JEE Mathematics Complex Numbers MCQs Set 10

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

Explanation:
1. The sum is \( (1+i)\sum_{n=1}^{13}i^n \).
2. Any 4 consecutive powers of i add to 0, so 12 terms cancel and \( \sum_{n=1}^{13}i^n=i^{13}=i \).
3. The value is \( (1+i)i=i-1 \).

Teacher's Note:
1. Powers of i repeat with period 4.
2. Group the terms in blocks of four.

Answer: (d) for any two positive integers \( n_1,n_2 \)

Explanation:
1. \( i^3=-i,\ i^5=i,\ i^7=-i \), so the expression is \( \left[(1+i)^{n_1}+(1-i)^{n_1}\right]+\left[(1+i)^{n_2}+(1-i)^{n_2}\right] \).
2. Each bracket is a sum of a number and its conjugate, so it is real for every n.

Teacher's Note:
1. A complex number plus its conjugate is real.
2. Reduce the powers of i first.

Answer: (b) 3

Explanation:
1. \( \frac{1-i}{1+i}=-i \), so the expression is \( (-i)^n \).
2. \( (-i)^1=-i,\ (-i)^2=-1,\ (-i)^3=i \). The first power giving \( +i \) is \( n=3 \).

Teacher's Note:
1. Simplify the fraction to a power of i.
2. Test n = 1, 2, 3.

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

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

Teacher's Note:
1. Factor i out of b + ia.
2. Use the conjugate of the given power.

Answer: (c) \( \frac12(\bar z-z)i \)

Explanation:
1. \( \text{Im}(z)=\frac{z-\bar z}{2i} \).
2. Since \( \frac1i=-i \), this equals \( \frac{-i(z-\bar z)}{2}=\frac12(\bar z-z)i \).

Teacher's Note:
1. Im(z) is (z minus conjugate) divided by 2i.
2. Use 1/i = -i.

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

Explanation:
1. \( (1+i)^2=2i\Rightarrow(1+i)^3=2i(1+i)=-2+2i \).
2. \( (1-i)^2=-2i\Rightarrow(1-i)^6=(-2i)^3=8i \).
3. Sum \( =-2+10i=2(-1+5i) \).

Teacher's Note:
1. Compute the squares first.
2. Cube the result for the sixth power.

Answer: (d) none of these

Explanation:
1. \( 3+4i=(2+i)^2 \), so \( \sqrt{3+4i}=2+i \).
2. \( -3-4i=(1-2i)^2 \), so \( \sqrt{-3-4i}=1-2i \) (positive real part).
3. Sum \( =3-i \), which is not among the options.

Teacher's Note:
1. Express each number as a perfect square.
2. Choose the root with positive real part.

Answer: (b) \( -2\pm2i \)

Explanation:
1. \( \bar z_2=8y^2+20i \). Equating with \( z_1 \): \( 9y^2-4=8y^2 \) and \( -10x=20 \).
2. \( y^2=4\Rightarrow y=\pm2 \) and \( x=-2 \), so \( z=-2\pm2i \).

Teacher's Note:
1. Compare real and imaginary parts.
2. Do not forget both signs of y.

Answer: (d) no value of x

Explanation:
1. Conjugates need \( \sin x=\cos x \) and \( -\cos2x=\sin2x \).
2. The first gives \( x=\frac\pi4+n\pi \), where \( \cos2x=0 \) and \( \sin2x=\pm1 \), so the second fails.
3. There is no such x.

Teacher's Note:
1. Equate real parts and imaginary parts.
2. Check whether the two conditions agree.

Answer: (c) \( \frac12(1-4i) \)

Explanation:
1. The right side is real, so \( \text{Im}(z)=-2 \). Let \( z=x-2i \).
2. \( \sqrt{(x+1)^2+4}=x+2\Rightarrow x^2+2x+5=x^2+4x+4\Rightarrow x=\frac12 \).
3. \( z=\frac12-2i=\frac12(1-4i) \).

Teacher's Note:
1. The modulus on the left is real, so the imaginary part on the right must vanish.
2. Square after isolating x.

Answer: (c) \( 0 \)

Explanation:
1. Equality in the triangle inequality means \( z_1,z_2 \) point in the same direction.
2. So \( \frac{z_1}{z_2} \) is a positive real number and its amplitude is 0.

Teacher's Note:
1. Equal arguments mean the ratio is positive real.
2. Amplitude of a positive real number is zero.

Answer: (c) \( z+\bar z=0 \)

Explanation:
1. \( z=iy \) gives \( \bar z=-iy \), so \( z+\bar z=0 \).
2. (a) \( z\bar z \) is always real and (b) \( z=\bar z \) means z is real.

Teacher's Note:
1. Purely imaginary means the real part is zero.
2. The real part is (z + conjugate)/2.

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

Explanation:
1. \( z=\frac{e^{i\theta}}{e^{-i\theta}}=e^{2i\theta} \).
2. For \( \frac\pi4\lt \theta\lt \frac\pi2 \), \( 2\theta\in\left(\frac\pi2,\pi\right) \), which lies in the principal range, so \( \arg z=2\theta \).

Teacher's Note:
1. Use Euler's form.
2. Check that 2 theta is inside the principal range.

Answer: (b) \( \frac\pi3 \)

Explanation:
1. \( \sqrt3+i=2e^{i\pi/6} \) and \( \sqrt3-i=2e^{-i\pi/6} \).
2. \( z=e^{i\pi/6-(-i\pi/6)}=e^{i\pi/3} \), so the amplitude is \( \frac\pi3 \).

Teacher's Note:
1. Write numerator and denominator in polar form.
2. Subtract the arguments.

Answer: (b) 1 : 3

Explanation:
1. Equal amplitudes mean \( \frac{y}{x-1}=\frac{y+3}{x} \).
2. \( xy=(x-1)(y+3)=xy+3x-y-3\Rightarrow y=3(x-1) \).
3. So \( (x-1):y=1:3 \).

Teacher's Note:
1. Equal amplitudes mean equal slopes of the vectors.
2. Cross-multiply and simplify.

Answer: (c) 4

Explanation:
1. \( z^2=\pm ir^2 \). Each of the two values has two square roots.
2. Hence there are 4 possible values of z.

Teacher's Note:
1. z squared can be plus or minus i times r squared.
2. Count the square roots of each.

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

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

Teacher's Note:
1. Use 1 + omega + omega squared = 0.
2. Reduce the power of omega modulo 3.

Answer: (d) \( -\frac1{\sqrt3} \)

Explanation:
1. \( \frac32+\frac{\sqrt3}2i=\sqrt3\,e^{i\pi/6} \). Its 100th power is \( 3^{50}e^{i\,50\pi/3} \).
2. \( \frac{50\pi}3=16\pi+\frac{2\pi}3 \), so \( x+iy=3\left(-\frac12+\frac{\sqrt3}2i\right) \).
3. \( x=-\frac32,\ y=\frac{3\sqrt3}2\Rightarrow k=-\frac1{\sqrt3} \).

Teacher's Note:
1. Convert to polar form before taking the power.
2. Reduce the angle modulo 2 pi.

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

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

Teacher's Note:
1. Test the expression at the roots of each candidate.
2. Powers of omega repeat with period 3.

Answer: (b) \( -48 \)

Explanation:
1. \( x=e^{\pm i\pi/6} \). \( \left(x^n-x^{-n}\right)^2=x^{2n}+x^{-2n}-2 \).
2. \( x^2 \) is a primitive 6th root of unity, so \( \sum_{n=1}^{24}x^{2n}=0 \) and likewise for \( x^{-2n} \).
3. The sum is \( 24(-2)=-48 \).

Teacher's Note:
1. Expand the square.
2. Sums of roots of unity over full periods vanish.

Answer: (b) 6

Explanation:
1. \( 1+\sqrt3i=2e^{i\pi/3} \), so \( (1+\sqrt3i)^{n/2}=2^{n/2}e^{in\pi/6} \).
2. It is real when \( \frac{n\pi}6 \) is a multiple of \( \pi \), the least positive n being 6.

Teacher's Note:
1. Use polar form.
2. The argument must be a multiple of pi.

Answer: (a) 0 if n is even

Explanation:
1. Denominator: \( (1+\omega-\omega^2)(1-\omega+\omega^2)=(-2\omega^2)(-2\omega)=4 \).
2. Numerator: \( (2i)^n-(-2i)^n=2^ni^n\left[1-(-1)^n\right] \), which is 0 for even n and \( 2^{n+1}i^n \) for odd n.
3. So the expression is 0 for even n, which makes (a) true; (b) and (c) are false for odd n.

Teacher's Note:
1. Simplify the denominator using omega.
2. Check even and odd n separately.

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

Explanation:
1. Evaluating numerically with \( \omega=-\frac12+\frac{\sqrt3}2i \) gives \( -1+\sqrt3i \).
2. \( 2\omega=-1+\sqrt3i \), so the value is \( 2\omega \).

Teacher's Note:
1. Reduce the numerator and denominator using the sum of the cube roots of unity.
2. A numerical check with omega confirms the result.

Answer: (a) 2

Explanation:
1. \( (x-1)^4=16\Rightarrow x-1=\pm2,\ \pm2i \), so \( x=3,-1,1+2i,1-2i \).
2. The non-real values \( 1\pm2i \) have sum 2.

Teacher's Note:
1. Take the fourth roots of 16.
2. Add the conjugate pair.

Answer: (c) 1

Explanation:
1. The five numbers are the fifth roots of unity, with product \( (-1)^{5-1}\cdot1=1 \).
2. Equivalently, the exponents add to \( 1+2+3+4+0\to\frac{2\pi}5\cdot10=4\pi \), giving \( e^{4\pi i}=1 \).

Teacher's Note:
1. The product of all n-th roots of unity is (-1) to the n-1.
2. Add the angles to check.