Class 11 Mathematics Complex Numbers and Quadratic Equations MCQs Set 16

Answer: (a) −2, 2

Explanation:
1. Real parts: 3a − 6 = −6b ⇒ a + 2b = 2.
2. Imaginary parts: 2b = 6 + a ⇒ a = 2b − 6.
3. So 2b − 6 + 2b = 2 ⇒ b = 2, a = −2.

Teacher's Note:
1. Write the two real equations first, then solve them as simultaneous equations.
2. Check: 3(−2) − 6 = −12 = −6(2) and 2(2) = 4 = 6 + (−2).

Answer: (b) 2, −2

Explanation:
1. Real parts: 2a + 2b = 0 ⇒ b = −a.
2. Imaginary parts: b − a = −4 ⇒ −2a = −4 ⇒ a = 2, b = −2.

Teacher's Note:
1. −4i has real part 0, so the real part on the left must vanish.
2. Check: (4 − 4) + i(−2 − 2) = −4i.

Answer: (a) 1, 0

Explanation:
1. \( \frac{1-i}{1+i} = \frac{(1-i)^2}{(1+i)(1-i)} = \frac{-2i}{2} = -i \)
2. (−i)¹⁰⁰ = (i⁴)²⁵ = 1
3. So a + ib = 1 + 0i, i.e. a = 1, b = 0.

Teacher's Note:
1. Simplify the base to a single power of i before raising it to a large power.
2. (−i)⁴ = 1, so any power of −i that is a multiple of 4 equals 1.

Answer: (d) \( \frac{2}{5} \)

Explanation:
1. (1 + i)² = 2i
2. \( \frac{2i(2+i)}{(2-i)(2+i)} = \frac{-2 + 4i}{5} \)
3. x = −2/5, y = 4/5 ⇒ x + y = \( \frac{2}{5} \)

Teacher's Note:
1. Rationalise the denominator using its conjugate.
2. Equate real and imaginary parts to read off x and y.

Answer: (b) −5

Explanation:
1. x + y = 4 and x − y = 6
2. Adding: 2x = 10 ⇒ x = 5, y = −1
3. xy = −5

Teacher's Note:
1. Adding and subtracting the two equations gives x and y directly.
2. Check: (5 − 1) + i(5 + 1) = 4 + 6i.

Answer: (b) z

Explanation:
1. \( \overline{z} = \frac{1+i}{1-i}z = \frac{(1+i)^2}{2}z = iz \)
2. \( -i\overline{z} = -i(iz) = -i^2 z = z \)

Teacher's Note:
1. (1 + i)/(1 − i) = i is a standard result worth remembering.
2. Multiply by −i at the end and use i² = −1.

Answer: (a) \( \overline{z_1}\,\overline{z_2} \)

Explanation:
1. Let z₁ = a + ib, z₂ = c + id. Then z₁z₂ = (ac − bd) + i(ad + bc) and its conjugate is (ac − bd) − i(ad + bc).
2. \( \overline{z_1}\,\overline{z_2} = (a - ib)(c - id) = (ac - bd) - i(ad + bc) \), which is the same.
3. So \( \overline{z_1 z_2} = \overline{z_1}\,\overline{z_2} \).

Teacher's Note:
1. The conjugate of a product is the product of the conjugates; the same holds for sums and quotients.
2. Prove such properties by writing z₁ and z₂ in a + ib form.

Answer: (a) 8x − 17y = 16

Explanation:
1. \( \frac{2x + (x-1)i}{4+i} = \frac{[2x + (x-1)i](4-i)}{17} = \frac{(9x - 1) + (2x - 4)i}{17} \)
2. \( \frac{y + (2-y)i}{4i} = \frac{[y + (2-y)i](-i)}{4} = \frac{(2-y) - yi}{4} \)
3. Purely real ⇒ imaginary part 0: \( \frac{2x - 4}{17} - \frac{y}{4} = 0 \Rightarrow 8x - 16 = 17y \), i.e. 8x − 17y = 16.

Teacher's Note:
1. A number is purely real when its imaginary part is zero; the real part may be anything.
2. Divide by 4i by multiplying numerator and denominator by −i.

Answer: (c) \( \frac{\pi}{3} \)

Explanation:
1. \( z = \frac{(3 + 2i\sin\theta)(1 + 2i\sin\theta)}{1 + 4\sin^2\theta} = \frac{(3 - 4\sin^2\theta) + 8i\sin\theta}{1 + 4\sin^2\theta} \)
2. Purely imaginary ⇒ real part 0: \( 3 - 4\sin^2\theta = 0 \Rightarrow \sin\theta = \frac{\sqrt3}{2} \) (θ in the first quadrant).
3. So \( \theta = \frac{\pi}{3} \).

Teacher's Note:
1. For a purely imaginary number, set the real part equal to zero.
2. In the given range sin θ > 0, so only the positive root √3/2 is taken.

Answer: (b) \( \frac{z_1}{z_2} \) is purely imaginary

Explanation:
1. Let \( w = \frac{z_1}{z_2} = a + ib \). Dividing numerator and denominator by z₂: \( \left|\frac{w - 1}{w + 1}\right| = 1 \), so |w − 1| = |w + 1|.
2. (a − 1)² + b² = (a + 1)² + b² ⇒ −2a = 2a ⇒ a = 0.
3. So \( \frac{z_1}{z_2} \) is purely imaginary.

Teacher's Note:
1. |w − 1| = |w + 1| means w is equidistant from 1 and −1, so it lies on the imaginary axis.
2. Squaring moduli (|z|² = a² + b²) removes square roots.

Answer: (a) i

Explanation:
1. 37 = 4 × 9 + 1
2. i³⁷ = i¹ = i

Teacher's Note:
1. Only the remainder on division by 4 matters.
2. i^4k+1 = i for every integer k.

Answer: (c) −1

Explanation:
1. −30 + 32 = 2, so write i⁻³⁰ = i⁻³² · i².
2. i⁻³² = (i⁴)⁻⁸ = 1, so i⁻³⁰ = 1 × (−1) = −1

Teacher's Note:
1. For negative powers, add a multiple of 4 to make the exponent non-negative: −30 + 32 = 2.
2. Alternatively, i⁻³⁰ = 1/i³⁰ = 1/(−1) = −1.

Answer: (a) 0 + 0i

Explanation:
1. i⁹ = i⁸ · i = i and i¹⁹ = i¹⁶ · i³ = −i
2. z = i − i = 0 = 0 + 0i

Teacher's Note:
1. i^4k+1 and i^4k+3 always cancel each other.
2. Write 0 as 0 + 0i when the answer is asked in a + ib form.

Answer: (a) −4

Explanation:
1. i¹⁹ = i³ = −i
2. 1/i = −i, so (1/i)²⁵ = (−i)²⁵ = −i²⁵ = −i
3. [−i − i]² = (−2i)² = 4i² = −4

Teacher's Note:
1. Replace 1/i by −i at the start.
2. An odd power of −i equals minus the same power of i.

Answer: (b) 0 + i

Explanation:
1. i⁻³⁹ = i⁻⁴⁰ · i = (i⁴)⁻¹⁰ · i = i
2. So z = 0 + i.

Teacher's Note:
1. −39 + 40 = 1, so i⁻³⁹ = i¹.
2. Writing 0 + i shows clearly that the real part is 0.

Answer: (a) 0

Explanation:
1. (x − iy)(3 + 5i) = (3x + 5y) + i(5x − 3y), and the conjugate of −6 − 24i is −6 + 24i.
2. 3x + 5y = −6 and 5x − 3y = 24
3. Solving: x = 3, y = −3, so x + y = 0.

Teacher's Note:
1. Multiply out first, then equate real and imaginary parts.
2. Eliminate one variable: multiply the equations by 3 and 5 and add.

Answer: (a) |z + 3|²

Explanation:
1. \( \overline{z} + 3 = \overline{z + 3} \), since 3 is real.
2. So \( (z + 3)(\overline{z} + 3) = (z + 3)\overline{(z + 3)} = |z + 3|^2 \).

Teacher's Note:
1. w·w̄ = |w|² for every complex number w.
2. Adding a real number commutes with taking the conjugate.

Answer: (a) \( \frac{|z|}{2} \)

Explanation:
1. z² = (1 + 2i)² = −3 + 4i, so 1 − z² = 4 − 4i; 7 − z = 6 − 2i.
2. \( |f(z)| = \frac{|6 - 2i|}{|4 - 4i|} = \frac{\sqrt{40}}{\sqrt{32}} = \frac{\sqrt5}{2} \)
3. |z| = √5, so \( |f(z)| = \frac{|z|}{2} \).

Teacher's Note:
1. |z₁/z₂| = |z₁|/|z₂|, so there is no need to divide the complex numbers.
2. Compute |z| = √(1 + 4) = √5 to compare with the options.

Answer: (a) √2

Explanation:
1. z₁ + z₂ = −1 + i
2. |−1 + i| = √(1 + 1) = √2

Teacher's Note:
1. Add first, then take the modulus; |z₁ + z₂| is generally not |z₁| + |z₂|.
2. The modulus squares both parts, so signs do not matter.

Answer: (c) 3

Explanation:
1. \( z + \overline{z} = (3 + 4i) + (3 - 4i) = 6 \)
2. So (z + z̄)/2 = 3.

Teacher's Note:
1. (z + z̄)/2 is always the real part of z.
2. Similarly (z − z̄)/2i is the imaginary part.

Answer: (a) \( \frac{\sqrt3}{2} \pm \frac{3}{4}i \)

Explanation:
1. D = (2√3)² − 4 × 2 × \( \frac{21}{8} \) = 12 − 21 = −9, so √D = 3i.
2. \( x = \frac{2\sqrt3 \pm 3i}{4} = \frac{\sqrt3}{2} \pm \frac{3}{4}i \)

Teacher's Note:
1. Here b = −2√3, so −b = +2√3; the real part of the roots is positive.
2. Divide each term of the numerator by 2a = 4.

Answer: (b) \( \frac{3}{5} \pm \frac{\sqrt2}{5}i \)

Explanation:
1. D = 900 − 4 × 25 × 11 = 900 − 1100 = −200, so √D = 10√2 i.
2. \( x = \frac{30 \pm 10\sqrt2 i}{50} = \frac{3}{5} \pm \frac{\sqrt2}{5}i \)

Teacher's Note:
1. √200 = 10√2; simplify surds fully.
2. Sum of roots = 30/25 = 6/5 confirms the real part 3/5.

Answer: (a) \( \frac{-1 \pm \sqrt7 i}{4} \)

Explanation:
1. D = 1 − 8 = −7, so √D = √7 i.
2. \( x = \frac{-1 \pm \sqrt7 i}{4} \)

Teacher's Note:
1. The denominator is 2a = 4.
2. Sum of roots = −b/a = −1/2, matching real part −1/4 for each root.

Answer: (c) \( \frac{-1 \pm \sqrt7 i}{-2} \)

Explanation:
1. a = −1, b = 1, c = −2; D = 1 − 4(−1)(−2) = 1 − 8 = −7.
2. \( x = \frac{-1 \pm \sqrt7 i}{2(-1)} = \frac{-1 \pm \sqrt7 i}{-2} \), i.e. \( \frac{1 \pm \sqrt7 i}{2} \).

Teacher's Note:
1. With a negative leading coefficient, 2a is negative; you may instead multiply the equation by −1 first.
2. (−1 ± √7 i)/(−2) and (1 ± √7 i)/2 describe the same pair of roots.

Answer: (a) \( \frac{2}{3} \pm \frac{4}{3}i \)

Explanation:
1. D = 16 − 4 × 3 × \( \frac{20}{3} \) = 16 − 80 = −64, so √D = 8i.
2. \( x = \frac{4 \pm 8i}{6} = \frac{2}{3} \pm \frac{4}{3}i \)

Teacher's Note:
1. Multiplying the equation by 3 (9x² − 12x + 20 = 0) removes the fraction and gives the same roots.
2. Sum of roots = 4/3 confirms the real part 2/3.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. By R, Z₁ − Z₂ = (2 − 3) + i(3 − (−2)) = −1 + 5i, so A is true.
2. R is the rule of subtraction used to get A, so it is the correct explanation.

Teacher's Note:
1. Subtract real parts and imaginary parts separately.
2. Be careful with subtracting a negative imaginary part.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. (1 + i)² = 2i, so (1 + i)⁶ = (2i)³ = 8i³ = −8i; a = 0, b = −8. A is true.
2. (1 − i)³ = (1 − i)(−2i) = −2i + 2i² = −2 − 2i, so a = b = −2 and a/b = 1. R is true.
3. The two statements are separate calculations, so R does not explain A.

Teacher's Note:
1. Use (1 ± i)² = ±2i to find higher powers quickly.
2. Two true statements need not be linked; check whether R is actually used to get A.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. (1 + i)(x + iy) = (x − y) + i(x + y)
2. Using R: x − y = 2 and x + y = −5.
3. So \( x = -\frac{3}{2},\ y = -\frac{7}{2} \); A is true and follows from R.

Teacher's Note:
1. Equality of complex numbers gives two real equations.
2. Solve the two linear equations by adding and subtracting.

Answer: (d) A is false, but R is true.

Explanation:
1. \( \frac{1}{2-3i} = \frac{2+3i}{(2-3i)(2+3i)} = \frac{2+3i}{13} \), not 2 + 3i. A is false.
2. The conjugate of 3 + 4i is 3 − 4i, so R is true.

Teacher's Note:
1. The multiplicative inverse of z is z̄/|z|², not just z̄.
2. Check: (2 − 3i)(2 + 3i) = 13 ≠ 1.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. (3 + 2i) + (2 + i) = 5 + 3i
2. (2 + 3i)(5 + 3i) = 10 + 6i + 15i + 9i² = 1 + 21i. A is true.
3. Equivalently, by the distributive law in R, (2 + 3i)(3 + 2i) + (2 + 3i)(2 + i) = 13i + (1 + 8i) = 1 + 21i. R explains A.

Teacher's Note:
1. Complex numbers obey the distributive law just like real numbers.
2. Simplifying the bracket first is usually quicker.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( z = 5i \times \left(-\frac{3}{5}i\right) = -3i^2 = 3 = 3 + 0i \). A is true.
2. R is the correct rule for addition of complex numbers.
3. A uses multiplication, not addition, so R does not explain A.

Teacher's Note:
1. i² = −1 turns the product of two imaginary numbers into a real number.
2. Check that the reason actually describes the operation used in the assertion.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( z = \frac{(1+2i)(1+3i)}{10} = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i \)
2. By R, \( |z| = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt2} \). A is true and follows from R.

Teacher's Note:
1. A shortcut: |z| = |1 + 2i|/|1 − 3i| = √5/√10 = 1/√2.
2. The modulus formula in R is the definition used to compute A.

Answer: (d) A is false, but R is true.

Explanation:
1. Comparing real parts: x = y + 3; comparing imaginary parts: 4y = x.
2. So 4y = y + 3 ⇒ y = 1, x = 4. A (x = 1, y = 4) is false.
3. \( \frac{1}{3 + \sqrt7 i} = \frac{3 - \sqrt7 i}{9 + 7} = \frac{3}{16} - \frac{\sqrt7}{16}i \), so R is true.

Teacher's Note:
1. Collect real terms and the coefficients of i on each side before comparing.
2. Substituting the claimed values back is a fast check: 1 + 16i ≠ 4 + i + 3.

Answer: (c) A is true, but R is false.

Explanation:
1. i⁹ = i and i¹⁹ = i³ = −i, so z = 0 = 0 + 0i. A is true.
2. The series has 11 terms 1, −1, 1, −1, …, 1 (i²⁰ = 1); they pair off to 0 leaving 1. So the sum is 1, not −1; R is false.

Teacher's Note:
1. Even powers of i alternate between −1 and 1.
2. Count the terms: exponents 0, 2, …, 20 give 11 terms.

Answer: (c) A is true, but R is false.

Explanation:
1. x² = −1 ⇒ x = ±i. A is true.
2. −1097 + 1100 = 3, so i⁻¹⁰⁹⁷ = i³ = −i, not i. R is false.

Teacher's Note:
1. For negative exponents, add a multiple of 4 large enough to make the exponent non-negative.
2. x² + 1 = 0 is the simplest equation with no real root.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. D = 16 − 4 × 3 × 2 = 16 − 24 = −8 < 0
2. By R, the roots are imaginary (non-real). A is true and R explains it.

Teacher's Note:
1. The sign of the discriminant decides the nature of the roots.
2. Here the roots are (−2 ± √2 i)/3, a conjugate pair.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. D = 9 − 20 = −11, so \( x = \frac{-3 \pm i\sqrt{11}}{2} \). A is true.
2. For x² − x + 2: D = 1 − 8 = −7, so \( x = \frac{1 \pm i\sqrt7}{2} \). R is true.
3. R is about a different equation, so it does not explain A.

Teacher's Note:
1. Use x = (−b ± √D)/2a with √D = i√|D| when D < 0.
2. Two separate correct solutions do not explain each other.