Class 11 Mathematics Complex Numbers and Quadratic Equations MCQs Set 15

Answer: (b) \( x = \frac{3}{4},\ y = \frac{33}{4} \)

Explanation:
1. Equating real parts: 4x = 3 ⇒ \( x = \frac{3}{4} \).
2. Equating imaginary parts: 3x − y = −6 ⇒ \( y = 3x + 6 = \frac{9}{4} + 6 = \frac{33}{4} \).

Teacher's Note:
1. Two complex numbers are equal only when their real parts and imaginary parts are separately equal.
2. This works only because x and y are given to be real.

Answer: (a) (2, −1)

Explanation:
1. (1 − i)x + (1 + i)y = (x + y) + i(y − x)
2. Comparing with 1 − 3i: x + y = 1 and y − x = −3.
3. Adding: 2y = −2 ⇒ y = −1, so x = 2. Hence (x, y) = (2, −1).

Teacher's Note:
1. Collect the real and imaginary parts on the left before comparing.
2. Check: (1 − i)2 + (1 + i)(−1) = 2 − 2i − 1 − i = 1 − 3i.

Answer: (b) −1

Explanation:
1. 103 = 4 × 25 + 3, so i¹⁰³ = (i⁴)²⁵ · i³ = −i.
2. −i = 0 + (−1)i ⇒ a = 0, b = −1
3. a + b = −1

Teacher's Note:
1. To find iⁿ, divide n by 4 and use the remainder: i⁰ = 1, i¹ = i, i² = −1, i³ = −i.
2. Write the answer in a + ib form before reading off a and b.

Answer: (a) Real number

Explanation:
1. i¹⁰ = i² = −1, i²⁰ = (i⁴)⁵ = 1, i³⁰ = i² = −1
2. Sum = 1 − 1 + 1 − 1 = 0, which is a real number.

Teacher's Note:
1. Even powers of i are always real (±1).
2. 0 is a real number but not a natural number.

Answer: (a) i

Explanation:
1. i^4x+1 = i^4x · i = i (for integer x)
2. i^4x−1 = i^4x · i⁻¹ = 1/i = −i
3. \( \frac{i - (-i)}{2} = \frac{2i}{2} = i \)

Teacher's Note:
1. 1/i = −i, obtained by multiplying numerator and denominator by i.
2. i^4x = 1 for every integer x.

Answer: (a) 5 + 5i

Explanation:
1. z₁ + z₂ = (2 + 3) + (3 + 2)i
2. = 5 + 5i

Teacher's Note:
1. Add real parts with real parts and imaginary parts with imaginary parts.
2. Addition of complex numbers is like adding two-dimensional vectors component-wise.

Answer: (a) −1 + 5i

Explanation:
1. z₁ − z₂ = (2 − 3) + (3 − (−2))i
2. = −1 + 5i

Teacher's Note:
1. Subtracting −2i gives +2i; watch the sign of the imaginary part.
2. Subtraction z₁ − z₂ is the same as z₁ + (−z₂).

Answer: (b) 0

Explanation:
1. (−i)(3i) = −3i² = 3
2. \( \left(-\frac{1}{6}i\right)^3 = -\frac{1}{216}i^3 = \frac{i}{216} \)
3. Product = \( 3 \times \frac{i}{216} = \frac{i}{72} = 0 + \frac{1}{72}i \), so a = 0.

Teacher's Note:
1. i³ = −i, so (−i/6)³ = −i³/216 = i/216.
2. A purely imaginary number has real part a = 0.

Answer: (a) 14 − 5i

Explanation:
1. (2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i²
2. = 2 − 5i + 12 = 14 − 5i

Teacher's Note:
1. Multiply as binomials and replace i² by −1.
2. (a + ib)(c + id) = (ac − bd) + i(ad + bc).

Answer: (a) 2

Explanation:
1. 3(7 + 7i) = 21 + 21i and i(7 + 7i) = 7i + 7i² = −7 + 7i
2. Sum = 14 + 28i, so a = 14, b = 28.
3. b/a = 28/14 = 2

Teacher's Note:
1. Expand each product separately, then add.
2. Replace i² by −1 before collecting real parts.

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

Explanation:
1. (1 + i)² = 1 + 2i + i² = 2i
2. \( \frac{2i}{2-i} = \frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i + 2i^2}{5} = \frac{-2 + 4i}{5} \)
3. x = −2/5, y = 4/5, so x + y = \( \frac{2}{5} \).

Teacher's Note:
1. (1 + i)² = 2i and (1 − i)² = −2i are worth remembering.
2. To divide, multiply numerator and denominator by the conjugate of the denominator.

Answer: (a) −4, 0

Explanation:
1. (1 − i)² = 1 − 2i + i² = −2i
2. (1 − i)⁴ = (−2i)² = 4i² = −4
3. So a = −4 and b = 0.

Teacher's Note:
1. Square in stages: first (1 − i)², then square the result.
2. (±2i)² = −4 in both cases.

Answer: (a) \( \frac{1}{5}(9 + 12i) \)

Explanation:
1. \( \frac{6+3i}{2-i} \times \frac{2+i}{2+i} = \frac{12 + 6i + 6i + 3i^2}{4 + 1} \)
2. \( = \frac{9 + 12i}{5} \)

Teacher's Note:
1. (a − ib)(a + ib) = a² + b², which makes the denominator real.
2. Check by multiplying back: (2 − i)(9 + 12i)/5 = (30 + 15i)/5 = 6 + 3i.

Answer: (a) \( -\frac{3}{34} - \frac{29i}{34} \)

Explanation:
1. The inverse is \( \frac{4-3i}{3+5i} \).
2. \( \frac{(4-3i)(3-5i)}{(3+5i)(3-5i)} = \frac{12 - 20i - 9i + 15i^2}{9 + 25} = \frac{-3 - 29i}{34} \)
3. \( = -\frac{3}{34} - \frac{29}{34}i \)

Teacher's Note:
1. The multiplicative inverse of p/q is simply q/p.
2. Use the conjugate of the new denominator to simplify.

Answer: (a) First

Explanation:
1. \( \frac{Z_1}{Z_2} = \frac{(\sqrt3 + \sqrt3 i)(\sqrt3 - i)}{3 + 1} = \frac{3 - \sqrt3 i + 3i + \sqrt3}{4} \)
2. \( = \frac{3 + \sqrt3}{4} + \frac{3 - \sqrt3}{4}i \)
3. Both the real part and the imaginary part are positive, so the point lies in the first quadrant.

Teacher's Note:
1. The quadrant is decided by the signs of the real part (x) and imaginary part (y).
2. √3 ≈ 1.73, so 3 − √3 > 0.

Answer: (d) −1 − i

Explanation:
1. Additive inverse of Z₂ = −2 − 3i
2. Z₁ + (−Z₂) = (1 − 2) + (2 − 3)i = −1 − i

Teacher's Note:
1. The additive inverse of a + ib is −a − ib; their sum is 0.
2. Do not confuse it with the multiplicative inverse (reciprocal).

Answer: (b) 7 + 2i

Explanation:
1. Conjugate of 9 − i is 9 + i.
2. 6 + i³ = 6 − i, whose conjugate is 6 + i; 9 + i² = 8, whose conjugate is 8.
3. Sum = 9 + i + 6 + i − 8 = 7 + 2i

Teacher's Note:
1. Simplify powers of i before taking the conjugate.
2. The conjugate of a real number is the number itself.

Answer: (a) 5 − i

Explanation:
1. Z₁ + Z₂ = 5 + i
2. Its conjugate is 5 − i.

Teacher's Note:
1. The conjugate changes only the sign of the imaginary part.
2. Conjugate of a sum = sum of the conjugates.

Answer: (a) 2

Explanation:
1. Z₁Z₂ = (1 + i)(2 − i) = 2 − i + 2i − i² = 3 + i
2. Its conjugate is 3 − i, so a = 3, b = −1.
3. a + b = 2

Teacher's Note:
1. Multiply first, then take the conjugate (or multiply the conjugates; both give the same).
2. Remember that b is the coefficient of i with its sign.

Answer: (b) \( -\frac{2}{25} - \frac{11}{25}i \)

Explanation:
1. (1 − 2i)² = 1 − 4i + 4i² = −3 − 4i
2. \( \frac{2-i}{-3-4i} = \frac{(2-i)(-3+4i)}{9+16} = \frac{-6 + 8i + 3i - 4i^2}{25} = \frac{-2 + 11i}{25} \)
3. Conjugate = \( -\frac{2}{25} - \frac{11}{25}i \)

Teacher's Note:
1. Simplify to a + ib form first, then change the sign of the imaginary part.
2. The conjugate of −3 − 4i is −3 + 4i, used here to rationalise.

Answer: (c) \( -\frac{9}{13} - \frac{19}{13}i \)

Explanation:
1. \( \frac{3+5i}{2-3i} = \frac{(3+5i)(2+3i)}{4+9} = \frac{6 + 9i + 10i + 15i^2}{13} = \frac{-9 + 19i}{13} \)
2. Conjugate = \( -\frac{9}{13} - \frac{19}{13}i \)

Teacher's Note:
1. Conjugate of a quotient equals the quotient of the conjugates.
2. Keep the real part unchanged and flip only the sign of the imaginary part.

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

Explanation:
1. Z₁Z₂ = (2 − i)(−2 + i) = −4 + 2i + 2i − i² = −3 + 4i
2. \( \overline{Z_1} = 2 + i \), so \( \frac{-3+4i}{2+i} = \frac{(-3+4i)(2-i)}{5} = \frac{-6 + 3i + 8i - 4i^2}{5} = \frac{-2 + 11i}{5} \)
3. So a = \( -\frac{2}{5} \).

Teacher's Note:
1. Z₂ = −Z₁ here, so Z₁Z₂ = −Z₁², which is a quick check.
2. Multiply by the conjugate of 2 + i, which is 2 − i.

Answer: (b) −6 + 5i

Explanation:
1. i⁻¹⁵ = i⁻¹⁶ · i = i, and i⁻⁸ = 1.
2. Z = −5i − 6 = −6 − 5i
3. \( \overline{Z} = -6 + 5i \)

Teacher's Note:
1. For negative powers, add a multiple of 4 to the exponent: i⁻¹⁵ = i⁻¹⁵⁺¹⁶ = i.
2. Then take the conjugate by changing the sign of the imaginary part.

Answer: (a) 5

Explanation:
1. i⁷ = i⁴ · i³ = −i, so 4 + 3i⁷ = 4 − 3i.
2. \( |4 - 3i| = \sqrt{16 + 9} = 5 \)

Teacher's Note:
1. The modulus is never negative, so −5 can be rejected at once.
2. |a + ib| = |a − ib|, so the sign of the imaginary part does not change the modulus.

Answer: (b) 2

Explanation:
1. Z₂ − Z₁ = 1 + i
2. |1 + i|² = 1² + 1² = 2

Teacher's Note:
1. |z|² = a² + b², so there is no need to take a square root.
2. |Z₂ − Z₁| is the distance between the points Z₁ and Z₂ in the Argand plane.

Answer: (a) 1

Explanation:
1. (1 + i)(2 + i) = 2 + i + 2i + i² = 1 + 3i
2. \( Z = \frac{1+3i}{3+i} \), so \( |Z| = \frac{|1+3i|}{|3+i|} = \frac{\sqrt{10}}{\sqrt{10}} = 1 \)

Teacher's Note:
1. Use |z₁/z₂| = |z₁|/|z₂| and |z₁z₂| = |z₁||z₂| to avoid long division.
2. Directly: |1 + i||2 + i|/|3 + i| = √2 · √5/√10 = 1.

Answer: (d) 0

Explanation:
1. (1 − i)² = −2i and (1 + i)² = 2i
2. \( \frac{1}{-2i} + \frac{1}{2i} = 0 \)
3. So the number is 0 and its modulus is 0.

Teacher's Note:
1. Squares of 1 ± i are ±2i; using them shortens the work a lot.
2. |0| = 0 is the only complex number with modulus 0.

Answer: (c) 66

Explanation:
1. z₁ + z₂ = 5 − 2i, so |z₁ + z₂|² = 25 + 4 = 29.
2. z₁ − z₂ = 1 + 6i, so |z₁ − z₂|² = 1 + 36 = 37.
3. Sum = 29 + 37 = 66

Teacher's Note:
1. Check with the identity |z₁ + z₂|² + |z₁ − z₂|² = 2(|z₁|² + |z₂|²) = 2(13 + 20) = 66.
2. Work with |z|² = a² + b² to avoid square roots.

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

Explanation:
1. 9x² = −16 ⇒ \( x^2 = -\frac{16}{9} \)
2. \( x = \pm\sqrt{-\frac{16}{9}} = \pm\frac{4}{3}i \)

Teacher's Note:
1. √(−a) = i√a for a > 0.
2. Check: 9(4i/3)² = 9 × (−16/9) = −16.

Answer: (a) \( \pm\sqrt2 i \)

Explanation:
1. x² = −2
2. \( x = \pm\sqrt{-2} = \pm\sqrt2 i \)

Teacher's Note:
1. A quadratic equation always has two roots; here they are a pair of conjugates.
2. √(−2) = √2 · √(−1) = √2 i.

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

Explanation:
1. D = b² − 4ac = 9 − 36 = −27
2. \( \sqrt{D} = \sqrt{-27} = 3\sqrt3 i \)
3. \( x = \frac{-3 \pm 3\sqrt3 i}{2} \)

Teacher's Note:
1. When D < 0, write √D = i√|D| and use the quadratic formula as usual.
2. The sum of the roots must be −b/a = −3, which rules out options with +3.

Answer: (a) \( \frac{-1 \pm \sqrt3 i}{2} \)

Explanation:
1. D = 1 − 4 = −3
2. \( x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt3 i}{2} \)

Teacher's Note:
1. Sum of roots = −b/a = −1 is a quick check on the real part.
2. Complex roots of a real quadratic always come in conjugate pairs.

Answer: (a) \( \frac{-1 \pm i\sqrt7}{2\sqrt2} \)

Explanation:
1. a = √2, b = 1, c = √2; D = 1 − 4 · √2 · √2 = 1 − 8 = −7
2. \( x = \frac{-1 \pm \sqrt{-7}}{2\sqrt2} = \frac{-1 \pm i\sqrt7}{2\sqrt2} \)

Teacher's Note:
1. The quadratic formula works even when the coefficients are irrational.
2. The denominator is 2a = 2√2, not 2.

Answer: (b) \( 1 \pm \sqrt3 i \)

Explanation:
1. (y + 1)(y − 3) + 7 = y² − 2y − 3 + 7 = y² − 2y + 4 = 0
2. D = 4 − 16 = −12, √D = 2√3 i
3. \( y = \frac{2 \pm 2\sqrt3 i}{2} = 1 \pm \sqrt3 i \)

Teacher's Note:
1. Expand to the standard form ay² + by + c = 0 before using the formula.
2. Sum of roots = 2, so the real part of each root is 1.

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

Explanation:
1. i⁴ = 1, so i^4k = 1 and the next powers are i, −1, −i. A is true.
2. The sum is 1 + i − 1 − i = 0, not 1. R is false.

Teacher's Note:
1. The sum of any four consecutive powers of i is 0.
2. Powers of i repeat in a cycle of 4: i, −1, −i, 1.

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

Explanation:
1. i⁻³⁵ = i⁻³⁶ · i = (i⁴)⁻⁹ · i = i, not −i. A is false.
2. −(1 − i) = −1 + i, so R is true.

Teacher's Note:
1. For a negative power, add the next multiple of 4 to the exponent: −35 + 36 = 1.
2. The additive inverse of z is −z, since z + (−z) = 0.

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

Explanation:
1. \( \frac{(5 + \sqrt2 i)(1 + \sqrt2 i)}{1 + 2} = \frac{5 + 5\sqrt2 i + \sqrt2 i - 2}{3} = \frac{3 + 6\sqrt2 i}{3} = 1 + 2\sqrt2 i \). A is false.
2. (1 + i)(1 − i) = 2, so (1 + i)⁵(1 − i)⁵ = 2⁵ = 32. R is true.

Teacher's Note:
1. Multiply numerator and denominator by the conjugate 1 + √2 i of the denominator.
2. Combine powers with the same exponent: aⁿbⁿ = (ab)ⁿ.