JEE Mathematics Complex Numbers MCQs Set 23

Answer: (b) \( i-1 \)

Explanation:
1. Any four consecutive powers of i add up to 0.
2. \( \sum_{n=1}^{13} i^{n}=i^{13}+\) (twelve terms that cancel) \(=i \).
3. \( \sum_{n=1}^{13} i^{n+1}=i^{2}+\dots+i^{14}=i^{14}=-1 \) after the cancelling block.
4. Total \( =i-1 \). Answer: \( i-1 \).

Teacher's Note:
1. Group powers of i in blocks of four, since each block sums to zero.
2. Then handle only the leftover terms.

Answer: (b) \( 3 \)

Explanation:
1. \( \dfrac{1-i}{1+i}=\dfrac{(1-i)^{2}}{2}=-i \).
2. We need \( (-i)^{n}=i \).
3. \( (-i)^{1}=-i,\ (-i)^{2}=-1,\ (-i)^{3}=i \). So n = 3.

Teacher's Note:
1. Reduce the fraction to a power of i first.
2. Then list the cycle of powers until the required value appears.

Answer: (a) \( \beta+i\alpha \)

Explanation:
1. \( b+ia=i(a-ib) \).
2. So \( (b+ia)^{5}=i^{5}(a-ib)^{5}=i\,\overline{(a+ib)^{5}}=i(\alpha-i\beta) \).
3. \( i(\alpha-i\beta)=\beta+i\alpha \). Answer: \( \beta+i\alpha \).

Teacher's Note:
1. Factor i out of b + ia to bring it to the conjugate of a + ib.
2. The conjugate of a power is the power of the conjugate.

Answer: (a) \( 3 \)

Explanation:
1. \( i^{n}+i^{-n}=2\cos\dfrac{n\pi}{2} \).
2. For n = 0, 1, 2, 3 this gives 2, 0, -2, 0, and it then repeats.
3. The distinct values are 2, 0, -2, so there are 3.

Teacher's Note:
1. Write the sum as a cosine to see the period immediately.
2. Count only distinct values, not distinct values of n.

Answer: (b) \( 2(-1+5i) \)

Explanation:
1. \( (1+i)^{2}=2i \), so \( (1+i)^{3}=2i(1+i)=-2+2i \).
2. \( (1-i)^{2}=-2i \), so \( (1-i)^{6}=(-2i)^{3}=-8i^{3}=8i \).
3. Sum \( =-2+10i=2(-1+5i) \).

Teacher's Note:
1. Use (1 + i)^2 = 2i and (1 - i)^2 = -2i to cut the powers down.
2. Keep track of the sign when cubing -2i.

Answer: (b) \( -2\pm 2i \)

Explanation:
1. \( \bar z_2=8y^{2}+20i \).
2. Equate real parts: \( 9y^{2}-4=8y^{2} \), so \( y^{2}=4,\ y=\pm2 \).
3. Equate imaginary parts: \( -10x=20 \), so \( x=-2 \). Hence \( z=-2\pm2i \).

Teacher's Note:
1. Equate real and imaginary parts separately after writing the conjugate.
2. y has two values because only its square is fixed.

Answer: (d) no value of x

Explanation:
1. Conjugate means \( \sin x=\cos x \) and \( -\cos2x=\sin2x \) (the imaginary parts are negatives).
2. The first gives \( \tan x=1 \), so \( \cos2x=\cos^{2}x-\sin^{2}x=0 \).
3. Then the second needs \( \sin2x=0 \), but \( \sin2x=2\sin x\cos x=\pm1 \) when \( \tan x=1 \).
4. The conditions contradict, so no x works.

Teacher's Note:
1. Two complex numbers are conjugates only if both real and imaginary conditions hold together.
2. Check that the solution of the first condition satisfies the second.

Answer: (c) \( 0 \)

Explanation:
1. Equality in the triangle inequality holds only when \( z_1 \) and \( z_2 \) point in the same direction.
2. Then \( \dfrac{z_1}{z_2} \) is a positive real number.
3. Its argument is 0.

Teacher's Note:
1. Equality in the triangle inequality means collinear vectors in the same direction.
2. A positive real number has argument 0.

Answer: (b) \( x=0,\ y=\sqrt2+1 \)

Explanation:
1. \( |z+1|=|z-1| \) means z is equidistant from -1 and 1, so \( x=0 \) and \( z=iy \).
2. \( \dfrac{iy-1}{iy+1}=\dfrac{(iy-1)(1-iy)}{1+y^{2}}=\dfrac{(y^{2}-1)+2iy}{1+y^{2}} \).
3. Argument \( \dfrac{\pi}{4} \) needs real part = imaginary part > 0: \( y^{2}-1=2y \), so \( y=1+\sqrt2 \) (the root with y > 0). Hence \( z=i(\sqrt2+1) \).

Teacher's Note:
1. The condition on moduli puts z on the imaginary axis.
2. An argument of 45 degrees means equal positive real and imaginary parts.

Answer: (a) \( 2\theta \)

Explanation:
1. \( z=\dfrac{e^{i\theta}}{e^{-i\theta}}=e^{2i\theta} \).
2. For \( \dfrac{\pi}{4}<\theta<\dfrac{\pi}{2} \), \( 2\theta \) lies in \( \left(\dfrac{\pi}{2},\pi\right) \), which is inside the principal range \( (-\pi,\pi] \).
3. So \( \arg z=2\theta \).

Teacher's Note:
1. Always check that the angle you get lies in the principal range.
2. Here 2 theta stays below pi, so no adjustment is needed.

Answer: (b) \( \dfrac{\pi}{3} \)

Explanation:
1. \( \arg(\sqrt3+i)=\dfrac{\pi}{6} \) and \( \arg(\sqrt3-i)=-\dfrac{\pi}{6} \).
2. \( \arg z=\dfrac{\pi}{6}-\left(-\dfrac{\pi}{6}\right)=\dfrac{\pi}{3} \), which is in the principal range.

Teacher's Note:
1. The argument of a quotient is the difference of the arguments.
2. Both numbers are in the first and fourth quadrants, so the arguments are easy.

Answer: (b) \( 1:3 \)

Explanation:
1. Equal arguments mean the points 1, -3i and z are collinear on a ray from the same direction, so \( \dfrac{y}{x-1}=\dfrac{y+3}{x} \).
2. Cross-multiplying, \( xy=(x-1)(y+3)=xy+3x-y-3 \), so \( y=3(x-1) \).
3. Therefore \( (x-1):y=1:3 \).

Teacher's Note:
1. Equal arguments give equal slopes of the two vectors.
2. Cross-multiply and cancel the common xy term.

Answer: (c) \( 4 \)

Explanation:
1. \( z^{2}=\pm ir^{2} \) (modulus of \( z^{2} \) is \( r^{2} \)).
2. Each of the two equations \( z^{2}=ir^{2} \) and \( z^{2}=-ir^{2} \) has two roots.
3. Total 4 values, at arguments \( \pm\dfrac{\pi}{4} \) and \( \pm\dfrac{3\pi}{4} \).

Teacher's Note:
1. Fix the modulus of z squared first; only the sign of the imaginary unit is free.
2. Each square-root equation gives two roots.

Answer: (d) \( -128\omega^{2} \)

Explanation:
1. \( 1+\omega=-\omega^{2} \), so \( 1+\omega-\omega^{2}=-2\omega^{2} \).
2. \( (-2\omega^{2})^{7}=-128\omega^{14} \).
3. \( \omega^{14}=\omega^{12}\omega^{2}=\omega^{2} \), so the value is \( -128\omega^{2} \).

Teacher's Note:
1. Replace 1 + w by -w^2 right away.
2. Reduce the exponent modulo 3 using w^3 = 1.

Answer: (d) \( -\dfrac{1}{\sqrt3} \)

Explanation:
1. \( \dfrac{3}{2}+\dfrac{\sqrt3}{2}i=\sqrt3\left(\cos30^{\circ}+i\sin30^{\circ}\right) \).
2. Raising to power 100: \( 3^{50}(\cos3000^{\circ}+i\sin3000^{\circ}) \), and \( 3000^{\circ}=8\times360^{\circ}+120^{\circ} \).
3. So \( x+iy=3\left(-\dfrac{1}{2}+\dfrac{\sqrt3}{2}i\right) \), giving \( x=-\dfrac{3}{2},\ y=\dfrac{3\sqrt3}{2} \).
4. \( k=\dfrac{x}{y}=-\dfrac{1}{\sqrt3} \).

Teacher's Note:
1. Use De Moivre's theorem after writing the base in polar form.
2. Reduce the angle modulo 360 degrees.

Answer: (b) \( x^{2}+x+1 \)

Explanation:
1. The roots of \( x^{2}+x+1 \) are \( \omega,\omega^{2} \).
2. At \( x=\omega \): \( \omega^{3m}+\omega^{3n-1}+\omega^{3r-2}=1+\omega^{-1}+\omega^{-2}=1+\omega^{2}+\omega=0 \).
3. The same holds at \( \omega^{2} \), so \( x^{2}+x+1 \) is a factor.

Teacher's Note:
1. Test the polynomial at the roots of the proposed divisor.
2. For roots of unity, reduce powers using w^3 = 1.

Answer: (b) \( -48 \)

Explanation:
1. \( x+\dfrac{1}{x}=\sqrt3 \), so \( x=e^{\pm i\pi/6} \).
2. \( x^{n}-x^{-n}=\pm2i\sin\dfrac{n\pi}{6} \), so each square is \( -4\sin^{2}\dfrac{n\pi}{6} \).
3. Over 24 terms (two full periods of 12), \( \sum\sin^{2}\dfrac{n\pi}{6}=12 \).
4. The sum is \( -4\times12=-48 \).

Teacher's Note:
1. The square of a difference of conjugates is negative here, since it is (2i sin)^2.
2. Use the fact that sin squared averages to one half over a full period.

Answer: (b) \( 6 \)

Explanation:
1. \( 1+\sqrt3 i=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right) \).
2. The argument of the power is \( \dfrac{n}{2}\cdot\dfrac{\pi}{3}=\dfrac{n\pi}{6} \), which must be a multiple of \( \pi \).
3. The smallest positive n is 6.

Teacher's Note:
1. A complex number is real when its argument is a multiple of pi.
2. Write the argument of the power and solve for the least n.

Answer: (b) \( 2\omega \)

Explanation:
1. Using \( \omega^{3}=1 \): \( 1+2\omega+3\omega^{2}=\omega\left(\omega^{2}+2+3\omega\right)=\omega(2+3\omega+\omega^{2}) \), so the first fraction equals \( \omega \).
2. Also \( \omega^{2}(2+3\omega+\omega^{2})=2\omega^{2}+3+\omega=3+\omega+2\omega^{2} \), so the second fraction equals \( \dfrac{1}{\omega^{2}}=\omega \).
3. Sum \( =2\omega \).

Teacher's Note:
1. Substituting the numerical value of omega is the safest way to check such sums.
2. Remember omega = -1/2 + (root 3 / 2) i.