JEE Mathematics Complex Numbers MCQs Set 15

Answer: (a) \( |z+3|^2 \)

Explanation:
1. Since \( \overline{z+3}=\bar z+3 \), the product is \( (z+3)\overline{(z+3)}=|z+3|^2 \).

Teacher's Note:
1. w times conjugate(w) is |w|².
2. Conjugation leaves real numbers unchanged.

Answer: (c) Four solutions

Explanation:
1. Taking moduli: \( |z|^2=|z| \), so \( |z|=0 \) or \( |z|=1 \).
2. If \( |z|=1 \), then \( \bar z=1/z \) and the equation becomes \( z^3=1 \), which has 3 roots.
3. With \( z=0 \) there are 4 solutions in all.

Teacher's Note:
1. Use the modulus to split into |z| = 0 and |z| = 1.
2. Do not forget the trivial solution z = 0.

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

Explanation:
1. \( \bar z_2=8y^2+20i \).
2. Equate real parts: \( 9y^2-4=8y^2 \), so \( y=\pm2 \).
3. Imaginary parts: \( -10x=20 \), so \( x=-2 \).
4. Hence \( z=-2\pm2i \).

Teacher's Note:
1. Equate real and imaginary parts separately.
2. y appears squared, so both signs are valid.

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

Explanation:
1. Options (a), (b) and (d) are standard identities.
2. \( z=\bar z \) holds only when z is real, so it is not true for every complex z.

Teacher's Note:
1. Test an identity with z = i to see whether it is universal.
2. Conjugate of a power equals the power of the conjugate.

Answer: (a) Any real number

Explanation:
1. The equation says z is equidistant from \( i \) and \( -i \).
2. That is the perpendicular bisector of the segment joining them, the real axis.

Teacher's Note:
1. Equidistance from two points gives the perpendicular bisector.
2. For z = x + iy the equation reduces to 2y = 0.

Answer: (c) \( \tfrac13 \)

Explanation:
1. Write the sum as \( 2|z-\tfrac12|+3|z-\tfrac23| \).
2. The larger weight 3 pulls the minimum to \( z=\tfrac23 \), where the value is \( 2\left|\tfrac23-\tfrac12\right|=\tfrac13 \).

Teacher's Note:
1. For a weighted sum of two distances the minimum is at the heavier point.
2. Check with the triangle inequality: the sum is at least the smaller weight times the distance between the points.

Answer: (b) \( 2|z_1|^2+2|z_2|^2 \)

Explanation:
1. Expand: \( |z_1\pm z_2|^2=|z_1|^2+|z_2|^2\pm2\,\mathrm{Re}(z_1\bar z_2) \).
2. Adding the two cases cancels the cross term: \( 2|z_1|^2+2|z_2|^2 \).

Teacher's Note:
1. This is the parallelogram law.
2. The sum of squares of the diagonals equals the sum of squares of the four sides.

Answer: (a) Equal to 1

Explanation:
1. For \( |z|=1 \), \( \frac1z=\bar z \).
2. So \( \left|\bar z_1+\bar z_2+\bar z_3\right|=1 \), and the modulus of a conjugate equals the original modulus.
3. Hence \( |z_1+z_2+z_3|=1 \).

Teacher's Note:
1. On the unit circle the reciprocal is the conjugate.
2. |conj(w)| = |w|.

Answer: (c) \( 6+8i,\ 6+17i \)

Explanation:
1. The second condition says z is equidistant from 4 and 8, so \( \mathrm{Re}(z)=6 \). Put \( z=6+iy \).
2. The first gives \( 9(36+y^2)=25(36+(y-8)^2) \), i.e. \( y^2-25y+136=0 \), so \( y=8 \) or \( y=17 \).
3. Both values check, so \( z=6+8i \) or \( 6+17i \).

Teacher's Note:
1. Use the second condition to fix the real part.
2. Verify both roots in the original equation.

Answer: (c) \( \alpha^2+\beta^2=1 \)

Explanation:
1. For real x the numerator and denominator are conjugates, so the ratio has modulus 1.
2. Hence \( \alpha^2+\beta^2=1 \).

Teacher's Note:
1. Modulus of the ratio of conjugates equals 1.
2. The condition is on α and β only, not on x.

Answer: (b) \( -\sec\alpha \)

Explanation:
1. \( |z|=\sqrt{1+\tan^2\alpha}=|\sec\alpha| \).
2. In the third quadrant \( \cos\alpha<0 \), so \( |\sec\alpha|=-\sec\alpha \).

Teacher's Note:
1. The square root of sec²α is |secα|, not secα.
2. Check the sign of cos in the given quadrant.

Answer: (c) \( z_1z_2 \) is unimodular

Explanation:
1. \( \frac1{z_1}+\frac1{z_2}=\frac{z_1+z_2}{z_1z_2} \).
2. So \( |z_1+z_2|=\frac{|z_1+z_2|}{|z_1z_2|} \) and, since \( z_1+z_2\ne0 \), \( |z_1z_2|=1 \).

Teacher's Note:
1. Combine the reciprocals into a single fraction.
2. Only the product has to have unit modulus.

Answer: (d) Infinitely many

Explanation:
1. Put \( z=x+iy \): \( x^2-y^2+2ixy+x^2+y^2=0 \) gives \( 2x^2+2ixy=0 \).
2. Real part: \( x=0 \). Then the imaginary part \( 2xy=0 \) holds for every y.
3. So every purely imaginary number is a solution: infinitely many.

Teacher's Note:
1. Separate real and imaginary parts.
2. x = 0 makes both equations true regardless of y.

Answer: (a) 1

Explanation:
1. By the triangle inequality, \( |z|+|1-z|\ge|z+(1-z)|=1 \).
2. Equality holds for every point on the segment from 0 to 1, so the minimum is 1.

Teacher's Note:
1. |z| + |z − 1| is the sum of distances from 0 and 1.
2. It is least when z lies on the segment joining them.

Answer: (b) 6, 0

Explanation:
1. Write \( z+1=(z+4)-3 \). Then \( |z+1|\le|z+4|+3\le6 \) and, the lower bound being attained at \( z=-1 \) (where \( |z+4|=3 \)), the least value is 0.
2. The greatest value 6 is attained at \( z=-7 \).

Teacher's Note:
1. The disc |z + 4| ≤ 3 contains the point −1 on its boundary.
2. A modulus is never negative, so −6 is impossible.

Answer: (c) 1 or -1

Explanation:
1. Since \( |z+iw|\le|z|+|w|\le2 \), equality needs \( |z|=|w|=1 \) and \( z=iw \).
2. Likewise \( |z-i\bar w|=2 \) needs \( z=-i\bar w \).
3. Hence \( iw=-i\bar w \), so \( w=-\bar w \) and w is purely imaginary: \( w=\pm i \).
4. Then \( z=iw=\mp1 \), i.e. z = 1 or -1.

Teacher's Note:
1. The triangle inequality is an equality only for vectors in the same direction.
2. Use the bounds on |z| and |w| to force unit moduli.

Answer: (a) Purely imaginary

Explanation:
1. Let \( z_1=re^{i\alpha},\ z_2=re^{i\beta} \). Then \( \frac{z_1+z_2}{z_1-z_2}=\frac{\cos\frac{\alpha-\beta}2}{i\sin\frac{\alpha-\beta}2}=-i\cot\frac{\alpha-\beta}2 \).
2. This is always purely imaginary (or zero).

Teacher's Note:
1. The sum and difference of two equal-length vectors are perpendicular (diagonals of a rhombus).
2. A purely imaginary quotient means perpendicular vectors.

Answer: (d) \( (a^2+b^2)(|z_1|^2+|z_2|^2) \)

Explanation:
1. Expand each term: \( |az_1-bz_2|^2=a^2|z_1|^2+b^2|z_2|^2-2ab\,\mathrm{Re}(z_1\bar z_2) \) and \( |bz_1+az_2|^2=b^2|z_1|^2+a^2|z_2|^2+2ab\,\mathrm{Re}(z_1\bar z_2) \).
2. The cross terms cancel, leaving \( (a^2+b^2)(|z_1|^2+|z_2|^2) \).

Teacher's Note:
1. Use |u ± v|² = |u|² + |v|² ± 2Re(u conj v).
2. The cross terms have opposite signs and cancel.

Answer: (a) \( \bar z_1,\bar z_2,\bar z_3,\bar z_4 \) are also roots of the equation

Explanation:
1. Conjugate the equation: since all coefficients are real, \( a_0\bar z^4+a_1\bar z^3+\cdots+a_4=0 \).
2. So the conjugate of every root is also a root, which is option (a).

Teacher's Note:
1. Non-real roots of a real polynomial come in conjugate pairs.
2. Option (b) fails when a root is non-real and its conjugate is a different root.

Answer: (a) \( |w+1|<|w-8| \)

Explanation:
1. \( |z+1|<|z-2| \) means \( \mathrm{Re}(z)<\tfrac12 \).
2. Then \( \mathrm{Re}(w)=3\,\mathrm{Re}(z)+2<\tfrac72 \).
3. The condition \( |w+1|<|w-8| \) means \( \mathrm{Re}(w)<\tfrac72 \), which is option (a). The other options give \( \mathrm{Re}(w)<3 \), \( \mathrm{Re}(w)>\tfrac72 \) and \( \mathrm{Re}(w)<-\tfrac12 \), none implied.

Teacher's Note:
1. Translate each modulus inequality into a half-plane.
2. The perpendicular bisector of −1 and 8 is Re(w) = 7/2.

Answer: (a) n

Explanation:
1. The disc has centre \( 4-3i \) (distance 5 from the origin) and radius 1, so \( m=4 \) and \( n=6 \).
2. Let \( f(x)=x^3+x+\frac4x \). Then \( f'(x)=3x^2+1-\frac4{x^2}=0\Rightarrow3x^4+x^2-4=0 \), so \( x^2=1 \).
3. The minimum is \( f(1)=6=n \).

Teacher's Note:
1. The extreme moduli on a disc are |centre| ± radius.
2. Differentiate and solve the quartic as a quadratic in x².

Answer: (a) No solution

Explanation:
1. The first is a circle with centre \( -1+i \) and radius \( \sqrt2 \); the second is a circle with centre 0 and radius 3.
2. The distance between centres is \( \sqrt2 \), and \( \sqrt2+\sqrt2=2\sqrt2<3 \), so the small circle lies entirely inside the big circle.
3. The circles do not meet: no solution.

Teacher's Note:
1. Compare the distance between centres with the sum and difference of radii.
2. If d + r1 < r2 the smaller circle lies inside the larger.

Answer: (d) \( -\tfrac{2\pi}3 \)

Explanation:
1. The point \( (-1,-\sqrt3) \) lies in the third quadrant with reference angle \( \tfrac\pi3 \).
2. The principal argument is \( -\left(\pi-\tfrac\pi3\right)=-\tfrac{2\pi}3 \).

Teacher's Note:
1. Principal arguments lie in (−π, π].
2. For the third quadrant subtract π from the reference-angle form.

Answer: (a) \( \tfrac\pi6 \)

Explanation:
1. \( \arg(1+\sqrt3i)=\tfrac\pi3 \) and \( \arg(\sqrt3+i)=\tfrac\pi6 \).
2. The amplitude of the quotient is \( \tfrac\pi3-\tfrac\pi6=\tfrac\pi6 \).

Teacher's Note:
1. Argument of a quotient is a difference of arguments.
2. Amplitude is another name for argument.