CBSE Class 11 Mathematics Complex Numbers and Quadratic Equations JEE Previous Year Questions Set 01

Answer: (d) x = 4n, where n is any positive integer

Explanation:
1. \( \frac{1 + i}{1 - i} = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{2i}{2} = i \).
2. So iˣ = 1.
3. iˣ = 1 exactly when x is a multiple of 4, so x = 4n.

Teacher's Note:
1. (1 + i)/(1 − i) = i is a standard simplification worth remembering.
2. Powers of i repeat in cycles of 4, so iˣ = 1 needs x to be a multiple of 4.

Answer: (d) −2

Explanation:
1. z = (p + iq)³ = p³ + 3p²(iq) + 3p(iq)² + (iq)³ = (p³ − 3pq²) + i(3p²q − q³).
2. Comparing with x − iy: x = p³ − 3pq², y = q³ − 3p²q.
3. \( \frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2(p^2 + q^2) \).
4. So the required value is −2.

Teacher's Note:
1. Cube both sides instead of working with the cube root.
2. Note the minus sign: z = x − iy, so y is the negative of the imaginary part.

Answer: (a) 6

Explanation:
1. |z + 1| = |(z + 4) − 3| ≤ |z + 4| + 3.
2. ≤ 3 + 3 = 6.
3. At z = −7, |z + 4| = 3 and |z + 1| = 6, so the maximum is 6.

Teacher's Note:
1. Rewrite the required modulus in terms of the given one and apply the triangle inequality.
2. Check that the bound is attained; z = −7 lies on the boundary of the given region.

Answer: (d) \( \frac{-1}{i + 1} \)

Explanation:
1. The required number is the conjugate of \( \frac{1}{i - 1} \).
2. \( \overline{\left(\frac{1}{i - 1}\right)} = \frac{1}{-i - 1} \).
3. \( = \frac{-1}{i + 1} \).

Teacher's Note:
1. The conjugate of a conjugate is the number itself, so just take the conjugate of the given number.
2. To conjugate a quotient, replace i by −i everywhere.

Answer: (b) \( \sqrt{5} + 1 \)

Explanation:
1. Let r = |z|. \( |z| = \left| \left(z - \frac{4}{z}\right) + \frac{4}{z} \right| \le 2 + \frac{4}{r} \).
2. So r² − 2r − 4 ≤ 0, giving r ≤ 1 + √5.
3. z = 1 + √5 gives z − 4/z = (1 + √5) − (√5 − 1) = 2, so the maximum |z| = √5 + 1.

Teacher's Note:
1. Write z as (z − 4/z) + 4/z and apply the triangle inequality to get an inequality in r = |z|.
2. Solve the quadratic inequality r² − 2r − 4 ≤ 0 and keep the positive bound.

Answer: (c) β ∈ (1, ∞)

Explanation:
1. Real coefficients and distinct roots with the same real part 1 mean the roots are 1 + iy and 1 − iy (y ≠ 0).
2. Product of roots β = (1 + iy)(1 − iy) = 1 + y².
3. Since y ≠ 0, β > 1, i.e. β ∈ (1, ∞).

Teacher's Note:
1. Non-real roots of a quadratic with real coefficients are conjugates of each other.
2. Two distinct real roots cannot both have real part 1, so the roots must be non-real.

Answer: (c) lies in the interval (1, 2)

Explanation:
1. \( \left| z + \frac{1}{2} \right| \ge |z| - \frac{1}{2} \ge 2 - \frac{1}{2} = \frac{3}{2} \).
2. At z = −2, \( \left| -2 + \frac{1}{2} \right| = \frac{3}{2} \), so the minimum is 3/2.
3. 3/2 lies in the interval (1, 2).

Teacher's Note:
1. Use |z₁ + z₂| ≥ |z₁| − |z₂| for a lower bound.
2. The minimum is reached at the point of |z| = 2 nearest to −1/2, which is z = −2.

Answer: (d) {z : |z| = 1, z ≠ 1}

Explanation:
1. Rearranging, \( z(k - \bar{w}) = k - w \), so \( z = \frac{k - w}{k - \bar{w}} \).
2. k is real, so \( k - \bar{w} = \overline{k - w} \); hence |z| = 1.
3. z = 1 would need w = \( \bar{w} \), i.e. Im w = 0, which is not allowed; so z ≠ 1.

Teacher's Note:
1. A quotient of a number and its own conjugate always has modulus 1.
2. Check the excluded value separately: z = 1 forces w to be real.

Answer: (c) a non-zero real number

Explanation:
1. \( \frac{z - i}{z + i} = \frac{(z - i)(\bar{z} - i)}{|z + i|^2} \) and \( (z - i)(\bar{z} - i) = |z|^2 - 1 - i(z + \bar{z}) \).
2. Purely imaginary means |z|² − 1 = 0, so |z| = 1 and \( \frac{1}{z} = \bar{z} \).
3. \( z + \frac{1}{z} = z + \bar{z} = 2\,\text{Re}(z) \), which is real.
4. Re z = 0 only for z = ±i, which make the fraction 0 or undefined; so z + 1/z is a non-zero real number.

Teacher's Note:
1. On the unit circle, 1/z = z̄, so z + 1/z = 2Re(z) is always real.
2. z + 1/z can equal 1 (at z = 1/2 + (√3/2)i), so option (b) is wrong.

Answer: (b) y² + 4x − 4 = 0

Explanation:
1. z² = (1 + iα)² = 1 − α² + 2iα, so x = 1 − α² and y = 2α.
2. α = y/2, so x = 1 − y²/4.
3. 4x = 4 − y², i.e. y² + 4x − 4 = 0.

Teacher's Note:
1. Expand the square and compare real and imaginary parts.
2. Eliminate the parameter α to get the relation between x and y.

Answer: (b) 2

Explanation:
1. |z₂| ≤ |z₂ − (3 + 4i)| + |3 + 4i| = 5 + 5 = 10.
2. |z₁ − z₂| ≥ |z₁| − |z₂| ≥ 12 − 10 = 2.
3. Equality holds for z₂ = 6 + 8i and z₁ = \( \frac{12}{10} \)(6 + 8i) = 7.2 + 9.6i, so the minimum is 2.

Teacher's Note:
1. First find the largest possible |z₂| using the triangle inequality.
2. Both points lie on the same ray from the origin when the minimum distance is reached.

Answer: (c) \( \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \)

Explanation:
1. Multiply by the conjugate: \( \frac{(2 + 3i\sin\theta)(1 + 2i\sin\theta)}{1 + 4\sin^2\theta} = \frac{2 - 6\sin^2\theta + 7i\sin\theta}{1 + 4\sin^2\theta} \).
2. Purely imaginary requires 2 − 6sin²θ = 0, so sin²θ = 1/3.
3. sin θ = 1/√3, i.e. θ = \( \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

Teacher's Note:
1. A number is purely imaginary when its real part is zero (and the imaginary part is non-zero).
2. Rationalise with the conjugate of the denominator to separate the real part.

Answer: (b) 1 + i

Explanation:
1. 1 unit east: 2 + i + 1 = 3 + i.
2. 2 units north: 3 + i + 2i = 3 + 3i.
3. \( 2\sqrt{2} \) units south-west means 2 units west and 2 units south: 3 + 3i − 2 − 2i = 1 + i.

Teacher's Note:
1. East adds to the real part and north adds to the imaginary part.
2. A distance d along a 45° direction splits into d/√2 along each axis.

Answer: (d) \( -1365\sqrt{3}\,i \)

Explanation:
1. z³ = 1 + 3ai − 3a² − a³i; imaginary part 3a − a³ = 0 gives a = √3 (a > 0), and z³ = 1 − 9 = −8.
2. Sum of the GP = \( \frac{z^{12} - 1}{z - 1} \), with z¹² = (z³)⁴ = 4096 and z − 1 = \( \sqrt{3}\,i \).
3. Sum = \( \frac{4095}{\sqrt{3}\,i} = \frac{-4095\,i}{\sqrt{3}} = -1365\sqrt{3}\,i \).

Teacher's Note:
1. The sum of a GP formula works for complex ratios too: (zⁿ − 1)/(z − 1).
2. Finding z³ = −8 first makes z¹² = 4096 easy.

Answer: (b) {0}

Explanation:
1. Write \( w = \frac{1 + z}{1 - z} - \frac{8\alpha z}{1 - z} \).
2. For |z| = 1, \( \frac{1 + z}{1 - z} \) is purely imaginary, and \( \text{Re}\left(\frac{z}{1 - z}\right) = -\frac{1}{2} \).
3. So Re(w) = 0 + 4α = 4α.
4. w purely imaginary for all such z needs 4α = 0, so α = 0.

Teacher's Note:
1. For |z| = 1, (1 + z)/(1 − z) is always purely imaginary; split w to use this fact.
2. Re(z/(1 − z)) = −1/2 on the unit circle follows from z z̄ = 1.

Answer: (c) \( 4 + \sqrt{13} \)

Explanation:
1. |z| ≤ |z − (3 − 2i)| + |3 − 2i| ≤ 4 + √13, with equality at a suitable z.
2. |0 − 3 + 2i| = √13 < 4, so z = 0 satisfies the condition; least |z| = 0.
3. Difference = (4 + √13) − 0 = 4 + √13.

Teacher's Note:
1. Check whether the origin itself satisfies the condition; if so, the least value of |z| is 0.
2. The greatest value comes from the triangle inequality with the fixed point 3 − 2i.

Answer: (d) None of these

Explanation:
1. \( \frac{1 + i}{1 - i} = \frac{(1 + i)^2}{2} = i \).
2. iⁿ = 1 first happens at n = 4 (i¹ = i, i² = −1, i³ = −i, i⁴ = 1).
3. 4 is not among 8, 16, 12, so the answer is None of these.

Teacher's Note:
1. Simplify the base to i before thinking about powers.
2. 8, 16 and 12 also satisfy iⁿ = 1, but the question asks for the smallest such n.

Answer: (d) no value of x

Explanation:
1. Conjugate of sin x + i cos 2x is sin x − i cos 2x; equating with cos x − i sin 2x gives sin x = cos x and cos 2x = sin 2x.
2. sin x = cos x gives cos 2x = cos²x − sin²x = 0.
3. Also sin²x = cos²x = 1/2, so sin 2x = 2 sin x cos x = 2 sin²x = 1.
4. Then cos 2x = sin 2x would mean 0 = 1, which is impossible; so there is no value of x.

Teacher's Note:
1. Equate real parts and imaginary parts separately to get two real equations.
2. Use the first equation to evaluate both sides of the second instead of solving it in general.

Answer: (c) 1 or −1

Explanation:
1. 2 = |z + iw| ≤ |z| + |w| ≤ 2, so |z| = |w| = 1 and z, iw point in the same direction, i.e. z = iw.
2. Similarly 2 = |z − i\( \bar{w} \)| forces z = −i\( \bar{w} \).
3. So iw = −i\( \bar{w} \), i.e. w = −\( \bar{w} \): w is purely imaginary with |w| = 1, so w = ±i.
4. Then z = iw = i(±i) = ∓1, so z = 1 or −1.

Teacher's Note:
1. Equality in |a + b| ≤ |a| + |b| holds only when a and b point in the same direction.
2. Since both moduli are at most 1, a sum of modulus 2 forces both to have modulus exactly 1.

Answer: (d) n₁ > 0, n₂ > 0

Explanation:
1. i³ = −i, i⁵ = i, i⁷ = −i.
2. Expression = [(1 + i)^n₁ + (1 − i)^n₁] + [(1 + i)^n₂ + (1 − i)^n₂].
3. Each bracket is of the form \( w + \bar{w} = 2\,\text{Re}(w) \), which is real.
4. So the expression is real for all positive integers n₁, n₂.

Teacher's Note:
1. (1 − i)ⁿ is the conjugate of (1 + i)ⁿ, and a number plus its conjugate is always real.
2. No relation between n₁ and n₂ is needed, which is why option (d) is correct.

Answer: (b) i − 1

Explanation:
1. \( \sum_{n=1}^{13} (i^n + i^{n+1}) = (1 + i)\sum_{n=1}^{13} i^n \).
2. Every 4 consecutive powers of i add to 0, so i¹ + ... + i¹² = 0 and the sum is i¹³ = i.
3. Value = (1 + i)i = i + i² = i − 1.

Teacher's Note:
1. Any four consecutive powers of i add up to zero: i + i² + i³ + i⁴ = 0.
2. Take the common factor iⁿ(1 + i) out of each term first.

Answer: (a) equal to 1

Explanation:
1. |z_k| = 1 gives \( \frac{1}{z_k} = \bar{z}_k \).
2. So \( 1 = \left| \bar{z}_1 + \bar{z}_2 + \bar{z}_3 \right| = |\overline{z_1 + z_2 + z_3}| \).
3. Hence |z₁ + z₂ + z₃| = 1.

Teacher's Note:
1. On the unit circle the reciprocal of a number is its conjugate.
2. |z̄| = |z| for every complex number z.

Answer: (a) 0

Explanation:
1. \( w = \frac{(z - 1)(\bar{z} + 1)}{|z + 1|^2} = \frac{z\bar{z} + z - \bar{z} - 1}{|z + 1|^2} \).
2. With \( z\bar{z} = 1 \), the numerator is \( z - \bar{z} = 2i\,\text{Im}(z) \), which is purely imaginary.
3. So Re(w) = 0.

Teacher's Note:
1. Multiply numerator and denominator by the conjugate of the denominator to make the denominator real.
2. z z̄ = |z|² = 1 cancels the real terms in the numerator.

Answer: (a) 48

Explanation:
1. \( z\bar{z}(\bar{z}^2 + z^2) = 350 \), with \( z\bar{z} = x^2 + y^2 \) and \( z^2 + \bar{z}^2 = 2(x^2 - y^2) \).
2. So (x² + y²)(x² − y²) = 175 = 25 × 7 (other factor pairs give no integer squares).
3. x² + y² = 25, x² − y² = 7 gives x² = 16, y² = 9, so z = ±4 ± 3i.
4. The rectangle has sides 8 and 6, so its area is 48.

Teacher's Note:
1. Write z z̄ and z² + z̄² in terms of x and y to get a real equation.
2. With integers, try factor pairs of 175 where both x² and y² come out as perfect squares.

Answer: (d) \( \frac{3}{4} \)

Explanation:
1. z satisfies the quadratic z² + z + (1 − a) = 0 with real coefficients.
2. Since z is non-real, the discriminant is negative: 1 − 4(1 − a) < 0.
3. 4a − 3 < 0, so a < 3/4; hence a cannot be 3/4.

Teacher's Note:
1. A quadratic with real coefficients has non-real roots exactly when its discriminant is negative.
2. At a = 3/4 the discriminant is zero and the root z = −1/2 is real, which is not allowed.

Answer: (a) |w₁| = 1; (b) |w₂| = 1; (c) \( \text{Re}(w_1\bar{w}_2) = 0 \)

Explanation:
1. Given a² + b² = 1, c² + d² = 1 and Re(z₁\( \bar{z}_2 \)) = ac + bd = 0.
2. Let a = cos α, b = sin α. ac + bd = 0 with c² + d² = 1 gives (c, d) = ±(−sin α, cos α).
3. Then a² + c² = cos²α + sin²α = 1 and b² + d² = 1, so |w₁| = |w₂| = 1.
4. Re(w₁\( \bar{w}_2 \)) = ab + cd = cos α sin α − sin α cos α = 0.

Teacher's Note:
1. Re(z₁z̄₂) = ac + bd, just like a dot product of (a, b) and (c, d).
2. Writing a = cos α, b = sin α turns the conditions into simple trigonometric identities.

Answer: (a) zero; (d) purely imaginary

Explanation:
1. \( \frac{z_1 + z_2}{z_1 - z_2} = \frac{(z_1 + z_2)(\bar{z}_1 - \bar{z}_2)}{|z_1 - z_2|^2} = \frac{|z_1|^2 - |z_2|^2 + (z_2\bar{z}_1 - z_1\bar{z}_2)}{|z_1 - z_2|^2} \).
2. |z₁| = |z₂| removes the real part, and \( z_2\bar{z}_1 - z_1\bar{z}_2 \) is purely imaginary; so the value is purely imaginary.
3. It is zero when z₂ = −z₁, e.g. z₁ = 1 + i, z₂ = −1 − i, which fits the given conditions.
4. Hence it may be zero or purely imaginary.

Teacher's Note:
1. A number minus its own conjugate (w − w̄) is always purely imaginary.
2. Test a simple example such as z₂ = −z₁ to see whether the value zero is possible.

Answer: (a) \( -1 + \sqrt{1 - y^2} \); (d) \( -1 - \sqrt{1 - y^2} \)

Explanation:
1. \( \frac{az + b}{z + 1} = \frac{(az + b)(\bar{z} + 1)}{|z + 1|^2} \); its imaginary part is \( \frac{(a - b)y}{(x + 1)^2 + y^2} = \frac{y}{(x + 1)^2 + y^2} \).
2. Setting this equal to y (y ≠ 0): (x + 1)² + y² = 1.
3. (x + 1)² = 1 − y², so \( x = -1 \pm \sqrt{1 - y^2} \).

Teacher's Note:
1. Multiply by the conjugate of z + 1 and keep only the imaginary part of the numerator.
2. The condition a − b = 1 is exactly what makes the imaginary part simplify to y/|z + 1|².

Answer: 5

Explanation:
1. 2z − 6 + 5i = 2(z − 3 − 2i) + 9i.
2. |2z − 6 + 5i| ≥ |9i| − 2|z − 3 − 2i| ≥ 9 − 2 × 2 = 5.
3. At z = 3 (here |z − 3 − 2i| = |−2i| = 2), |2z − 6 + 5i| = |5i| = 5.
4. So the minimum value is 5.

Teacher's Note:
1. Express the required quantity in terms of the given one, z − 3 − 2i, plus a constant.
2. Then use |A + B| ≥ |B| − |A| to get the lower bound and check it is attained.