CBSE Class 11 Mathematics Complex Numbers and Quadratic Equations Competitive Exam MCQs Set 01

Answer: (b) 1, −1, i, −i

Explanation:
1. x⁴ − 1 = (x² − 1)(x² + 1) = (x − 1)(x + 1)(x − i)(x + i).
2. Setting each factor to zero gives x = 1, −1, i, −i.
3. Hence the roots are 1, −1, i, −i.

Teacher's Note:
1. A polynomial of degree 4 has exactly four roots, so factorise completely using x² + 1 = (x − i)(x + i).
2. Option (a) repeats 1 and misses −1; always check that the listed roots actually satisfy the equation.

Answer: (b) 0

Explanation:
1. i² = −1, so the factor 1 + i² = 1 − 1 = 0.
2. A product with one zero factor is zero.
3. Hence the value is 0.

Teacher's Note:
1. Before multiplying long products, look for a factor that becomes zero.
2. Remember i² = −1, i³ = −i and i⁴ = 1; these four values repeat for higher powers.

Answer: (d) cannot be determined

Explanation:
1. The terms are 1, −1, 1, −1, ... since i² = −1 and i⁴ = 1.
2. The series has n + 1 terms; the sum is 1 if n is even and 0 if n is odd.
3. As n is not specified, the value cannot be determined.

Teacher's Note:
1. Even powers of i are i² = −1, i⁴ = 1, i⁶ = −1, ..., so consecutive pairs of terms cancel.
2. When the answer depends on whether n is odd or even and n is not given, the value cannot be fixed.

Answer: (b) \( -\sqrt{6} \)

Explanation:
1. \( \sqrt{-2} = i\sqrt{2} \) and \( \sqrt{-3} = i\sqrt{3} \).
2. Product = \( i^2\sqrt{6} = -\sqrt{6} \).
3. Hence the value is \( -\sqrt{6} \).

Teacher's Note:
1. The rule √a √b = √(ab) fails when both a and b are negative, so √6 is a trap.
2. Always write the square root of a negative number as i times the square root of a positive number first.

Answer: (a) 16

Explanation:
1. \( \frac{1}{i} = -i \), so \( 1 + \frac{1}{i} = 1 - i \).
2. Product = \( [(1 + i)(1 - i)]^4 = (1 - i^2)^4 = 2^4 \).
3. Hence the value is 16.

Teacher's Note:
1. Use 1/i = −i; multiplying numerator and denominator by i shows this at once.
2. Combine the two fourth powers into one, since aⁿbⁿ = (ab)ⁿ, instead of expanding each.

Answer: (b) 8

Explanation:
1. \( \frac{2i}{1 + i} = \frac{2i(1 - i)}{(1 + i)(1 - i)} = \frac{2i + 2}{2} = 1 + i \).
2. (1 + i)² = 2i, (1 + i)⁴ = (2i)² = −4, (1 + i)⁸ = 16.
3. (1 + i)² and (1 + i)⁶ = −8i are not real, so the least n is 8.

Teacher's Note:
1. Simplify the base first; here it reduces to 1 + i.
2. Square repeatedly: (1 + i)² = 2i is the key result for powers of 1 + i.

Answer: (b) b = 0, d = 0

Explanation:
1. Order relations (>, <) are not defined for non-real complex numbers.
2. a + ib > c + id has meaning only when both numbers are real.
3. So the imaginary parts must vanish: b = 0 and d = 0.

Teacher's Note:
1. Complex numbers cannot be compared by size; only their moduli can be compared.
2. Both numbers must be real, so both imaginary parts b and d must be zero.

Answer: (c) b + ia

Explanation:
1. Test option (c): (b + ia)(1 + b − ia) = b + b² − iab + ia + iab − i²a².
2. = b + b² + a² + ia = b + 1 + ia, using a² + b² = 1.
3. So \( \frac{1 + b + ia}{1 + b - ia} = b + ia \).

Teacher's Note:
1. Cross-multiplying with the suggested answer is often faster than rationalising.
2. Watch for where the given condition a² + b² = 1 can replace a sum of squares.

Answer: (b) \( \frac{1}{2} \)

Explanation:
1. Let θ = π/5. \( \frac{1}{1 + \cos\theta + i\sin\theta} = \frac{1 + \cos\theta - i\sin\theta}{(1 + \cos\theta)^2 + \sin^2\theta} \).
2. Denominator = 1 + 2cos θ + cos²θ + sin²θ = 2(1 + cos θ).
3. Real part = \( \frac{1 + \cos\theta}{2(1 + \cos\theta)} = \frac{1}{2} \).

Teacher's Note:
1. To find the real part of 1/(a + ib), multiply by the conjugate: the real part is a/(a² + b²).
2. The answer does not depend on the angle, which is why cos(π/5) disappears.

Answer: (c) Second

Explanation:
1. \( \frac{1 + 2i}{1 - i} = \frac{(1 + 2i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 3i + 2i^2}{2} = \frac{-1 + 3i}{2} \).
2. Real part −1/2 < 0 and imaginary part 3/2 > 0.
3. So the point (−1/2, 3/2) lies in the second quadrant.

Teacher's Note:
1. Write the number as a + ib first; the point (a, b) then fixes the quadrant.
2. Second quadrant means real part negative and imaginary part positive.

Answer: (c) \( \frac{\sqrt{5}}{5} \) unit

Explanation:
1. \( \frac{1 - i}{3 + i} = \frac{(1 - i)(3 - i)}{10} = \frac{3 - 4i + i^2}{10} = \frac{2 - 4i}{10} = \frac{1 - 2i}{5} \).
2. Adding \( \frac{4i}{5} \): \( \frac{1 - 2i + 4i}{5} = \frac{1 + 2i}{5} \).
3. Modulus = \( \frac{\sqrt{1 + 4}}{5} = \frac{\sqrt{5}}{5} \).

Teacher's Note:
1. Simplify to a single a + ib before taking the modulus; |z₁ + z₂| is not |z₁| + |z₂| in general.
2. Rationalise by multiplying with the conjugate of the denominator.

Answer: (c) \( \left| \frac{1}{z_1} + \frac{1}{z_2} + \ldots + \frac{1}{z_n} \right| \)

Explanation:
1. |z_k| = 1 gives \( z_k\bar{z}_k = 1 \), so \( \frac{1}{z_k} = \bar{z}_k \).
2. \( \left| \frac{1}{z_1} + \ldots + \frac{1}{z_n} \right| = |\bar{z}_1 + \ldots + \bar{z}_n| = |\overline{z_1 + \ldots + z_n}| \).
3. Since \( |\bar{w}| = |w| \), this equals |z₁ + ... + z_n|.

Teacher's Note:
1. For a number on the unit circle, the reciprocal equals the conjugate: 1/z = z̄.
2. A number and its conjugate always have the same modulus.

Answer: (b) (a² + b²)(|z₁|² + |z₂|²)

Explanation:
1. \( |az_1 - bz_2|^2 = a^2|z_1|^2 + b^2|z_2|^2 - 2ab\,\text{Re}(z_1\bar{z}_2) \).
2. \( |bz_1 + az_2|^2 = b^2|z_1|^2 + a^2|z_2|^2 + 2ab\,\text{Re}(z_1\bar{z}_2) \).
3. Adding, the middle terms cancel: (a² + b²)(|z₁|² + |z₂|²).

Teacher's Note:
1. Use |z₁ ± z₂|² = |z₁|² + |z₂|² ± 2Re(z₁z̄₂) for each term.
2. A quick check with z₁ = 1, z₂ = 0, a = b = 1 gives 2, matching (a² + b²)|z₁|².

Answer: (b) \( \frac{1}{5} \)

Explanation:
1. Squaring, \( (x + iy)^2 = \frac{1 + 2i}{3 + 4i} \).
2. Taking modulus, \( (x^2 + y^2) = \frac{|1 + 2i|}{|3 + 4i|} = \frac{\sqrt{5}}{5} \).
3. So \( (x^2 + y^2)^2 = \frac{5}{25} = \frac{1}{5} \).

Teacher's Note:
1. Taking modulus on both sides avoids finding x and y separately.
2. Use |z²| = |z|² and |z₁/z₂| = |z₁|/|z₂|.

Answer: (d) x − iy

Explanation:
1. Squaring, a + ib = (x + iy)² = x² − y² + 2ixy, so a = x² − y² and b = 2xy.
2. Then (x − iy)² = x² − y² − 2ixy = a − ib.
3. Hence x − iy is a square root of a − ib.

Teacher's Note:
1. Taking conjugates on both sides of an equation keeps it true.
2. The conjugate of a square root of z is a square root of z̄.

Answer: (a) \( \left| \frac{\bar{z}}{z} \right| \)

Explanation:
1. \( |\bar{z}|^2 = |z|^2 = z\bar{z} \), so the given expression equals 1.
2. \( \left| \frac{\bar{z}}{z} \right| = \frac{|\bar{z}|}{|z|} = 1 \).
3. Hence the expression equals \( \left| \frac{\bar{z}}{z} \right| \).

Teacher's Note:
1. The identity z z̄ = |z|² = |z̄|² is used again and again in modulus problems.
2. Options (b) and (c) equal |z|, which is 1 only for special z.

Answer: (a) \( \frac{1}{13} \)

Explanation:
1. |2 + 3i| = \( \sqrt{4 + 9} = \sqrt{13} \).
2. \( |z| = \frac{1}{|2 + 3i|^2} = \frac{1}{13} \).
3. Hence |z| = 1/13.

Teacher's Note:
1. Use |z₁/z₂| = |z₁|/|z₂| and |zⁿ| = |z|ⁿ; there is no need to expand (2 + 3i)².
2. A common slip is to write 1/√13 by forgetting the square.

Answer: (b) \( 1/\sqrt{26} \)

Explanation:
1. |1 − i| = √2 and |2 + 3i| = √13.
2. \( |z| = \frac{1}{\sqrt{2}\cdot\sqrt{13}} = \frac{1}{\sqrt{26}} \).
3. Hence |z| = 1/√26.

Teacher's Note:
1. The modulus of a product is the product of the moduli.
2. Work with moduli directly instead of multiplying out the denominator.

Answer: (c) 100

Explanation:
1. x² + y² = |x + iy|² = |1 + i|² |1 + 2i|² |1 + 3i|².
2. = 2 × 5 × 10.
3. = 100.

Teacher's Note:
1. x² + y² is |x + iy|², so use the product rule for moduli.
2. Expanding gives (1 + i)(1 + 2i)(1 + 3i) = −10, and (−10)² = 100 confirms it.

Answer: (c) 17

Explanation:
1. |z| is real, so Im(z) = 8; let z = x + 8i.
2. Then x + \( \sqrt{x^2 + 64} \) = 2, so x² + 64 = (2 − x)² = 4 − 4x + x².
3. 4x = −60, x = −15, so |z| = \( \sqrt{225 + 64} = 17 \).

Teacher's Note:
1. Since |z| is real, compare imaginary parts first to fix Im(z).
2. After squaring, check the root: 2 − x = 17 is positive, so x = −15 is valid.

Answer: (c) 10

Explanation:
1. Let z = x + iy; \( z + \bar{z} = 2x \), \( z - \bar{z} = 2iy \).
2. (3 + i)2x − (2 + i)2iy + 14i = (6x + 2y) + i(2x − 4y + 14) = 0.
3. 6x + 2y = 0 and 2x − 4y + 14 = 0 give x = −1, y = 3.
4. \( z\bar{z} = x^2 + y^2 = 1 + 9 = 10 \).

Teacher's Note:
1. Replace z + z̄ by 2x and z − z̄ by 2iy to turn the equation into two real equations.
2. A complex number is zero only when both its real and imaginary parts are zero.

Answer: (d) 5

Explanation:
1. Taking modulus, |z|³ = |z|, so |z| = 0 or |z| = 1.
2. |z| = 0 gives z = 0.
3. If |z| = 1, then \( \bar{z} = \frac{1}{z} \), so z³ = 1/z, i.e. z⁴ = 1, giving z = 1, −1, i, −i.
4. Total number of solutions = 1 + 4 = 5.

Teacher's Note:
1. Taking modulus of both sides is a quick way to find the possible sizes of z.
2. Do not forget z = 0, which is lost if you divide by z too early.

Answer: (d) 1

Explanation:
1. Since |β| = 1, \( \beta\bar{\beta} = 1 \).
2. \( |1 - \bar{\alpha}\beta| = |\beta\bar{\beta} - \bar{\alpha}\beta| = |\beta|\,|\bar{\beta} - \bar{\alpha}| = |\overline{\beta - \alpha}| = |\beta - \alpha| \).
3. So the given modulus = \( \frac{|\beta - \alpha|}{|\beta - \alpha|} = 1 \).

Teacher's Note:
1. Replace 1 by β β̄ whenever |β| = 1; it lets you take β out as a common factor.
2. The conjugate of β − α has the same modulus as β − α.

Answer: (b) 2

Explanation:
1. |z| + |z − 2| = |z| + |2 − z| ≥ |z + (2 − z)| (triangle inequality).
2. = |2| = 2.
3. Equality holds for z = 1 (1 + 1 = 2), so the minimum value is 2.

Teacher's Note:
1. Write |z − 2| as |2 − z| so that the sum inside the triangle inequality cancels z.
2. Always check that the lower bound is actually attained by some z.

Answer: (b) A² + B²

Explanation:
1. Take modulus of both sides and square: |a + ib|²|c + id|²|e + if|²|g + ih|² = |A + iB|².
2. |a + ib|² = a² + b², and similarly for the other factors.
3. Hence the product equals A² + B².

Teacher's Note:
1. The modulus of a product equals the product of the moduli.
2. Squaring the modulus removes the square roots: |x + iy|² = x² + y².