JEE Mathematics Complex Numbers MCQs Set 29

Answer: (c) A, B, C and D (all of them)

Explanation:
1. Since 1 + ki = i(k − i), each factor of the numerator over the matching factor of the denominator equals i.
2. So Z = iⁿ, which takes the values i, −1, −i, 1 for n = 1, 2, 3, 4 (and repeats).
3. Principal arguments π/2, π, −π/2, 0 all occur.

Teacher's Note:
1. Note 1 + ki = i(k − i).
2. Powers of i have period 4.

Answer: (d) A, B and C only

Explanation:
1. |z − α|² = k²|z − β|² gives (k² − 1)zz̄ − (k²β̄ − ᾱ)z − (k²β − α)z̄ + … = 0, a circle with centre (k²β − α)/(k² − 1) (A).
2. Its radius is k|α − β|/|1 − k²| (B).
3. For k = 1 the locus is the perpendicular bisector (C). (D) swaps α and β and is wrong.

Teacher's Note:
1. Expand |z − α|² = k²|z − β|².
2. Complete the square.

Answer: (a) \( b+ia \)

Explanation:
1. Multiply by (1 + b + ia)/(1 + b + ia): denominator (1 + b)² + a² = 2 + 2b.
2. Numerator (1 + b)² − a² + 2ia(1 + b) = 2b(1 + b) + 2ia(1 + b), using a² = 1 − b².
3. The ratio is b + ia.

Teacher's Note:
1. Rationalise using the conjugate of the denominator.
2. Replace a² by 1 − b².

Answer: (b) \( \cos\frac\pi4+i\sin\frac\pi4 \)

Explanation:
1. z⁴ = −1 = e^{iπ}, so z = e^{i(π + 2kπ)/4}, i.e. angles π/4, 3π/4, 5π/4, 7π/4.
2. Only the angle π/4 appears among the options.

Teacher's Note:
1. Write −1 in polar form.
2. Divide the argument by 4.

Answer: (c) A, C and D only

Explanation:
1. The foci are ±i, 2c = 2. For 0 < k < 2 the locus is a hyperbola (A); for k = 0 it is the perpendicular bisector y = 0, a line (C).
2. For k = 2 the two rays along the y-axis, |y| ≥ 1 (D).
3. For k > 2 no point can satisfy it, as a difference of distances cannot exceed the distance between the foci (B false).

Teacher's Note:
1. The difference of the distances is at most |z_1 − z_2|.
2. Treat the boundary cases k = 0 and k = 2 separately.

Answer: (d) B, C and D only

Explanation:
1. (P²)_{ik} = ∑_{j=1}^{n} ω^{i+j}ω^{j+k} = ω^{i+k}∑_{j=1}^{n} ω^{2j}.
2. ∑_{j=1}^{n} ω^{2j} = 0 exactly when 3 divides n (ω² is also a primitive cube root of unity).
3. 57 is a multiple of 3, so P² = 0; 55, 56, 58 are not, so P² ≠ 0.

Teacher's Note:
1. Compute a general entry of P².
2. Sum the geometric progression in ω².

Answer: (a) 1

Explanation:
1. z = ω is a primitive cube root of unity. The (1,2) entry of P² is z^{2s}[(−z)^r + z^r], which is 0 only when r is odd (r = 1 or 3).
2. The (1,1) entry is (−ω)^{2r} + ω^{4s} = ω^{2r} + ω^{s}. For r = 3 this is 1 + ω^s, which is never −1 (it would need ω^s = −2). For r = 1 it is ω² + ω^s = −1 only when ω^s = ω, i.e. s = 1.
3. The (2,2) entry is the same as the (1,1) entry, so only (r, s) = (1, 1) works: 1 pair (confirmed numerically for all nine pairs).

Teacher's Note:
1. Compute P² entry by entry.
2. Test all nine pairs.

Answer: (b) 0

Explanation:
1. Equal moduli and a relative argument π give z_1 = −z_2.
2. Hence z_1 + z_2 = 0.

Teacher's Note:
1. z_1/z_2 has modulus 1 and argument π.
2. That ratio is −1.

Answer: (c) the interior and boundary of an ellipse

Explanation:
1. |z − 1| + |z + 1| = 4 is an ellipse with foci ±1 and 2a = 4 > 2 (distance between foci).
2. ≤ includes the interior and the boundary.

Teacher's Note:
1. The sum of distances from two fixed points defines an ellipse.
2. The inequality is not strict.

Answer: (d) \( \frac53 \)

Explanation:
1. Comparing imaginary parts, Im z = 1. Let z = x + i; then √(x² + 1) + x = 3.
2. √(x² + 1) = 3 − x gives x² + 1 = 9 − 6x + x², so x = 4/3.
3. |z| = 3 − 4/3 = 5/3.

Teacher's Note:
1. |z| is real, so the imaginary part of z is 1.
2. Isolate the square root and square.

Answer: (a) 54

Explanation:
1. x is a primitive cube root of unity, so x^k + x^{−k} = 2 when 3 divides k and −1 otherwise.
2. Among k = 1, …, 27 there are 9 multiples of 3 (each term 4) and 18 others (each term 1).
3. Sum = 36 + 18 = 54.

Teacher's Note:
1. Use x³ = 1.
2. Count the multiples of 3 up to 27.

Answer: (b) \( |w|_{\min}=\frac1e,\ |w|_{\max}=e \)

Explanation:
1. For z = x + iy, |e^{iz}| = |e^{ix − y}| = e^{−y}.
2. On the segment DM the imaginary part y runs from −1 (at M) to 1 (at D).
3. So |w| ranges from e^{−1} to e^{1}.

Teacher's Note:
1. Write z = x + iy.
2. Find the range of y on the segment.

Answer: (c) 2

Explanation:
1. Put C at the origin: z_3 = 0, z_1 = a, z_2 = ia (a ≠ 0). Then (z_1 − z_2)² = a²(1 − i)² = −2ia².
2. (z_1 − z_3)(z_3 − z_2) = a(−ia) = −ia².
3. So k = 2 (the other orientation z_2 = −ia also gives 2).

Teacher's Note:
1. The legs CA and CB are perpendicular and equal, so z_2 − z_3 = ±i(z_1 − z_3).
2. Substitute into both sides.

Answer: (d) B and D only

Explanation:
1. With b = a − 1: (az + b)/(z + 1) = a − 1/(z + 1).
2. Im(−1/(z + 1)) = y/|z + 1|², so the condition gives |z + 1|² = 1 (y ≠ 0), i.e. (x + 1)² + y² = 1.
3. So x = −1 ± √(1 − y²): options B and D.

Teacher's Note:
1. Rewrite (az + b)/(z + 1) using b = a − 1.
2. The imaginary part of 1/w is −Im(w)/|w|².

Answer: (a) 3

Explanation:
1. |a + bω + cω²|² = a² + b² + c² − ab − bc − ca = (1/2)[(a − b)² + (b − c)² + (c − a)²].
2. For distinct integers the differences are at least 1 in absolute value and cannot all be 1; the least value is for (a, b, c) = (1, 2, 3): (1/2)(1 + 1 + 4) = 3.

Teacher's Note:
1. Use the identity a² + b² + c² − ab − bc − ca.
2. Two of the differences can be 1 but then the third is 2.

Answer: (b) the single point z = 0

Explanation:
1. |z − 1|² + |z + 1|² = 2|z|² + 2 (parallelogram law).
2. So 2|z|² + 2 = 2 gives |z| = 0: only the origin.

Teacher's Note:
1. Expand using |z ± 1|² = |z|² ± 2Re z + 1.
2. The cross terms cancel.

Answer: (c) \( b^2-b=30 \)

Explanation:
1. The roots are −b/2 ± iy with |z|² = 45, so y² = 45 − b²/4.
2. |z + 1|² = (1 − b/2)² + y² = 1 − b + b²/4 + 45 − b²/4 = 46 − b = 40, so b = 6.
3. b² − b = 36 − 6 = 30.

Teacher's Note:
1. Use |z|² = product of conjugate roots = 45.
2. Expand |z + 1|².

Answer: (d) the x-axis

Explanation:
1. |z − 5i| = |z + 5i| means z is equidistant from 5i and −5i.
2. That is the perpendicular bisector, the real axis (y = 0).

Teacher's Note:
1. Equal distances from two points give a perpendicular bisector.
2. The points 5i and −5i are symmetric about the real axis.

Answer: (a) 0

Explanation:
1. z_2 lies on a circle with centre 3 + 4i (distance 5 from O) and radius 4, so |z_2| ranges from 1 to 9.
2. The point z_2 with |z_2| = 9 (at z_2 = (9/5)(3 + 4i)) lies on the circle |z_1| = 9, so z_1 can equal z_2.
3. The minimum is 0.

Teacher's Note:
1. Find the range of |z_2|.
2. The two circles touch internally.

Answer: (b) 10

Explanation:
1. Let z = r e^{iα} and ω = (1/r)e^{iβ} with α − β = π/2.
2. z̄ω = e^{−i(α − β)} = e^{−iπ/2} = −i.
3. 10i · (−i) = 10.

Teacher's Note:
1. Write z and ω in polar form.
2. Use |ω| = 1/|z|.

Answer: (c) does not exist

Explanation:
1. Equal arguments mean the vectors z − 1 and z + 3 point in the same direction from the origin.
2. Then (x − 1)/y = (x + 3)/y, which is impossible for y ≠ 0; for y = 0 the ratio (x − 1) : y is undefined.
3. So the ratio does not exist.

Teacher's Note:
1. Write tan of both arguments.
2. Check the case y = 0 separately.

Answer: (d) exactly one solution

Explanation:
1. |z + 1 − i| = 2 is the circle with centre (−1, 1) and radius 2.
2. Its largest real part is −1 + 2 = 1, attained only at the point (1, 1).
3. So Re z ≥ 1 holds at exactly one point of the circle.

Teacher's Note:
1. Find the largest x on the circle.
2. Compare it with the condition x ≥ 1.

Answer: (a) 2

Explanation:
1. 2x = 2|z − 1| gives x² = (x − 1)² + y², i.e. y² = 2x − 1, a parabola.
2. The chord with slope tan(π/4) = 1 between (x_1, y_1) and (x_2, y_2): (y_1 − y_2)/(x_1 − x_2) = 2/(y_1 + y_2) = 1.
3. So y_1 + y_2 = 2: the imaginary part of z_1 + z_2 is 2.

Teacher's Note:
1. Find the locus first.
2. Use x = (y² + 1)/2 to get the slope of the chord.

Answer: (b) \( \frac{5\pi}{6} \)

Explanation:
1. The line through the points is y = 4x + 17; it crosses the negative real axis at x = −17/4, inside the segment.
2. The argument of z_1 is about 121° and that of z_2 is about −149° (i.e. 211°); points of the segment have arguments between 121° and 211°.
3. 5π/6 = 150° lies in this range; −135°, −45° and 30° do not.

Teacher's Note:
1. Find the arguments of the end points.
2. The argument varies continuously along the segment.

Answer: (c) A, B and C only

Explanation:
1. ∑rα^r = n/(α − 1) for r = 1, …, n − 1 and ∑α^r = −1, so ∑(n − r)α^r = −n − n/(α − 1) = nα/(1 − α) (A).
2. With α = e^{2πi/n}: α/(1 − α) = −1/2 + (i/2)cot(π/n), so real part −n/2 (C) and imaginary part (n/2)cot(π/n) (B).
3. (D) differs from (A) by the factor α and is false. (Checked numerically for n = 7.)

Teacher's Note:
1. Use the sum of r α^r for roots of unity.
2. Take real and imaginary parts.