JEE Mathematics Complex Numbers MCQs Set 24

Answer: (c) \( 1 \)

Explanation:
1. \( z_1,\dots,z_4 \) are the four non-real roots of \( z^5=1 \) and \( z_5=\cos2\pi+i\sin2\pi=1 \).
2. \( z^5-1=(z-1)(z-z_1)\cdots(z-z_4) \), so the product of the four non-real roots is \( 1 \) (constant term \( -1 \) divided by \( -1 \)).
3. Hence \( z_1z_2z_3z_4z_5=1\cdot1=1 \).

Teacher's Note:
1. Use the sum and product of roots of \( z^n=1 \).
2. Do not forget that \( z_5 \) repeats \( z_0=1 \).

Answer: (c) \( 4\sqrt3\,i \)

Explanation:
1. \( -\tfrac12+\tfrac{i\sqrt3}{2}=\omega \) and \( \tfrac12+\tfrac{i\sqrt3}{2}=-\omega^2 \).
2. \( \omega^{334}=\omega^{1}\) (as \( 334=3\cdot111+1 \)) and \( (-\omega^2)^{365}=-\omega^{730}=-\omega \) (as \( 730=3\cdot243+1 \)).
3. Expression \( =4+5\omega+3\omega=4+8\omega=4+8\left(-\tfrac12+\tfrac{i\sqrt3}{2}\right)=4\sqrt3\,i \).

Teacher's Note:
1. Write each number as a power of \( \omega \) (with a sign) and reduce exponents mod 3.
2. Watch the sign when \( \tfrac12+\tfrac{i\sqrt3}{2}=-\omega^2 \).

Answer: (b) \( \dfrac12\left(|\alpha+\beta|^2+|\alpha-\beta|^2\right) \)

Explanation:
1. \( |\alpha+\beta|^2=|\alpha|^2+|\beta|^2+2\,\mathrm{Re}(\alpha\bar\beta) \).
2. \( |\alpha-\beta|^2=|\alpha|^2+|\beta|^2-2\,\mathrm{Re}(\alpha\bar\beta) \).
3. Adding gives \( 2(|\alpha|^2+|\beta|^2) \), so the required value is half the sum (parallelogram law).

Teacher's Note:
1. Expand both squared moduli with \( |z|^2=z\bar z \).
2. The cross terms cancel when you add.

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

Explanation:
1. For conjugate roots \( p\pm iq \) the sum of roots is real.
2. The sum of roots is \( -i(a-1) \), which is real only if \( a=1 \).
3. For \( a=1 \) the equation is \( x^2+5=0 \) with roots \( \pm i\sqrt5 \), a conjugate pair.

Teacher's Note:
1. Conjugate roots force both the sum and the product to be real.
2. Check the surviving value in the original equation.

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

Explanation:
1. \( z=-1 \) is a root, so \( z^3+2z^2+3z+2=(z+1)(z^2+z+2) \).
2. The quadratic \( z^2+z+2=0 \) has discriminant \( 1-8=-7 \).
3. The non-real roots are \( z=\dfrac{-1\pm\sqrt{-7}}{2} \).

Teacher's Note:
1. Look for a small integer root first.
2. Then factor and use the quadratic formula.

Answer: (c) is less than 12

Explanation:
1. \( |z_1|<|z_1-1|+1<2,\ |z_2|<4,\ |z_3|<6 \).
2. So \( |z_1+z_2+z_3|\le|z_1|+|z_2|+|z_3|<2+4+6=12 \).
3. The value is less than 12; the other statements need not hold (for example \( z_1,z_2,z_3 \) near 0 gives a small sum, and values above 6 are possible).

Teacher's Note:
1. Bound each modulus with the triangle inequality.
2. Test the other options with a clever choice of points.

Answer: (b) \( 7 \)

Explanation:
1. \( |iz+z_0|=|i|\,|z+z_0/i|=|z+(-iz_0)|=|z+3-5i| \) (as \( -i(5+3i)=3-5i \)).
2. \( |z+3-5i|\le|z-i|+|i+3-5i|\le2+|3-4i|=2+5=7 \).
3. Equality is attainable, so the maximum is 7.

Teacher's Note:
1. Factor out i to turn the expression into a distance from a fixed point.
2. Use the triangle inequality and check that equality can occur.

Answer: (d) none of these

Explanation:
1. The points closer to \( 4 \) than to \( 2 \) lie on one side of the perpendicular bisector \( \mathrm{Re}(z)=3 \).
2. Squaring: \( (x-4)^2<(x-2)^2 \Rightarrow -8x+16<-4x+4 \Rightarrow x>3 \).
3. The region is \( \mathrm{Re}(z)>3 \), which is none of the listed options.

Teacher's Note:
1. Compare distances by squaring both sides.
2. The boundary is the perpendicular bisector of the two points.

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

Explanation:
1. Base \( \tfrac12<1 \), so the log is negative when the argument is greater than 1.
2. \( |z|^2+2|z|+4>2|z|^2+1\Rightarrow |z|^2-2|z|-3<0\Rightarrow(|z|-3)(|z|+1)<0 \).
3. Since \( |z|+1>0 \), we get \( |z|<3 \).

Teacher's Note:
1. Remember that a logarithm with base less than 1 reverses the inequality.
2. Put \( r=|z| \) and solve the quadratic inequality.

Answer: (c) a straight line

Explanation:
1. \( |z-1|=|z+1| \) means z is equidistant from \( 1 \) and \( -1 \).
2. This is the perpendicular bisector of the segment joining them, namely the imaginary axis \( x=0 \).
3. It is a straight line.

Teacher's Note:
1. A ratio of distances equal to 1 gives a perpendicular bisector.
2. A ratio different from 1 would give a circle.

Answer: (d) none of these

Explanation:
1. \( 1+z+z^3+z^4=(1+z)(1+z^3) \), so the roots are \( -1 \) (twice), \( e^{i\pi/3} \) and \( e^{-i\pi/3} \).
2. Only three distinct points occur, one of them repeated.
3. Four distinct vertices are needed for a square or a rhombus, and the repeated root means the roots are not four vertices of an equilateral triangle either, so none of the options is right.

Teacher's Note:
1. Factor first, then count the distinct roots.
2. A repeated root rules out a quadrilateral.

Answer: (a) concyclic

Explanation:
1. Equal moduli mean each point is at the same distance from the origin.
2. So all four points lie on a circle centred at the origin.
3. They are concyclic; nothing more can be said.

Teacher's Note:
1. Equal modulus means the same distance from the origin.
2. Square or rhombus needs extra conditions.

Answer: (c) \( 2 \)

Explanation:
1. Substituting shows \( z=i \) is a root; \( z^3+iz^2+2i=(z-i)(z^2+2iz-2) \), giving roots \( i,\ 1-i,\ -1-i \).
2. Two vertices \( 1-i \) and \( -1-i \) lie on the line \( y=-1 \), distance 2 apart.
3. The third vertex \( i \) is at height \( 2 \) above that line, so area \( =\tfrac12\cdot2\cdot2=2 \).

Teacher's Note:
1. Find the roots, then use base times height.
2. A horizontal base makes the area easy.

Answer: (a) the real axis

Explanation:
1. \( |z-3i|=|z+3i| \) means z is equidistant from \( 3i \) and \( -3i \).
2. The perpendicular bisector of these two points is the line \( y=0 \).
3. So z lies on the real axis.

Teacher's Note:
1. A modulus ratio of 1 gives a perpendicular bisector.
2. The two points lie on the imaginary axis, so the bisector is the real axis.

Answer: (b) \( \dfrac23 \)

Explanation:
1. For an equilateral triangle with vertices \( 0,z_1,z_2 \): \( z_1^2+z_2^2=z_1z_2 \), i.e. \( (z_1+z_2)^2=3z_1z_2 \).
2. Here \( z_1+z_2=-1 \) and \( z_1z_2=\lambda/2 \).
3. \( 1=\dfrac{3\lambda}{2}\Rightarrow\lambda=\dfrac23 \). The discriminant \( 4-8\lambda=-\tfrac43<0 \), so the roots are indeed non-real.

Teacher's Note:
1. Use the equilateral triangle condition with one vertex at the origin.
2. Check that the roots are indeed non-real.

Answer: (c) \( 4 \)

Explanation:
1. The foci are \( \pm i \) and the distance between them is 2.
2. The locus is an ellipse only if \( k>2 \) (for \( k=2 \) it is the segment joining the foci, for \( k<2 \) it is empty).
3. Among the options, \( k=4 \) works.

Teacher's Note:
1. The sum of focal distances must exceed the distance between the foci.
2. Equality gives a degenerate segment.

Answer: (c) \( \dfrac{1}{\bar z} \)

Explanation:
1. P lies on ray OQ, so \( p=kz \) with \( k>0 \), and \( |p||z|=1 \).
2. This gives \( k|z|^2=1\Rightarrow k=\dfrac1{z\bar z} \).
3. \( p=\dfrac{z}{z\bar z}=\dfrac1{\bar z} \).

Teacher's Note:
1. Use direction (same argument as z) and modulus (reciprocal of \( |z| \)).
2. This is the inversion of z in the unit circle.

Answer: (a) a hyperbola

Explanation:
1. \( x=1-t,\ y=\sqrt{t^2+t+2} \). With \( t=1-x \): \( y^2=(1-x)^2+(1-x)+2=x^2-3x+4 \).
2. \( y^2=\left(x-\tfrac32\right)^2+\tfrac74 \), i.e. \( y^2-\left(x-\tfrac32\right)^2=\tfrac74 \).
3. This is a hyperbola (upper branch).

Teacher's Note:
1. Eliminate the parameter t.
2. A difference of squares equal to a constant is a hyperbola.

Answer: (d) \( \dfrac{\sqrt3}{4} \)

Explanation:
1. The vertices are \( (0,1),\ \left(-\tfrac12,\tfrac{\sqrt3}{2}\right),\ \left(-\tfrac12,-\tfrac{\sqrt3}{2}\right) \).
2. The side joining \( \alpha,\beta \) is vertical with length \( \sqrt3 \); the distance of \( (0,1) \) from the line \( x=-\tfrac12 \) is \( \tfrac12 \).
3. Area \( =\tfrac12\cdot\sqrt3\cdot\tfrac12=\dfrac{\sqrt3}{4} \).

Teacher's Note:
1. Use base times perpendicular distance.
2. The cube roots of unity give a vertical base.

Answer: (b) \( -1 \)

Explanation:
1. Since \( \alpha^4=1 \), the sum equals \( \alpha^3+\alpha^2+\alpha \).
2. \( 1+\alpha+\alpha^2+\alpha^3=0 \) for a non-real fourth root of unity.
3. So \( \alpha+\alpha^2+\alpha^3=-1 \).

Teacher's Note:
1. Reduce each exponent mod 4.
2. Use the sum of powers of a root of unity.

Answer: (c) purely imaginary

Explanation:
1. Put \( z=\cos\theta+i\sin\theta \). Then \( \dfrac{z+i}{z-i}=\dfrac{\cos\theta+i(\sin\theta+1)}{\cos\theta+i(\sin\theta-1)} \).
2. Multiplying by the conjugate of the denominator, the numerator becomes \( \cos^2\theta+\sin^2\theta-1+2i\cos\theta=2i\cos\theta \) and the denominator is \( 2-2\sin\theta \).
3. The ratio is \( \dfrac{i\cos\theta}{1-\sin\theta} \), purely imaginary.

Teacher's Note:
1. Use the trigonometric form of z on the unit circle.
2. Alternatively \( \dfrac{z+i}{z-i} \) has argument \( \pm\tfrac{\pi}{2} \) since the angle in a semicircle is a right angle.

Answer: (a) \( |z_1|=|z_2| \) and \( 3\,\mathrm{amp}\,z_1+\mathrm{amp}\,z_2=0 \)

Explanation:
1. \( |z_1|=\dfrac{4\cdot2}{\sqrt2}=4\sqrt2 \) and \( |z_2|=\dfrac{4\cdot2}{\sqrt2}=4\sqrt2 \), so the moduli are equal.
2. \( \mathrm{amp}\,z_1=\tfrac{\pi}{3}-\tfrac{\pi}{3}-\tfrac{\pi}{4}=-\tfrac{\pi}{4} \), \( \mathrm{amp}\,z_2=\tfrac{2\pi}{3}-\tfrac{\pi}{6}+\tfrac{\pi}{4}=\tfrac{3\pi}{4} \).
3. The sum is \( \tfrac{\pi}{2}\ne0 \) but \( 3(-\tfrac{\pi}{4})+\tfrac{3\pi}{4}=0 \), so option (a) is correct.

Teacher's Note:
1. Find modulus and amplitude of each factor.
2. Add amplitudes in a product and subtract them in a quotient.