JEE Mathematics Complex Numbers MCQs Set 12

Answer: (c) \( \frac1{\bar z} \)

Explanation:
1. P lies on the ray OQ, with \( OP=\frac1{|z|} \).
2. So \( P=\frac1{|z|}\cdot\frac z{|z|}=\frac z{|z|^2}=\frac1{\bar z} \).

Teacher's Note:
1. P is on the same ray as Q (inversion).
2. Use z z-bar =
3. z
4. squared.

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

Explanation:
1. \( \alpha,\beta=-\frac12\pm\frac{\sqrt3}2i \) form a vertical segment of length \( \sqrt3 \) on the line \( x=-\frac12 \).
2. The point \( (0,1) \) is at horizontal distance \( \frac12 \) from this line.
3. Area \( =\frac12\cdot\sqrt3\cdot\frac12=\frac{\sqrt3}4 \).

Teacher's Note:
1. Plot the three points.
2. Use base times height over two.

Answer: (c) 2

Explanation:
1. Multiplying by \( \frac{1+ix}{1+ix} \): the numerator is \( (1+ix)^2(1+2ix)=(1-5x^2)+i(4x-2x^3) \) over \( 1+x^2 \).
2. The imaginary part vanishes when \( x(4-2x^2)=0 \), so \( x=0 \) or \( x=\pm\sqrt2 \).
3. Two non-zero real values.

Teacher's Note:
1. Rationalise the denominator.
2. Set the imaginary part to zero.

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

Explanation:
1. Taking conjugates of \( p(z_k)=0 \) with real coefficients gives \( p(\bar z_k)=0 \).
2. So the conjugates are also roots. The negatives or reciprocals need not be roots.

Teacher's Note:
1. Real coefficients make non-real roots occur in conjugate pairs.
2. Test the other options with a simple example.

Answer: (b) 1

Explanation:
1. \( |\bar z|^2=|z|^2=z\bar z \), so the fraction equals 1.
2. Its modulus is 1.

Teacher's Note:
1. The modulus of the conjugate equals the modulus of z.
2. z times z-bar equals the modulus squared.

Answer: (a) \( \frac{n^2(n+1)^2}{4}-n \)

Explanation:
1. \( (k-\omega)(k-\omega^2)=k^2+k+1 \), so the general term is \( (k-1)(k^2+k+1)=k^3-1 \) for \( k=2,\dots,n \).
2. Sum \( =\sum_{k=2}^nk^3-(n-1)=\left\{\frac{n(n+1)}2\right\}^2-1-(n-1)=\frac{n^2(n+1)^2}4-n \).

Teacher's Note:
1. Simplify each term using omega.
2. Use the formula for the sum of cubes.

Answer: (b) \( -1 \)

Explanation:
1. If n is not a multiple of 3, then \( \alpha^{-n} \) and \( \alpha^{-2n} \) are the two non-real cube roots of unity (in some order).
2. Their sum is \( \omega+\omega^2=-1 \).

Teacher's Note:
1. The powers cycle through the non-real roots.
2. The two non-real cube roots add to -1.

Answer: (b) \( -1 \)

Explanation:
1. \( \alpha^{4n}=1 \), so the sum is \( \alpha^{-1}+\alpha^{-2}+\alpha^{-3}=\alpha^3+\alpha^2+\alpha \).
2. Since \( 1+\alpha+\alpha^2+\alpha^3=0 \), the sum is \( -1 \).

Teacher's Note:
1. Reduce the exponents using alpha to the fourth equals 1.
2. Use the sum of the fourth roots of unity.

Answer: (b) \( \sqrt5 \)

Explanation:
1. \( z=\frac{(1+3i)(1-i)}2=\frac{4+2i}2=2+i \).
2. \( |z|=\sqrt{4+1}=\sqrt5 \).

Teacher's Note:
1. Rationalise the denominator.
2. Find the modulus of 2 + i.

Answer: (c) purely imaginary

Explanation:
1. For \( |z|=1 \), \( \bar z=\frac1z \). Let \( w=\frac{z+i}{z-i} \); then \( \bar w=\frac{\bar z-i}{\bar z+i}=\frac{1-iz}{1+iz}=-\frac{z+i}{z-i}=-w \).
2. \( \bar w=-w \) means w is purely imaginary.

Teacher's Note:
1. Take the conjugate and use the conjugate equals 1/z.
2. A number equal to minus its conjugate is purely imaginary.

Answer: (a) \( |z_1+z_2|\le|z_1|+|z_2| \)

Explanation:
1. (a) is the triangle inequality, always true.
2. (b) fails for \( z_1=1,z_2=-1 \) since \( 0\ge1 \) is false. (c) fails for \( z_1=1,z_2=-1 \) since \( 2\le0 \) is false. (d) fails for \( z_1=z_2 \).

Teacher's Note:
1. The triangle inequality is the standard fact.
2. Disprove the others with simple examples.

Answer: (a) \( iz \)

Explanation:
1. The diagonals intersect at the centre O. B is obtained by rotating A about O through \( 90^\circ \) anticlockwise.
2. Rotation by \( 90^\circ \) multiplies by i: \( B=iz \).

Teacher's Note:
1. A square's vertices are 90 degrees apart about the centre.
2. Anticlockwise rotation multiplies by i.

Answer: (b) a circle

Explanation:
1. Expanding gives \( 2z\bar z+\bar\alpha z+\alpha\bar z=0 \), i.e. \( |z|^2+\text{Re}(\bar\alpha z)=0 \), a circle through the origin.

Teacher's Note:
1. A z z-bar term with real coefficients and linear terms gives a circle.
2. Write it in the form
3. z + a
4. squared = constant.

Answer: (a) \( 1+i(z_0-1) \)

Explanation:
1. The centre is \( z=1 \). PQ subtends a right angle at the centre, so \( z_1-1=\pm i(z_0-1) \).
2. \( z_1=1+i(z_0-1) \) (or \( 1-i(z_0-1) \)).

Teacher's Note:
1. Rotate the radius vector by 90 degrees.
2. Add back the centre.

Answer: (a) \( z_1-z_2+z_3-z_4=0 \)

Explanation:
1. The diagonals of a rhombus (parallelogram) bisect each other: \( z_1+z_3=z_2+z_4 \).
2. So \( z_1-z_2+z_3-z_4=0 \).

Teacher's Note:
1. Diagonals of a parallelogram have the same midpoint.
2. Rearrange the midpoint equality.

Answer: (b) purely imaginary

Explanation:
1. Squaring: \( z_1\bar z_2+\bar z_1z_2=0 \), so \( \text{Re}(z_1\bar z_2)=0 \).
2. Then \( \frac{z_1}{z_2}=\frac{z_1\bar z_2}{|z_2|^2} \) is purely imaginary.

Teacher's Note:
1. Expand both squares of moduli.
2. Cross terms must add to zero.

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

Explanation:
1. \( z+\bar z>2 \) means \( \text{Re}(z)>1 \), the half plane to the right of the centre \( 1 \).
2. The ring \( 1\le|z-1|\le4 \) has area \( \pi(16-1)=15\pi \); the right half is half of it.
3. Area \( =\frac{15\pi}{2} \).

Teacher's Note:
1. Translate the conditions into geometry about the centre 1.
2. The line Re(z)=1 passes through the centre and halves the ring.

Answer: (b) \( z=3+\omega \)

Explanation:
1. With centroid 1, \( A+B+C=3 \) and, for anticlockwise order, \( C-1=(A-1)e^{i4\pi/3} \).
2. \( C-1=\frac{-3-\sqrt3 i}{2} \), so \( A-1=(C-1)e^{-i4\pi/3} \), which gives \( A=\frac{5}{2}-\frac{\sqrt3}{2}i \).
3. And \( 3+\omega=3+\frac{-1-\sqrt3 i}{2}=\frac52-\frac{\sqrt3}{2}i \), so \( A=3+\omega \).

Teacher's Note:
1. Rotate the vector from the centroid by 120 degrees to get the other vertices.
2. Verify the orientation with the sign of the cross product.

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

Explanation:
1. The expression must be defined, so \( z\ne1 \).
2. For it to be real, it equals its conjugate: \( \frac{w-\bar wz}{1-z}=\frac{\bar w-w\bar z}{1-\bar z} \).
3. Cross-multiplying and simplifying gives \( (w-\bar w)(1-z\bar z)=0 \); since w is non-real, \( |z|=1 \). So \( |z|=1,\ z\ne1 \).

Teacher's Note:
1. A complex number is real when it equals its conjugate.
2. The factor w minus w-bar is non-zero, so the other factor must vanish.

Answer: (d) a straight line

Explanation:
1. \( |w|=1\Rightarrow|z|=\left|z-\frac i3\right| \).
2. z is equidistant from the origin and the point \( \frac i3 \), so it lies on their perpendicular bisector, a straight line.

Teacher's Note:
1. Equal moduli means equal distances from two fixed points.
2. The locus of equidistant points is the perpendicular bisector.

Answer: (b) 1

Explanation:
1. \( |a+bw+cw^2|^2=a^2+b^2+c^2-ab-bc-ca=\frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \).
2. The integers are not all equal, so at least one of the three differences is a non-zero integer, and then at least two differences are non-zero. The smallest possible value of the bracket is \( 1+1+0 \)-type arrangements, e.g. \( (a,b,c)=(1,0,0) \) gives \( \frac12(1+0+1)=1 \).
3. So the least value of \( |a+bw+cw^2|^2 \) is 1 and the minimum modulus is 1.

Teacher's Note:
1. Use the identity for a squared plus b squared plus c squared minus ab minus bc minus ca.
2. Try small integers such as 1, 0, 0.

Answer: (b) \( \frac{-\sqrt3+\sqrt{-1}}{\sqrt{-4}} \)

Explanation:
1. Option (b): \( \frac{-\sqrt3+i}{2i}=\frac12+\frac{\sqrt3}2i=e^{i\pi/3} \).
2. Its cube is \( e^{i\pi}=-1 \), so it is a cube root of \( -1 \).
3. For the others: (a) \( \frac{\sqrt3+i}{2}=e^{i\pi/6} \) has cube \( i \); (c) \( \frac{\sqrt3-i}{2i}=e^{-2i\pi/3} \) has cube \( 1 \); (d) \( -i \) has cube \( i \). None equals \( -1 \).

Teacher's Note:
1. Convert to polar form and cube.
2. A cube root of minus one has argument 60, 180 or 300 degrees.

Answer: (c) \( \frac{3\pi}4 \)

Explanation:
1. \( \bar z=-i\bar\omega \Rightarrow z=i\omega \Rightarrow \omega=-iz \).
2. \( z\omega=-iz^2 \), so \( \arg(z\omega)=-\frac\pi2+2\arg z=\pi \).
3. Hence \( \arg z=\frac{3\pi}4 \).

Teacher's Note:
1. Take the conjugate of the given relation to express omega in terms of z.
2. Add the arguments when multiplying complex numbers.

Answer: (a) 1

Explanation:
1. Write \( z_1=e^{i\theta} \); then \( \arg z_2=-\theta \) and \( z_2=e^{-i\theta} \) (same modulus 1).
2. \( z_1z_2=e^{i\theta}e^{-i\theta}=1 \).

Teacher's Note:
1. With unit modulus, a complex number is determined by its argument.
2. Arguments of a product add up to zero.