JEE Mathematics Complex Numbers MCQs Set 22

Answer: (c) 1

Explanation:
1. Put \( x=\cot\theta \). Then \( xi+1=\frac{\sin\theta+i\cos\theta}{\sin\theta}=\frac{ie^{-i\theta}}{\sin\theta} \) and \( xi-1=\frac{ie^{i\theta}}{\sin\theta} \).
2. So \( \frac{xi+1}{xi-1}=e^{-2i\theta} \) and the power gives \( e^{-2im\theta} \).
3. Multiplying by \( e^{2im\theta} \) gives 1.

Teacher's Note:
1. Substitute x = cot θ so that x i ± 1 become exponentials.
2. The exponents cancel.

Answer: (d) none of these

Explanation:
1. Let z = x + iy. Then \( z-ix=x+i(y-x) \) and \( z-y=(x-y)+iy \).
2. Equate the squared moduli: \( x^2+(y-x)^2=(x-y)^2+y^2\Rightarrow x^2=y^2 \).
3. So z lies on the lines y = ±x, which is none of (a)-(c).

Teacher's Note:
1. Write z = x + iy and square both sides.
2. x² = y² represents a pair of lines.

Answer: (d) \( (0,3) \)

Explanation:
1. The locus is an ellipse with foci a² and 2a (on the real axis) if \( |a^2-2a|<3 \).
2. \( -33. With a > 0, a ∈ (0, 3).

Teacher's Note:
1. An ellipse needs the sum of distances to exceed the distance between the foci.
2. Solve |a² − 2a| < 3.

Answer: (c) −i

Explanation:
1. \( x_r=e^{-i\pi/3^r} \).
2. The infinite product is \( e^{-i\pi\left(\frac13+\frac19+\cdots\right)}=e^{-i\pi/2}=-i \).

Teacher's Note:
1. Write each factor in exponential form.
2. Sum the geometric series in the exponent: 1/3 + 1/9 + ... = 1/2.

Answer: (b) \( |z-(5-i)|=\sqrt5 \)

Explanation:
1. The angle between z − z₁ and z₂ − z is a right angle, so z lies on the circle with z₁z₂ as diameter.
2. Centre \( \frac{z_1+z_2}2=5-i \) and radius \( \frac{|z_1-z_2|}2=\frac{|2+4i|}2=\sqrt5 \).
3. So \( |z-(5-i)|=\sqrt5 \).

Teacher's Note:
1. A right angle subtended at z means z is on the circle with z₁z₂ as diameter.
2. Compute the midpoint and half the distance.

Answer: (a) 0

Explanation:
1. |z| is a positive real number, whose argument is 0.
2. So the integrand [0] = 0 and the integral is 0.

Teacher's Note:
1. The modulus is a positive real number.
2. The argument of a positive real is 0.

Answer: (d) no point

Explanation:
1. The first curve is the ray from 3i making 135° with the positive real axis: \( z=-t+i(3+t),\ t>0 \).
2. Since \( \arg(2z+1-2i)=\arg\left(z-\left(-\frac12+i\right)\right) \), the second is the ray from \( -\frac12+i \) at 45°: \( z=-\frac12+s+i(1+s),\ s>0 \).
3. Equating: \( -t=-\frac12+s \) and \( 3+t=1+s \) give \( t=-\frac34<0 \), which is not allowed: the rays do not meet.

Teacher's Note:
1. Write each argument condition as a ray from a point.
2. Check that the parameters are positive.

Answer: (b) 2

Explanation:
1. z₁ lies on the circle of radius 12 about the origin. z₂ lies on the circle of radius 5 about 3 + 4i, whose centre is at distance 5 from the origin.
2. The farthest point of the second circle from the origin is at distance 5 + 5 = 10 < 12, so the second circle lies inside the first.
3. The minimum distance between points of the two circles is \( 12-10=2 \).

Teacher's Note:
1. Find the distance from the origin to the centre of the second circle.
2. The smaller circle is entirely inside the larger one.

Answer: (b) 7

Explanation:
1. \( |iz+z_1|=|i|\,|z+z_1/i|=|z-iz_1| \) and \( iz_1=-3+5i \).
2. So we need the maximum distance from a point of the disc \( |z-i|\le2 \) to the point \( -3+5i \).
3. The distance from the centre i to −3 + 5i is \( |3-4i|=5 \), so the maximum is 5 + 2 = 7.

Teacher's Note:
1. Factor out i to rewrite the expression as a distance.
2. Maximum distance to a disc = distance to centre + radius.

Answer: (c) \( [1,9] \)

Explanation:
1. The region is the interior of the ellipse with foci 1 and −3 and major axis 8: centre −1, semi-major axis 4, vertices at 3 and −5.
2. The point 4 lies on the major axis, so the nearest point of the region is the vertex 3 (distance 1) and the farthest is the vertex −5 (distance 9).
3. Hence \( 1\le|z-4|\le9 \).

Teacher's Note:
1. Identify the region as an ellipse with foci 1 and −3.
2. The extreme distances from a point on the major axis occur at the vertices.

Answer: (b) 2

Explanation:
1. Put z = x + iy: \( x^2+y^2+7x-7iy=0 \). The imaginary part gives y = 0.
2. Then \( x^2+7x=0 \), so x = 0 or x = −7.
3. Two solutions: z = 0 and z = −7.

Teacher's Note:
1. Separate real and imaginary parts.
2. y = 0 from the imaginary part.

Answer: (d) 0

Explanation:
1. \( (1+i)^2=2i \), so \( (1+i)^6=(2i)^3=-8i \) and \( (1-i)^6=(-2i)^3=8i \).
2. Their sum is 0.

Teacher's Note:
1. Square first: (1 + i)² = 2i.
2. The two terms are conjugates.

Answer: (c) \( \frac{9^n-1}8 \)

Explanation:
1. \( x^n-1=(x-1)(x-\omega)\cdots(x-\omega^{n-1}) \).
2. So \( (x-\omega)\cdots(x-\omega^{n-1})=\frac{x^n-1}{x-1} \). Put x = 9: \( \frac{9^n-1}{8} \).

Teacher's Note:
1. Factorise xⁿ − 1 over the roots of unity.
2. Remove the factor (x − 1) and put x = 9.

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

Explanation:
1. The principal arguments satisfy \( -\pi<\arg z\le\pi \) (R), so \( \arg z_1+\arg z_2\le2\pi \) with equality only when both are π.
2. Hence both z₁ and z₂ are negative real numbers, in particular purely real: A is true, and R is the reason.

Teacher's Note:
1. Use the range of the principal argument.
2. The sum of two principal arguments reaches 2π only at the maximum π + π.

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

Explanation:
1. \( \left|\frac1{z_1}+\frac1{z_2}\right|=\frac{|z_1+z_2|}{|z_1z_2|} \), so \( |z_1z_2|=1 \) because |z₁ + z₂| ≠ 0: A is true.
2. R is false: only the product has modulus 1, e.g. z₁ = 2 and z₂ = 1/2 satisfy the condition but are not unimodular.

Teacher's Note:
1. Rewrite 1/z₁ + 1/z₂ as (z₁ + z₂)/(z₁z₂).
2. A counter-example disproves R.