JEE Mathematics Complex Numbers MCQs Set 21

Answer: (d) -1

Explanation:
1. From \( z^2-z+1=0 \), \( z=e^{\pm i\pi/3} \).
2. So \( z^{100}+z^{-100}=2\cos\tfrac{100\pi}3 \) and \( \tfrac{100\pi}3=32\pi+\tfrac{4\pi}3 \).
3. The value is \( 2\cos\tfrac{4\pi}3=-1 \).

Teacher's Note:
1. z + 1/z = 2cosθ identifies θ = π/3.
2. Reduce the angle modulo 2π.

Answer: (b) 2

Explanation:
1. Since \( \alpha^3=1 \): \( \alpha^{-2}=\alpha \) and \( \alpha^{-1}=\alpha^2 \).
2. The exponent is \( |1-\alpha+\alpha^2+\alpha-\alpha^2|=|1|=1 \).
3. So the value is \( 2^1=2 \).

Teacher's Note:
1. Reduce negative powers using α³ = 1.
2. The terms cancel in pairs.

Answer: (a) -64

Explanation:
1. \( -1\pm i\sqrt3=2e^{\pm2\pi i/3} \), so \( (-1\pm i\sqrt3)^{15}=2^{15}e^{\pm10\pi i}=2^{15} \).
2. Also \( (1\mp i)^{20}=(\mp2i)^{10}=2^{10}i^{10}=-2^{10} \).
3. Each fraction is \( -2^5=-32 \), so the sum is \( -64 \).

Teacher's Note:
1. Convert to polar form and reduce 15 × 2π/3 = 10π.
2. (2i)^10 and (−2i)^10 are both equal to −1024.

Answer: (b) \( 2^{2n} \)

Explanation:
1. With \( \omega^3=1 \) the factors cycle: \( 1-\omega+\omega^2=-2\omega \), \( 1-\omega^2+\omega^4=1-\omega^2+\omega=-2\omega^2 \), then they repeat.
2. Each pair of consecutive factors multiplies to \( 4\omega^3=4 \).
3. With n pairs the product is \( 4^n=2^{2n} \).

Teacher's Note:
1. Reduce powers of ω modulo 3 before multiplying.
2. Group the 2n factors in pairs.

Answer: (c) n

Explanation:
1. Since \( x^n-1=(x-1)(x-\omega)\cdots(x-\omega^{n-1}) \), we have \( (x-\omega)\cdots(x-\omega^{n-1})=1+x+\cdots+x^{n-1} \).
2. Put \( x=1 \): the product equals \( n \).

Teacher's Note:
1. Divide x^n − 1 by x − 1 and substitute x = 1.
2. Check n = 3: (1 − ω)(1 − ω²) = 3.

Answer: (b) \( \tfrac14(n-1)n(n^2+3n+4) \)

Explanation:
1. The kth term is \( k\,(k+1-\omega)(k+1-\omega^2)=k\left[(k+1)^2+(k+1)+1\right]=k^3+3k^2+3k \), using \( (m-\omega)(m-\omega^2)=m^2+m+1 \).
2. Note that \( (k+1)^3-1=k^3+3k^2+3k \).
3. So the sum is \( \sum_{k=1}^{n-1}\left[(k+1)^3-1\right]=\left(\frac{n(n+1)}2\right)^2-1-(n-1)=\frac{n^2(n+1)^2}4-n \).
4. This simplifies to \( \tfrac14(n-1)n(n^2+3n+4) \).

Teacher's Note:
1. (m − ω)(m − ω²) = m² − m(ω + ω²) + ω³ = m² + m + 1.
2. The sum of cubes is (n(n + 1)/2)².

Answer: (a) \( \omega,\ \omega^2 \)

Explanation:
1. \( z^3+2z^2+2z+1=(z+1)(z^2+z+1) \), so the roots are \( -1,\ \omega,\ \omega^2 \).
2. For \( z=-1 \): \( -1+1+1=1\ne0 \), so it is not common.
3. For \( z=\omega \): \( \omega^{1985}=\omega^{2} \) (as \( 1985\equiv2\pmod3 \)) and \( \omega^{100}=\omega \), so the sum is \( \omega^2+\omega+1=0 \). By conjugation, \( \omega^2 \) also satisfies it.

Teacher's Note:
1. Factor the cubic first to find its roots.
2. Reduce exponents modulo 3.

Answer: (d) \( |z_k|=|z_{k+1}| \)

Explanation:
1. Every nth root of unity lies on the unit circle: \( |z_k|=1 \) for all k.
2. So any two of them have equal modulus: \( |z_k|=|z_{k+1}| \).

Teacher's Note:
1. The roots of unity form a regular polygon inscribed in the unit circle.
2. Option (c) would force |z_k| = 0, which is impossible.

Answer: (d) \( 4k \)

Explanation:
1. The right angle means \( z_2/z_1=\pm i \).
2. Since \( z_2/z_1 \) is also an nth root of unity, \( (\pm i)^n=1 \).
3. This holds exactly when \( 4\mid n \), i.e. \( n=4k \).

Teacher's Note:
1. The ratio of two nth roots of unity is again an nth root of unity.
2. i has order 4.

Answer: (c) 1

Explanation:
1. Factorise: \( z^4+z^3+2z^2+z+1=(z^2+1)(z^2+z+1) \).
2. The roots are \( \pm i \) and \( \omega,\omega^2 \), all of modulus 1.
3. So \( |z|=1 \).

Teacher's Note:
1. Try grouping as (z² + 1)(z² + z + 1).
2. Both factors have roots on the unit circle.

Answer: (b) 0

Explanation:
1. The fourth roots are \( 1,i,-1,-i \) and their squares are \( 1,-1,1,-1 \).
2. The sum is 0.

Teacher's Note:
1. The squares of the 4th roots are the 2nd roots of unity, each counted twice.
2. The sum of the roots of unity of any order above 1 is zero.

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

Explanation:
1. Let \( f(x)=x^n-1 \). Then \( \frac{f'(x)}{f(x)}=\sum_{i=0}^{n-1}\frac1{x-\alpha^i} \).
2. At \( x=2 \): \( \frac{n2^{n-1}}{2^n-1}=\frac1{2-1}+\sum_{i=1}^{n-1}\frac1{2-\alpha^i} \).
3. So \( \sum_{i=1}^{n-1}\frac1{2-\alpha^i}=\frac{n2^{n-1}}{2^n-1}-1=\frac{(n-2)2^{n-1}+1}{2^n-1} \).

Teacher's Note:
1. Logarithmic differentiation of x^n − 1 turns the sum into a quotient.
2. Test n = 3: both forms give 5/7.

Answer: (a) \( 3(|z_1|^2+|z_2|^2+|z_3|^2) \)

Explanation:
1. Expand each square modulus: \( |A|^2=\sum|z_k|^2+2\,\mathrm{Re}\sum_{j<k}z_j\bar z_k \); for B and C the cross terms get the factors \( \omega^{k-j} \) and \( \omega^{-(k-j)} \).
2. Adding the three expressions, each cross term is multiplied by \( 1+\omega^{m}+\omega^{2m}=0 \) (with \( m=1,2 \)).
3. Only the diagonal terms remain, each counted 3 times: \( 3\sum|z_k|^2 \).

Teacher's Note:
1. This is Parseval's identity for the three-point discrete Fourier transform.
2. Sums of the form 1 + ω^m + ω^(2m) vanish unless m is a multiple of 3.

Answer: (a) \( \sin^{-1}\left(\tfrac{2\sqrt2}3\right) \)

Explanation:
1. The roots are \( z=-1\pm i\sqrt2 \). Take \( z_1=-1+i\sqrt2,\ z_2=-1-i\sqrt2 \).
2. Numerator: \( x_1y_2-x_2y_1=(-1)(-\sqrt2)-(-1)(\sqrt2)=2\sqrt2 \).
3. Denominator: \( \sqrt3\cdot\sqrt3=3 \), so \( \sin\theta=\tfrac{2\sqrt2}3 \).

Teacher's Note:
1. The roots are conjugates, so the angle between them is twice the argument of either.
2. Check: sin 2φ = 2 sinφ cosφ = 2(√2/√3)(1/√3) = 2√2/3.

Answer: (b) \( i\sin x \)

Explanation:
1. \( \sinh y=\frac{e^y-e^{-y}}2 \), so \( \sinh(ix)=\frac{e^{ix}-e^{-ix}}2=i\sin x \).

Teacher's Note:
1. sinh(ix) = i sin x and cosh(ix) = cos x.
2. These are the bridges between hyperbolic and circular functions.

Answer: (a) -1

Explanation:
1. \( \cosh(i\pi)=\cos\pi=-1 \), so \( \mathrm{sech}(i\pi)=\frac1{-1}=-1 \).

Teacher's Note:
1. cosh(ix) = cos x.
2. sech is the reciprocal of cosh.

Answer: (b) \( \sinh\alpha\sin\beta \)

Explanation:
1. \( \cosh(\alpha+i\beta)=\cosh\alpha\cosh(i\beta)+\sinh\alpha\sinh(i\beta)=\cosh\alpha\cos\beta+i\sinh\alpha\sin\beta \).
2. The imaginary part is \( \sinh\alpha\sin\beta \).

Teacher's Note:
1. Use the addition formula and cosh(iβ) = cos β, sinh(iβ) = i sin β.
2. Real part is coshα cosβ.

Answer: (c) \( 2i\sinh\alpha\sin\beta \)

Explanation:
1. From the previous expansion, \( \cosh(\alpha\pm i\beta)=\cosh\alpha\cos\beta\pm i\sinh\alpha\sin\beta \).
2. The difference is \( 2i\sinh\alpha\sin\beta \).

Teacher's Note:
1. The real parts cancel and the imaginary parts double.
2. Compare cos(a + b) − cos(a − b) = −2 sin a sin b.

Answer: (c) \( \cos^2u+\cosh^2v \)

Explanation:
1. \( \cos(u+iv)=\cos u\cosh v-i\sin u\sinh v \), so \( \alpha=\cos u\cosh v,\ \beta=-\sin u\sinh v \).
2. Then \( \alpha^2+\beta^2=\cos^2u(1+\sinh^2v)+\sin^2u\sinh^2v=\cos^2u+\sinh^2v \).
3. Adding 1 and using \( 1+\sinh^2v=\cosh^2v \): \( \alpha^2+\beta^2+1=\cos^2u+\cosh^2v \).

Teacher's Note:
1. Expand cos(u + iv) with cos(iv) = cosh v and sin(iv) = i sinh v.
2. Use cos²u + sin²u = 1.

Answer: (a) \( \tfrac12\tan^{-1}\left(\dfrac{2\alpha}{1-\alpha^2-\beta^2}\right) \)

Explanation:
1. Let \( w=\alpha+i\beta \). Then \( \tan(x+iy)=w \) and \( \tan(x-iy)=\bar w \).
2. So \( \tan2x=\tan[(x+iy)+(x-iy)]=\frac{w+\bar w}{1-w\bar w}=\frac{2\alpha}{1-\alpha^2-\beta^2} \).
3. Hence \( x=\tfrac12\tan^{-1}\left(\frac{2\alpha}{1-\alpha^2-\beta^2}\right) \).

Teacher's Note:
1. Use the conjugate of the equation to get tan(x − iy).
2. Apply the addition formula for tan.

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

Explanation:
1. \( \omega^3=1 \), so \( \omega^{14}=\omega^{12}\cdot\omega^2=\omega^2 \).

Teacher's Note:
1. Reduce the exponent modulo 3: 14 = 3 × 4 + 2.
2. ω² is the other imaginary cube root of unity.

Answer: (c) \( \omega,\ \omega^2 \)

Explanation:
1. Let the numbers be a and b with \( a=b^2 \) and \( b=a^2 \). Then \( a=a^4 \), so \( a^3=1 \) for a non-zero a.
2. For non-real values \( a=\omega \) and \( b=\omega^2 \): \( \omega^2=(\omega)^2 \) and \( (\omega^2)^2=\omega^4=\omega \).

Teacher's Note:
1. Cube roots of unity are closed under squaring.
2. The pair 1, 1 is the trivial real case, so the question asks for the non-real pair.

Answer: (c) A circle centred at the origin

Explanation:
1. \( z-zi=z(1-i) \) and \( |1-i|=\sqrt2 \).
2. So \( |z|\sqrt2=1 \), i.e. \( |z|=\tfrac1{\sqrt2} \).
3. The locus is a circle with centre at the origin.

Teacher's Note:
1. Factor z out of the expression.
2. The modulus of a product is the product of the moduli.

Answer: (c) -1

Explanation:
1. The cube roots of \( -1 \) are the roots of \( z^3+1=0 \).
2. The product of the roots of this cubic is \( -\frac{1}{1}=-1 \).

Teacher's Note:
1. Product of roots of a cubic az³ + ... + d is −d/a.
2. The roots are −1 and (1 ± i√3)/2.