JEE Mathematics Complex Numbers MCQs Set 20

Answer: (c) 3

Explanation:
1. The ratio equals \( \frac{2e^{i\pi/3}}{2e^{-i\pi/3}}=e^{2\pi i/3}=\omega \), a primitive cube root of unity.
2. Its nth power is an integer only when \( \omega^n=1 \), i.e. n is a multiple of 3.
3. Among the options, \( n=3 \).

Teacher's Note:
1. The ratio of a number to its conjugate is a point on the unit circle.
2. ω^n is real only for n divisible by 3 (value 1) in the given options.

Answer: (a) -2

Explanation:
1. \( \frac{\sqrt3+i}2=e^{i\pi/6} \) and \( \frac{-\sqrt3+i}2=e^{5i\pi/6} \).
2. Sixth powers: \( e^{i\pi}=-1 \) and \( e^{5i\pi}=-1 \).
3. Sum: \( -2 \).

Teacher's Note:
1. Convert both to polar form before taking the power.
2. The second number lies in the second quadrant, argument 5π/6.

Answer: (c) 1, 1

Explanation:
1. Use \( 1+\omega=-\omega^2 \): \( (1+\omega)^7=-\omega^{14}=-\omega^{2} \) (since \( \omega^{12}=1 \)).
2. Also \( -\omega^2=1+\omega \), so \( A=1,\ B=1 \).

Teacher's Note:
1. Reduce ω powers modulo 3.
2. 1 + ω + ω² = 0 lets you write −ω² as 1 + ω.

Answer: (d) 4

Explanation:
1. Use \( 1+\omega=-\omega^2 \) and \( 1+\omega^2=-\omega \): \( 1+\omega-\omega^2=-2\omega^2 \) and \( 1-\omega+\omega^2=-2\omega \).
2. Product: \( 4\omega^3=4 \).

Teacher's Note:
1. Replace each trinomial by a single multiple of ω or ω².
2. ω³ = 1.

Answer: (a) 64

Explanation:
1. \( \omega^4=\omega \), so \( 3+\omega^2+\omega=3-1=2 \).
2. Sixth power: \( 2^6=64 \).

Teacher's Note:
1. Reduce ω⁴ to ω first.
2. 1 + ω + ω² = 0.

Answer: (b) -1

Explanation:
1. Note \( b+c\omega+a\omega^2=\omega^2(a+b\omega+c\omega^2) \), so the first fraction is \( \omega^{-2}=\omega \).
2. Similarly \( c+a\omega+b\omega^2=\omega(a+b\omega+c\omega^2) \), so the second fraction is \( \omega^{-1}=\omega^2 \).
3. Sum: \( \omega+\omega^2=-1 \).

Teacher's Note:
1. Spot that the denominators are ω² and ω times the numerator.
2. ω + ω² = −1.

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

Explanation:
1. \( 1+\omega-\omega^2=-2\omega^2 \).
2. Seventh power: \( (-2)^7\omega^{14}=-128\omega^{2} \) since \( \omega^{14}=\omega^{12}\omega^2 \).

Teacher's Note:
1. Replace 1 + ω by −ω².
2. Reduce ω^14 modulo 3.

Answer: (c) -4

Explanation:
1. \( 3+5\omega+3\omega^2=2\omega \) and \( 3+3\omega+5\omega^2=2\omega^2 \).
2. The sum is \( 4\omega^2+4\omega^4=4(\omega^2+\omega)=-4 \).

Teacher's Note:
1. 3 + 3ω + 3ω² = 0 simplifies each trinomial.
2. ω⁴ = ω.

Answer: (a) \( 27abc \)

Explanation:
1. Let \( p=a+b\omega+c\omega^2 \) and \( q=a+b\omega^2+c\omega \).
2. Then \( p+q=2a-b-c=3a \) (using \( b+c=-a \)) and \( pq=a^2+b^2+c^2-ab-bc-ca=-3(ab+bc+ca) \) (using \( a+b+c=0 \)).
3. So \( p^3+q^3=(p+q)^3-3pq(p+q)=27a^3+27a(ab+bc+ca)=27a(a+b)(a+c)=27abc \), since \( a+b=-c \) and \( a+c=-b \).

Teacher's Note:
1. Use p³ + q³ = (p + q)³ − 3pq(p + q).
2. Check with a = 1, b = 2, c = −3: both sides equal −162.

Answer: (c) \( -128\omega \)

Explanation:
1. \( 1-\omega+\omega^2=-2\omega \) and \( 1+\omega-\omega^2=-2\omega^2 \).
2. So the product is \( (-2\omega)(-2\omega^2)^6=-2\omega\cdot64\omega^{12}=-128\omega \).

Teacher's Note:
1. Replace each trinomial by −2ω or −2ω².
2. ω^12 = 1.

Answer: (d) 248

Explanation:
1. Expand: \( (3\omega+8\omega^2)^2=9\omega^2+48\omega^3+64\omega^4=9\omega^2+48+64\omega \).
2. Likewise \( (3\omega^2+8\omega)^2=9\omega^4+48+64\omega^2=9\omega+48+64\omega^2 \).
3. Sum: \( 96+73(\omega+\omega^2)=96-73=23 \), and \( 225+23=248 \).

Teacher's Note:
1. Use ω³ = 1 and ω + ω² = −1.
2. Add the two expansions before substituting.

Answer: (b) \( 3(a^3+b^3) \)

Explanation:
1. Here \( x+y+z=(a+b)(1+\omega+\omega^2)=0 \), so \( x^3+y^3+z^3=3xyz \).
2. Now \( xyz=(a+b)(a\omega+b\omega^2)(a\omega^2+b\omega)=(a+b)(a^2+b^2-ab)=a^3+b^3 \).
3. So the value is \( 3(a^3+b^3) \).

Teacher's Note:
1. If x + y + z = 0 then x³ + y³ + z³ = 3xyz.
2. (aω + bω²)(aω² + bω) = a² + b² + ab(ω² + ω⁴) = a² + b² − ab.

Answer: (a) 0

Explanation:
1. \( 1+2\omega+\omega^2=\omega \) and \( 1+\omega+2\omega^2=\omega^2 \).
2. So the expression is \( \omega^{3n}-\omega^{6n}=1-1=0 \).

Teacher's Note:
1. Use 1 + ω + ω² = 0 to simplify each base.
2. Any power of ω that is a multiple of 3 equals 1.

Answer: (c) 0

Explanation:
1. Take \( \alpha=\omega,\ \beta=\omega^2 \). Then \( \alpha^4=\omega \), \( \beta^{28}=\omega^{56}=\omega^2 \) and \( \alpha\beta=1 \).
2. The sum is \( \omega+\omega^2+1=0 \).

Teacher's Note:
1. The two imaginary cube roots multiply to 1.
2. Reduce exponents modulo 3.

Answer: (d) \( x=0 \)

Explanation:
1. At \( x=0 \) the matrix is \( \begin{pmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&\omega\end{pmatrix} \) in which row 2 equals \( \omega \) times row 1, so the determinant is 0.
2. For \( x=1,\ \omega,\ \omega^2 \) a direct evaluation gives the determinant 1 (not zero).
3. So \( x=0 \) is a root.

Teacher's Note:
1. A determinant vanishes when two rows are proportional.
2. Test each option by substitution.

Answer: (a) 0

Explanation:
1. Row 2 equals \( \omega^n \) times row 1: \( \omega^n\cdot1=\omega^n,\ \omega^n\omega^n=\omega^{2n},\ \omega^n\omega^{2n}=\omega^{3n}=1 \).
2. Two proportional rows make the determinant zero.

Teacher's Note:
1. Look for proportional rows before expanding.
2. ω^(3n) = 1.

Answer: (a) -1

Explanation:
1. The exponent series is geometric with first term \( \tfrac12 \) and ratio \( \tfrac34 \): its sum is \( \frac{1/2}{1-3/4}=2 \).
2. So the expression is \( \omega+\omega^2=-1 \).

Teacher's Note:
1. Sum the geometric series in the exponent.
2. ω + ω² = −1.

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

Explanation:
1. The nth roots of unity are the roots of \( z^n-1=0 \).
2. The product of the roots of a monic polynomial of degree n is \( (-1)^n\times\text{constant term}=(-1)^n(-1)=(-1)^{n+1}=(-1)^{n-1} \).

Teacher's Note:
1. Check with n = 2: the roots 1 and −1 have product −1.
2. Check with n = 3: product is 1.

Answer: (a) 0

Explanation:
1. The roots are \( \alpha^k \) with \( \alpha=e^{2\pi i/n} \). The sum of pth powers is \( \sum_{k=0}^{n-1}(\alpha^p)^k=\frac{1-\alpha^{pn}}{1-\alpha^p}=0 \) since \( \alpha^p\neq1 \) and \( \alpha^{pn}=1 \).

Teacher's Note:
1. If p were a multiple of n every term would be 1 and the sum n.
2. Use the geometric series formula.

Answer: (a) \( \mathrm{Re}(z)<0 \)

Explanation:
1. Taking moduli: \( |z|^n=|z+1|^n \), so \( |z|=|z+1| \).
2. z is equidistant from 0 and \( -1 \), so \( \mathrm{Re}(z)=-\tfrac12 \).
3. Hence \( \mathrm{Re}(z)<0 \).

Teacher's Note:
1. Taking the nth root of moduli removes the exponent.
2. Equidistance from 0 and −1 gives the line Re z = −1/2.

Answer: (b) 0

Explanation:
1. Since \( \alpha^3=1 \): \( \alpha^{3n+1}=\alpha,\ \alpha^{3n+3}=1,\ \alpha^{3n+5}=\alpha^2 \).
2. Sum: \( \alpha+1+\alpha^2=0 \).

Teacher's Note:
1. Reduce exponents modulo 3.
2. 1 + α + α² = 0.

Answer: (b) \( -\dfrac{2n}{1-\alpha} \)

Explanation:
1. The kth term is \( (2k-1)\alpha^{k-1} \). Compute \( (1-\alpha)S=1+2\alpha+2\alpha^2+\cdots+2\alpha^{n-1}-(2n-1)\alpha^n \).
2. Using \( \alpha^n=1 \) and \( \alpha+\alpha^2+\cdots+\alpha^{n-1}=-1 \): \( (1-\alpha)S=1-2-(2n-1)=-2n \).
3. So \( S=-\frac{2n}{1-\alpha} \).

Teacher's Note:
1. Multiply the series by (1 − α) to telescope.
2. For nth roots of unity, 1 + α + ... + α^(n−1) = 0.

Answer: (b) \( \mathrm{cis}\left(\tfrac\pi{12}\right) \)

Explanation:
1. \( \frac12+\frac{i\sqrt3}2=\mathrm{cis}\tfrac\pi3 \).
2. A number \( \mathrm{cis}\theta \) is a fourth root if \( 4\theta\equiv\tfrac\pi3\pmod{2\pi} \), so \( \theta=\tfrac\pi{12}+\tfrac{k\pi}2 \).
3. Among the options only \( \tfrac\pi{12} \) fits.

Teacher's Note:
1. Fourth roots divide the argument by 4 and add multiples of π/2.
2. Test each option by multiplying its angle by 4.

Answer: (b) \( \omega^{14}=\omega^2 \)

Explanation:
1. \( \omega^3=1 \), so \( \omega^4=\omega \) and \( \omega^6=1 \) and \( \omega^5=\omega^2 \), making (a), (c), (d) false.
2. Also \( \omega^{14}=\omega^{12}\omega^2=\omega^2 \), so (b) is true.

Teacher's Note:
1. Reduce each exponent modulo 3.
2. Only the exponent that leaves remainder 2 matches ω².

Answer: (a) 2

Explanation:
1. The two ratios are \( \omega \) and \( \omega^2 \) (conjugates of each other, with product 1).
2. Sixth powers: \( \omega^6=1 \) and \( \omega^{12}=1 \), so the sum is \( 2 \).

Teacher's Note:
1. Each ratio is a cube root of unity.
2. Any sixth power of a cube root of unity is 1.