JEE Mathematics Complex Numbers MCQs Set 11

Answer: (c) 6

Explanation:
1. \( 2^ne^{in\pi/6}=2^ne^{-in\pi/6}\Rightarrow e^{in\pi/3}=1 \).
2. So \( \frac{n\pi}3=2k\pi\Rightarrow n=6k \), the least being 6.

Teacher's Note:
1. Write both numbers in polar form.
2. The ratio must equal one.

Answer: (b) 0

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

Teacher's Note:
1. List the fourth roots of unity.
2. Square each and add.

Answer: (a) \( -7 \)

Explanation:
1. Take \( \alpha=\omega,\ \beta=\omega^2 \). \( 1+2\omega+\omega^2=\omega \), so its cube is 1.
2. \( 3+3\omega+5\omega^2=2\omega^2 \), so its cube is 8.
3. The value is \( 1-8=-7 \).

Teacher's Note:
1. Use 1 + omega + omega squared = 0 to simplify.
2. Cube roots of unity cube to 1.

Answer: (d) 12

Explanation:
1. \( (\sqrt3-i)^n=2^ne^{-in\pi/6}=2^n \) needs \( \frac{n\pi}6=2k\pi \).
2. So \( n=12k \), a multiple of 12.

Teacher's Note:
1. Write the number in polar form.
2. The argument must be a multiple of 2 pi.

Answer: (b) \( -\frac\pi6 \)

Explanation:
1. \( 2-2\sqrt3i=4e^{-i\pi/3} \), so its square has argument \( -\frac{2\pi}3 \). \( (\sqrt3+i)^4 \) has argument \( \frac{2\pi}3 \).
2. \( \arg z=\frac\pi2+\frac{2\pi}3-\left(-\frac{2\pi}3\right)=\frac{11\pi}6\equiv-\frac\pi6 \).

Teacher's Note:
1. Take arguments on both sides.
2. Reduce the angle to the principal range.

Answer: (b) \( \frac12\left(|\alpha+\beta|^2+|\alpha-\beta|^2\right) \)

Explanation:
1. By the parallelogram law, \( |\alpha+\beta|^2+|\alpha-\beta|^2=2\left(|\alpha|^2+|\beta|^2\right) \).
2. Hence \( |\alpha|^2+|\beta|^2=\frac12\left(|\alpha+\beta|^2+|\alpha-\beta|^2\right) \).

Teacher's Note:
1. Expand both squares of moduli.
2. The cross terms cancel.

Answer: (b) \( \{1\} \)

Explanation:
1. Conjugate roots occur only when all coefficients are real.
2. So \( i(a-1) \) must be real, which forces \( a=1 \). Then \( x^2+5=0 \) has roots \( \pm\sqrt5i \).

Teacher's Note:
1. A polynomial with a pair of conjugate roots has real coefficients.
2. Set the imaginary coefficient to zero.

Answer: (a) \( \frac{-1\pm\sqrt{-7}}2 \)

Explanation:
1. \( z^3+2z^2+3z+2=(z+1)(z^2+z+2) \).
2. The quadratic \( z^2+z+2=0 \) gives \( z=\frac{-1\pm\sqrt{-7}}2 \), which are the non-real roots.

Teacher's Note:
1. Test z = -1 to find a factor.
2. Solve the remaining quadratic.

Answer: (a) \( z \)

Explanation:
1. With \( |z|=1 \), \( \bar z=\frac1z \). Then \( 1+\bar z=\frac{z+1}z \).
2. \( \frac{1+z}{1+\bar z}=\frac{(1+z)z}{1+z}=z \).

Teacher's Note:
1. Use the conjugate equals 1/z for a unit complex number.
2. Cancel (1 + z).

Answer: (a) 1

Explanation:
1. \( |\alpha|=1 \), and \( |\alpha^n|=|\alpha|^n=1 \).

Teacher's Note:
1. Roots of unity have modulus 1.
2. The modulus of a power is the power of the modulus.

Answer: (d) \( \text{Im}(\omega_1\bar\omega_2)=1 \)

Explanation:
1. \( z_1\bar z_2=(a+ib)(p-iq)=(ap+bq)+i(bp-aq) \), so \( bp-aq=1 \).
2. \( \omega_1\bar\omega_2=(a+ip)(b-iq)=(ab+pq)+i(pb-aq) \), so \( \text{Im}(\omega_1\bar\omega_2)=bp-aq=1 \).

Teacher's Note:
1. Expand both products.
2. Compare the imaginary parts.

Answer: (c) is less than 12

Explanation:
1. \( |z_1|\lt |z_1-1|+1\lt 2,\ |z_2|\lt 4,\ |z_3|\lt 6 \).
2. \( |z_1+z_2+z_3|\le|z_1|+|z_2|+|z_3|\lt 12 \).

Teacher's Note:
1. Use the triangle inequality for each variable.
2. Add the three bounds.

Answer: (b) 7

Explanation:
1. \( |iz+z_0|=|z+z_0/i|=|z-iz_0|=|z-(-3+5i)| \).
2. z lies in the disc of radius 2 about \( i \). The distance from \( i \) to \( -3+5i \) is \( |3-4i|=5 \), so the maximum is \( 5+2=7 \).

Teacher's Note:
1. Factor i out of the expression.
2. Maximum distance equals centre distance plus radius.

Answer: (a) \( |z|\lt 3 \)

Explanation:
1. Base \( \frac12 \): the log is negative when the argument exceeds 1.
2. \( |z|^2+2|z|+4>2|z|^2+1\Rightarrow|z|^2-2|z|-3\lt 0\Rightarrow(|z|-3)(|z|+1)\lt 0 \).
3. So \( |z|\lt 3 \).

Teacher's Note:
1. For base less than 1 the log is negative when the argument is greater than 1.
2. Factor the quadratic in
3. z
4. .

Answer: (c) a straight line

Explanation:
1. \( |z-1|=|z+1| \): the points equidistant from 1 and -1.
2. This is the perpendicular bisector, the imaginary axis, a straight line.

Teacher's Note:
1. Equal distances from two fixed points give a line.
2. The line is the perpendicular bisector.

Answer: (d) 0

Explanation:
1. The formula is the section formula, so A lies on the line BC.
2. The distance of A from BC is 0.

Teacher's Note:
1. Recognise the section formula.
2. A point on a line is at distance zero from it.

Answer: (b) an equilateral triangle

Explanation:
1. \( 1+z+z^3+z^4=(1+z)(1+z^3) \), so the roots are the cube roots of \( -1 \) together with \( -1 \).
2. The distinct roots are \( -1,e^{i\pi/3},e^{-i\pi/3} \), which are the vertices of an equilateral triangle on the unit circle.

Teacher's Note:
1. Factor by grouping.
2. Plot the cube roots of minus one.

Answer: (d) \( \frac\pi6 \)

Explanation:
1. \( \sqrt3-i=2e^{-i\pi/6} \), so \( (\sqrt3-i)^{25}=2^{25}e^{-i25\pi/6} \).
2. The reciprocal has argument \( \frac{25\pi}6=4\pi+\frac\pi6\equiv\frac\pi6 \).

Teacher's Note:
1. Take the reciprocal by negating the argument.
2. Reduce the angle modulo 2 pi.

Answer: (a) concyclic

Explanation:
1. All four points are at the same distance from the origin.
2. So they lie on a circle centred at the origin: they are concyclic.

Teacher's Note:
1. Equal moduli mean equal distances from the origin.
2. Points at equal distance from a point lie on a circle.

Answer: (a) \( z_2=1-\sqrt3i,\ z_3=-2 \)

Explanation:
1. \( z_1=2e^{i\pi/3} \). Clockwise rotation by \( 120^\circ \): \( z_2=2e^{i(\pi/3-2\pi/3)}=2e^{-i\pi/3}=1-\sqrt3i \).
2. \( z_3=2e^{-i\pi}=-2 \).

Teacher's Note:
1. Rotate by 120 degrees about the centre.
2. Clockwise means subtract the angle.

Answer: (c) 2

Explanation:
1. \( z=i \) is a root; dividing gives \( z^2+2iz-2=0 \) with roots \( -1-i,\ 1-i \).
2. Vertices \( (0,1),(1,-1),(-1,-1) \): base 2, height 2.
3. Area \( =\frac12\cdot2\cdot2=2 \).

Teacher's Note:
1. Find one root by trial.
2. Use base times height for the area.

Answer: (b) \( 2\sqrt5 \)

Explanation:
1. Complete the square: \( |z+(4+3i)|^2=|4+3i|^2-5=25-5=20 \).
2. Radius \( =\sqrt{20}=2\sqrt5 \).

Teacher's Note:
1. For z z-bar + a-bar z + a z-bar + c = 0, the centre is -a and the radius squared is
2. a
3. squared minus c.
4. Here
5. a
6. squared is 25.

Answer: (b) 3

Explanation:
1. For a circle on diameter \( z_1z_2 \): \( |z-z_1|^2+|z-z_2|^2=|z_1-z_2|^2 \).
2. \( |\omega-\omega^2|^2=|\sqrt3i|^2=3 \), so \( \lambda=3 \).

Teacher's Note:
1. The sum of squares of distances to the ends of a diameter is the square of the diameter.
2. Compute the distance between the two cube roots.

Answer: (b) \( \frac23 \)

Explanation:
1. For an equilateral triangle with vertex 0 and other vertices \( z_1,z_2 \): \( z_1^2+z_2^2=z_1z_2 \), i.e. \( (z_1+z_2)^2=3z_1z_2 \).
2. \( z_1+z_2=-1,\ z_1z_2=\frac\lambda2\Rightarrow1=\frac{3\lambda}2\Rightarrow\lambda=\frac23 \).

Teacher's Note:
1. Use the condition for an equilateral triangle with a vertex at the origin.
2. Express it with the sum and product of the roots.

Answer: (c) 4

Explanation:
1. The foci are \( \pm i \), at distance 2 apart. An ellipse needs \( k>2 \).
2. Among the options only \( k=4 \) satisfies this.

Teacher's Note:
1. The constant sum must exceed the distance between the foci.
2. Here the focal distance is 2.