JEE Mathematics Complex Numbers MCQs Set 14

Answer: (c) \( \cos\theta\pm i\sin\theta \)

Explanation:
1. The equation is \( x^2-2x\cos\theta+1=0 \).
2. By the quadratic formula \( x=\cos\theta\pm\sqrt{\cos^2\theta-1}=\cos\theta\pm i\sin\theta \).

Teacher's Note:
1. Both roots are valid, and they are reciprocals of each other.
2. This is the standard way of seeing that e^(iθ) + e^(−iθ) = 2cosθ.

Answer: (a) \( x=0.5,\ y=3.5 \)

Explanation:
1. Equate real parts: \( 9^x=3 \), so \( 3^{2x}=3^1 \) and \( x=\tfrac12 \).
2. Equate imaginary parts: \( -2y=-7 \), so \( y=3.5 \).

Teacher's Note:
1. Equality of complex numbers means equality of real and imaginary parts separately.
2. Both unknowns are real, so option (d) cannot be right.

Answer: (b) \( x=3,\ y=-1 \)

Explanation:
1. Multiply through by \( (3+i)(3-i)=10 \): \( (3-i)[(1+i)x-2i]+(3+i)[(2-3i)y+i]=10i \).
2. Solving the resulting real and imaginary equations, or testing the options, is quickest.
3. Substituting \( x=3,\ y=-1 \): first fraction \( \frac{3+3i-2i}{3+i}=\frac{3+i}{3+i}=1 \) and second \( \frac{-(2-3i)+i}{3-i}=\frac{-2+4i}{3-i}=-1+i \), total \( i \).

Teacher's Note:
1. When elimination is clumsy, test the offered options.
2. Option (b) satisfies the equation; the others do not.

Answer: (b) 4x-3

Explanation:
1. \( x^2+y^2=|x+iy|^2=\frac{9}{(2+\cos\theta)^2+\sin^2\theta}=\frac{9}{5+4\cos\theta} \).
2. Also \( x=\frac{3(2+\cos\theta)}{5+4\cos\theta} \), so \( 4x-3=\frac{12(2+\cos\theta)-3(5+4\cos\theta)}{5+4\cos\theta}=\frac{9}{5+4\cos\theta} \).
3. Hence \( x^2+y^2=4x-3 \).

Teacher's Note:
1. x² + y² is the square of the modulus.
2. Express both x² + y² and x in terms of cosθ and compare.

Answer: (b) -160

Explanation:
1. \( x=-5+4i \), so \( (x+5)^2=-16 \) and \( x^2+10x+41=0 \).
2. Divide: \( x^4+9x^3+35x^2-x+4=(x^2+10x+41)(x^2-x+4)-160 \).
3. At the given x the bracket vanishes, leaving \( -160 \).

Teacher's Note:
1. Build the quadratic with x as a root and divide the polynomial by it.
2. The remainder is the required value.

Answer: (b) \( a=1,\ b=0 \)

Explanation:
1. \( \frac{1-i}{1+i}=-i \).
2. \( (-i)^{100}=1 \) since 100 is a multiple of 4.
3. So \( a=1,\ b=0 \).

Teacher's Note:
1. Reduce the base first.
2. (−i)^4 = 1.

Answer: (c) \( x=\tfrac5{13},\ y=\tfrac{14}{13} \)

Explanation:
1. \( x+iy=\frac{4+i}{2-3i}=\frac{(4+i)(2+3i)}{13}=\frac{8+12i+2i-3}{13}=\frac{5+14i}{13} \).
2. So \( x=\tfrac5{13},\ y=\tfrac{14}{13} \).

Teacher's Note:
1. Divide by (2 − 3i) by multiplying with its conjugate.
2. |2 − 3i|² = 13.

Answer: (d) Inequality

Explanation:
1. Complex numbers can be added, subtracted, multiplied and divided (by a non-zero number).
2. There is no ordering on complex numbers consistent with the field operations, so inequalities such as \( 2i>1 \) are meaningless.

Teacher's Note:
1. Only real numbers (or moduli of complex numbers) can be compared by size.
2. A complex number can be zero or not, but it is never positive or negative.

Answer: (b) \( \dfrac{x}{x^2+y^2}-\dfrac{y}{x^2+y^2}i \)

Explanation:
1. \( \frac1{x+iy}=\frac{x-iy}{(x+iy)(x-iy)}=\frac{x-iy}{x^2+y^2} \).
2. This is option (b).

Teacher's Note:
1. The inverse equals the conjugate divided by the square of the modulus.
2. Check by multiplying back: the product is 1.

Answer: (c) \( 2n\pi\pm\tfrac\pi2 \)

Explanation:
1. Numerator after multiplying by the conjugate: \( (1+ic)(1+2ic)=1-2c^2+3ic \) where \( c=\cos\theta \).
2. Imaginary part zero gives \( \cos\theta=0 \), i.e. \( \theta=\pm\tfrac\pi2+2n\pi \).

Teacher's Note:
1. The set θ = π/2 + nπ is the same as 2nπ ± π/2.
2. Real means the imaginary part vanishes.

Answer: (b) -1024

Explanation:
1. \( z=\frac{3+i}{2-i}=\frac{(3+i)(2+i)}{5}=\frac{5+5i}{5}=1+i \).
2. \( z^2=2i \), so \( z^{20}=(2i)^{10}=1024\,i^{10}=-1024 \).

Teacher's Note:
1. Square first: (1 + i)² = 2i.
2. i^10 = i^2 = −1.

Answer: (a) \( x^2+y^2+6x-8y=0 \)

Explanation:
1. Let \( z=x+iy \). Then \( \frac{x+i(y-8)}{(x+6)+iy} \) has real part \( \frac{x(x+6)+y(y-8)}{(x+6)^2+y^2} \).
2. Setting it to zero gives \( x^2+y^2+6x-8y=0 \), a circle through 8i and −6.

Teacher's Note:
1. Re(z1/z2) = 0 means the vectors from −6 and 8i meet at a right angle.
2. This is the circle on the join of −6 and 8i as diameter.

Answer: (c) Zero or purely imaginary

Explanation:
1. \( |z_1+z_2|=|z_1-z_2| \) squares to \( 4\,\mathrm{Re}(z_1\bar z_2)=0 \).
2. \( \frac{z_1}{z_2}=\frac{z_1\bar z_2}{|z_2|^2} \) therefore has zero real part.
3. So the ratio is purely imaginary, or zero when \( z_1=0 \).

Teacher's Note:
1. Equal diagonals of a parallelogram make it a rectangle, so the sides are perpendicular.
2. Re(z1 conj z2) = 0 is the orthogonality condition.

Answer: (d) \( pqs=s^2+q^2r \)

Explanation:
1. For a real root z, the real and imaginary parts vanish separately: \( z^2+pz+r=0 \) and \( qz+s=0 \).
2. From the second, \( z=-s/q \).
3. Substitute: \( \frac{s^2}{q^2}-\frac{ps}{q}+r=0\Rightarrow s^2-pqs+q^2r=0 \), i.e. \( pqs=s^2+q^2r \).

Teacher's Note:
1. Split the equation into real and imaginary parts when z is known to be real.
2. Eliminate z between the two equations.

Answer: (c) \( x=1,\ y=0 \)

Explanation:
1. The terms \( i^0,\dots,i^{99} \) form 25 blocks of four, each summing to 0.
2. Only \( i^{100}=1 \) remains.
3. So \( x=1,\ y=0 \).

Teacher's Note:
1. Group the terms in blocks of four.
2. Count the leftover terms carefully: the sum has 101 terms.

Answer: (d) \( \dfrac{(p^2+1)^2}{4p^2+1} \)

Explanation:
1. \( \mu^2+\lambda^2=\left|\frac{(p+i)^2}{2p-i}\right|^2=\frac{(p^2+1)^2}{4p^2+1} \).

Teacher's Note:
1. Take modulus squared rather than rationalising.
2. |p + i|² = p² + 1 and |2p − i|² = 4p² + 1.

Answer: (b) \( a^2+b^2 \)

Explanation:
1. Take the squared modulus on both sides: \( (1+1)(1+4)(1+9)\cdots(1+n^2)=a^2+b^2 \).
2. The left side is \( 2\cdot5\cdot10\cdots(1+n^2) \).

Teacher's Note:
1. |z1 z2 ... zn|² = |z1|² |z2|² ... |zn|².
2. |1 + ki|² = 1 + k².

Answer: (b) \( \dfrac{-7-26i}{25} \)

Explanation:
1. \( \frac{(2+5i)(4+3i)}{25}=\frac{8+6i+20i-15}{25}=\frac{-7+26i}{25} \).
2. The conjugate is \( \frac{-7-26i}{25} \).

Teacher's Note:
1. Rationalise first, then conjugate.
2. The conjugate of a quotient is the quotient of the conjugates.

Answer: (c) \( \dfrac{13}{10}+i\left(\dfrac{-9}{10}\right) \)

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

Teacher's Note:
1. Expand the numerator before dividing.
2. The conjugate flips the sign of the imaginary part only.

Answer: (a) \( A^2+B^2 \)

Explanation:
1. Take the squared modulus of both sides.
2. The left side becomes \( (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) \) and the right side becomes \( A^2+B^2 \).

Teacher's Note:
1. Modulus is multiplicative.
2. This is the product identity behind the Brahmagupta–Fibonacci identity.

Answer: (b) \( |z|=1 \)

Explanation:
1. \( \bar z=\frac1z\iff z\bar z=1\iff|z|^2=1 \).
2. So the condition is \( |z|=1 \).

Teacher's Note:
1. The inverse of a unit complex number is its conjugate.
2. z = 1 is only one special case of |z| = 1.

Answer: (c) \( 3+5i \)

Explanation:
1. \( z^2=9-25+30i=-16+30i \).
2. \( z^3=(-16+30i)(3+5i)=-48-80i+90i-150=-198+10i \).
3. Adding \( \bar z=3-5i \) and 198: \( -198+10i+3-5i+198=3+5i \).

Teacher's Note:
1. Compute z³ in two steps.
2. Conjugate flips the sign of the imaginary part.

Answer: (b) II

Explanation:
1. A third-quadrant number has \( x<0,\ y<0 \).
2. Its conjugate has \( x<0,\ y>0 \), which is the second quadrant.

Teacher's Note:
1. Conjugation reflects the point in the real axis.
2. The sign of x stays the same while y changes sign.

Answer: (a) \( t(1-i),\ t\in R \)

Explanation:
1. Put \( z=x+iy \): \( (1+i)(x+iy)=(x-y)+i(x+y) \) and \( (1-i)(x-iy)=(x-y)-i(x+y) \).
2. Equality requires \( x+y=0 \), so \( z=x-ix=x(1-i) \).
3. This is the form \( t(1-i) \).

Teacher's Note:
1. Expand both sides and equate imaginary parts.
2. The real parts are automatically equal.