JEE Mathematics Complex Numbers MCQs Set 26

Answer: (b) \( z=-2\omega \)

Explanation:
1. Rotating \( C \) about the centroid through \( 120^\circ \) anticlockwise gives \( A \) (order A, B, C anticlockwise): \( A-1=(C-1)\omega \).
2. Choose \( \omega=-\frac12+\frac{\sqrt3}2i \). Then \( A=1+(\omega-1)\omega=1+\omega^2-\omega=1-\sqrt3i \).
3. Since \( -2\omega=1-\sqrt3i \), \( A=-2\omega \).

Teacher's Note:
1. Vertices of an equilateral triangle are related by rotations of 120 degrees about the centroid.
2. Express the result using 1 + omega + omega squared = 0.

Answer: (d) 1

Explanation:
1. With \( z=x+iy \): \( \operatorname{Re}((1-i)z)=x+y=\sqrt2 \) (a line); \( B \) is the circle \( (x-2)^2+(y-1)^2=9 \).
2. Substituting \( x=\sqrt2-y \): \( 2y^2-2(\sqrt2-1)y-(8-(2-\sqrt2)^2)=0 \), with roots \( y\approx2.17 \) and \( y\approx-1.76 \).
3. Only \( y\approx2.17\ge1 \) is in \( A \): exactly one element.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 48

Explanation:
1. \( z\bar z^3+\bar zz^3=|z|^2(z^2+\bar z^2)=2(x^2+y^2)(x^2-y^2)=350 \), so \( (x^2+y^2)(x^2-y^2)=175 \).
2. Integer solutions: \( x^2+y^2=25,\ x^2-y^2=7 \) giving \( (\pm4,\pm3) \) (other factorisations give non-square values).
3. Rectangle sides \( 8 \) and \( 6 \): area 48.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( z=-2 \)

Explanation:
1. \( \omega=-\frac12+\frac{\sqrt3}2i \) and \( \omega^2=-\frac12-\frac{\sqrt3}2i \) are at distance \( \sqrt3 \) apart on the line \( x=-\frac12 \).
2. The third vertex lies on the real axis at distance \( \frac{\sqrt3}{2}\cdot\sqrt3=\frac32 \) from this line: \( x=-\frac12\pm\frac32 \), i.e. \( 1 \) or \( -2 \).
3. Among the options only \( z=-2 \) is possible.

Teacher's Note:
1. The third vertex is on the perpendicular bisector of the given side.
2. Height of an equilateral triangle = (root 3 / 2) x side.

Answer: (b) 35 and 39

Explanation:
1. The point is \( z\approx-0.760+2.175i \) (from the previous step).
2. Then \( |z+1-i|^2\approx(0.240)^2+(1.175)^2=1.44 \) and \( |z-5-i|^2\approx(5.760)^2+(1.175)^2=34.56 \).
3. Sum \( \approx36 \), which lies between 35 and 39.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) (i), (iii) and (iv) only

Explanation:
1. \( z-z_1=t(z_2-z_1) \) and \( z-z_2=(1-t)(z_1-z_2) \): \( z \) lies on the segment between \( z_1,z_2 \).
2. So (i) holds, (iii) holds (\( t>0 \)) and (iv) holds (collinearity), but \( z-z_1 \) and \( z-z_2 \) point in opposite directions, so (ii) is false.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \operatorname{Im}(z)=0 \)

Explanation:
1. Let \( |z|=r \). The points \( ir \) and \( -ir \) are symmetric about the real axis, so \( z \) is equidistant from them only when \( z \) lies on the perpendicular bisector, the real axis: \( \operatorname{Im}z=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) equilateral

Explanation:
1. The modulus of the ratio is \( \left|\frac{1-\sqrt3i}2\right|=1 \), so \( |z_1-z_3|=|z_2-z_3| \).
2. Its argument is \( -60^\circ \), so the angle at \( z_3 \) is \( 60^\circ \).
3. An isosceles triangle with apex angle \( 60^\circ \) is equilateral.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 1

Explanation:
1. Expanding and using \( 1+\omega+\omega^2=0 \), \( \omega^3=1 \) the determinant simplifies to \( z^3 \).
2. The only root is \( z=0 \): 1 distinct number.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \{z:|z|=1,\ z\ne1\} \)

Explanation:
1. Let \( E=\frac{w-\bar wz}{1-z} \). \( E=\bar E \) gives \( (w-\bar wz)(1-\bar z)=(\bar w-w\bar z)(1-z) \).
2. Expanding: \( w-w\bar z-\bar wz+\bar w|z|^2=\bar w-\bar wz-w\bar z+w|z|^2 \), so \( (w-\bar w)(1-|z|^2)=0 \).
3. Since \( w\ne\bar w \), \( |z|=1 \), and \( z\ne1 \) for the expression to be defined.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) the imaginary axis

Explanation:
1. \( |z^2-1|\le|z^2|+1 \) with equality only when \( z^2 \) and \( -1 \) point in the same direction, i.e. \( z^2 \) is real and \( \le0 \).
2. That means \( z \) is purely imaginary (or 0): the imaginary axis.

Teacher's Note:
1. Equality in the triangle inequality needs the two vectors in the same direction.
2. Write z = x + iy and check: x must be 0.

Answer: (c) the imaginary axis (the y-axis)

Explanation:
1. \( \frac{z}{1-z^2}=\frac{1}{\frac1z-z}=\frac1{\bar z-z} \) since \( |z|=1 \).
2. With \( z=x+iy \): \( \bar z-z=-2iy \), so the value is \( \frac{i}{2y} \), purely imaginary: it lies on the y-axis.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) (i) and (iii) only

Explanation:
1. Write \( z_1=e^{i\theta} \); then \( z_2=e^{-i\theta} \).
2. So \( z_1z_2=1 \) and \( z_1=\bar z_2 \). (ii) and (iv) hold only for special values of \( \theta \).

Teacher's Note:
1. Equal moduli plus opposite arguments gives conjugates.
2. Test (ii) and (iv) with theta = 30 degrees.

Answer: (a) \( -\operatorname{cosec}\alpha \)

Explanation:
1. \( |z|=\sqrt{1+\cot^2\alpha}=|\operatorname{cosec}\alpha| \).
2. For \( -\frac{\pi}{2}<\alpha<0 \), \( \sin\alpha<0 \), so \( |\operatorname{cosec}\alpha|=-\operatorname{cosec}\alpha \).

Teacher's Note:
1. A modulus is never negative, so check the sign of cosec in the given interval.
2. Here alpha is in the fourth quadrant.

Answer: (d) \( \frac{15}{2}\pi \)

Explanation:
1. \( 1\le|z-1|\le4 \) is an annulus centred at 1 with area \( \pi(16-1)=15\pi \).
2. The condition \( z+\bar z>2 \) means \( x>1 \): the half-plane through the centre.
3. So the region is half the annulus: \( \frac{15\pi}{2} \).

Teacher's Note:
1. The line x = 1 passes through the centre of the annulus.
2. Half the area remains.

Answer: (a) \( 4k \)

Explanation:
1. The roots are \( e^{2\pi ir/n} \); the angle between two of them is a multiple of \( \frac{2\pi}n \).
2. A right angle needs \( \frac{2\pi m}{n}=\frac\pi2 \), so \( n=4m \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \pi \)

Explanation:
1. Principal arguments lie in \( (-\pi,\pi] \). If \( \arg z=\theta<0 \), then \( -z \) has argument \( \theta+\pi \) (which lies in \( (0,\pi) \)).
2. The difference is \( \pi \).

Teacher's Note:
1. Multiplying by -1 rotates by pi; choose the representative in the principal range.
2. For negative arg z, adding pi stays in range.

Answer: (d) 2

Explanation:
1. \( z_2 \) lies on a circle of radius 5 centred at \( 3+4i \) (at distance 5 from the origin), so \( |z_2|\le10 \).
2. Then \( |z_1-z_2|\ge|z_1|-|z_2|\ge12-10=2 \), attained when both lie on the same ray from the origin.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( (3+4i)e^{i\pi/4} \)

Explanation:
1. North-east is the direction \( e^{i\pi/4} \) and north-west is \( e^{3i\pi/4}=i\,e^{i\pi/4} \).
2. Position: \( 3e^{i\pi/4}+4ie^{i\pi/4}=(3+4i)e^{i\pi/4} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( -6+7i \)

Explanation:
1. Horizontally away: \( 1+2i\to6+2i \); vertically away: \( 6+5i=z_1 \).
2. Moving \( \sqrt2 \) along \( \hat i+\hat j \) adds \( 1+i \): \( 7+6i \).
3. Rotating anticlockwise by \( \frac\pi2 \) about the origin multiplies by \( i \): \( i(7+6i)=-6+7i \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \frac1{4\sin2^\circ} \)

Explanation:
1. The sum is \( \sin\theta+\sin3\theta+\dots+\sin29\theta=\frac{\sin^2 15\theta}{\sin\theta} \).
2. At \( \theta=2^\circ \): \( \frac{\sin^230^\circ}{\sin2^\circ}=\frac1{4\sin2^\circ} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac45 \)

Explanation:
1. Put \( \omega=2e^{i\theta} \): \( z=\frac32\cos\theta+\frac52i\sin\theta \).
2. Semi-axes \( \frac32 \) and \( \frac52 \): \( e^2=1-\frac{9/4}{25/4}=\frac{16}{25} \), so \( e=\frac45 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.