Note:
1. Equal modulus in the triangle inequality means the same direction.
2. Use coordinates once z_1 is placed on the real axis.

Answer: (d) 0

Explanation:
1. x = 2e^{iπ/3} and y = 2e^{−iπ/3}, so x^p + y^p = 2^{p+1}cos(pπ/3).
2. For p = 5: 2⁶cos(5π/3) = 64 × 1/2 = 32 = 2⁵ = z^p.
3. So x^p + y^p − z^p = 0 (the same holds for every prime p > 3, since cos(pπ/3) = 1/2).

Teacher's Note:
1. Write x and y in polar form.
2. cos(5π/3) = 1/2.

Answer: (a) 1

Explanation:
1. 1/2 + i√3/2 = e^{iπ/3}, so the values are exp[(3/4)(π/3 + 2kπ)i] = e^{iπ/4}·i^{3k}, k = 0, 1, 2, 3.
2. These are four distinct numbers e^{iπ/4}, e^{i7π/4}, e^{i5π/4}, e^{i3π/4}, whose product is e^{i(π/4 + 7π/4 + 5π/4 + 3π/4)} = e^{4πi} = 1.

Teacher's Note:
1. Write the base as e^{iπ/3} and use all four values of the fractional power.
2. Add the arguments.

Answer: (b) 3

Explanation:
1. 1 + ω⁴ = 1 + ω = −ω² and 1 + ω² = −ω.
2. (−ω²)^n = (−ω)^n gives ω^{2n} = ω^n, i.e. ω^n = 1, so n is a multiple of 3.
3. The least positive n is 3.

Teacher's Note:
1. Reduce ω⁴ to ω.
2. Use 1 + ω = −ω².

Answer: (c) 8

Explanation:
1. Multiply by the conjugate: z = (1 + 2is)(1 + is)/(1 + s²) = (1 − 2s² + 3is)/(1 + s²), with s = sin θ.
2. Purely imaginary requires 1 − 2s² = 0, so sin θ = ±1/√2.
3. In [0, 4π] each sign gives 4 values of θ: 8 in all.

Teacher's Note:
1. Rationalise the denominator.
2. Count the solutions of sin θ = ±1/√2 in two full turns.

Answer: (d) \( \text{centre }(-4,-3),\ \text{radius }2\sqrt5 \)

Explanation:
1. Write it as |z + (4 + 3i)|² = |4 + 3i|² − 5, because |z + c|² = z z̄ + c̄ z + c z̄ + |c|² with c = 4 + 3i.
2. So |z + 4 + 3i|² = 25 − 5 = 20: centre −4 − 3i and radius √20 = 2√5.

Teacher's Note:
1. Complete the square in z and z̄.
2. The coefficient of z̄ is the number c.

Answer: (a) 5

Explanation:
1. Arg = ±π/2 means the angle at z subtended by the segment joining 8i and −6 is a right angle, so z lies on the circle with that segment as a diameter.
2. The diameter ends are (0, 8) and (−6, 0): centre (−3, 4), radius half of √(36 + 64) = 5.
3. The circle is |z + 3 − 4i| = 5.

Teacher's Note:
1. The angle in a semicircle is a right angle.
2. Find the midpoint and half-length of the segment.

Answer: (b) 96

Explanation:
1. z̄_1 = 4/z_1, z̄_2 = 9/z_2 and z̄_3 = 16/z_3.
2. The expression is |z_1z_2z_3||8/z_1 + 27/z_2 + 64/z_3| = 24|2z̄_1 + 3z̄_2 + 4z̄_3|.
3. The modulus of the conjugate sum equals |2z_1 + 3z_2 + 4z_3| = 4, so the value is 24 × 4 = 96.

Teacher's Note:
1. Use z z̄ = |z|².
2. Factor out z_1z_2z_3.

Answer: (c) \( \frac{9\sqrt2\pi}{2} \)

Explanation:
1. The chord z_1z_2 has length 6 and subtends π/4 at points of the locus, so the circle has radius 6/(2 sin π/4) = 3√2.
2. Points seeing the chord at the smaller angle π/4 lie on the major arc, whose central angle is 2π − 2(π/4) = 3π/2.
3. Arc length = 3√2 × 3π/2 = 9√2π/2.

Teacher's Note:
1. Use the inscribed angle theorem to get the radius.
2. The locus is the major arc.

Answer: (d) \( -77-108i \)

Explanation:
1. (2 + 3i)² = −5 + 12i, (2 + 3i)³ = −46 + 9i, (2 + 3i)⁴ = −119 − 120i.
2. x⁴ − x³ + x² + 3x − 5 = (−119 − 120i) − (−46 + 9i) + (−5 + 12i) + (6 + 9i) − 5.
3. Real: −119 + 46 − 5 + 6 − 5 = −77. Imaginary: −120 − 9 + 12 + 9 = −108.

Teacher's Note:
1. Compute the powers of 2 + 3i step by step.
2. Add the real and imaginary parts separately.

Answer: (a) 1

Explanation:
1. Put z = iy (y real, non-zero): y⁴ − a_2y² + a_4 + i(−a_1y³ + a_3y) = 0.
2. Imaginary part: y² = a_3/a_1. Real part: y⁴ − a_2y² + a_4 = 0, so a_3²/a_1² − a_2a_3/a_1 + a_4 = 0, i.e. a_3² + a_1²a_4 = a_1a_2a_3.
3. The required sum is (a_3² + a_1²a_4)/(a_1a_2a_3) = 1.

Teacher's Note:
1. Separate the real and imaginary parts of the equation at z = iy.
2. Eliminate y² between the two equations.

Answer: (b) \( 10\pi \)

Explanation:
1. z = (1 − 2s² + 3is)/(1 + s²); it is real when sin θ = 0.
2. θ = 0, π, 2π, 3π, 4π, with sum 10π.

Teacher's Note:
1. Set the imaginary part 3 sin θ to zero.
2. List all the multiples of π in [0, 4π].

Answer: (c) \( 7A_0+7A_7x^7+7A_{14}x^{14} \)

Explanation:
1. The sum equals ∑_k A_k x^k ∑_{j=0}^{6} α^{jk}.
2. ∑_{j=0}^{6} α^{jk} = 7 if 7 divides k and 0 otherwise.
3. Within k ≤ 20 the multiples of 7 are 0, 7, 14, so the result is 7A_0 + 7A_7x⁷ + 7A_14x¹⁴.

Teacher's Note:
1. Interchange the sums.
2. A geometric sum of 7th roots of unity vanishes unless the exponent is a multiple of 7.