JEE Mathematics Complex Numbers MCQs Set 32

Answer: (c) -1

Explanation:
1. \( z=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}=e^{i\pi/6} \) and \( i=e^{i\pi/2} \).
2. \( iz=e^{i2\pi/3} \), \( z^5=e^{i5\pi/6} \), \( iz^8=e^{i(\pi/2+4\pi/3)}=e^{i11\pi/6} \).
3. Adding the real and imaginary parts: \( 1-\frac12-\frac{\sqrt3}{2}+\frac{\sqrt3}{2}+\ldots=\frac12+\frac{\sqrt3}{2}i=e^{i\pi/3} \).
4. \( \left(e^{i\pi/3}\right)^9=e^{i3\pi}=-1 \). Answer: (c)

Teacher's Note:
1. Convert to polar form first; powers then become multiplication of angles.
2. Use De Moivre's theorem for the final power.

Answer: (b) \( 5\,\mathrm{Re}(\omega)\gt1 \)

Explanation:
1. Solve for z: \( 5\omega-5\omega z=5+3z\Rightarrow z=\frac{5\omega-5}{3+5\omega} \).
2. \( |z|\lt1\Rightarrow |5\omega-5|\lt|5\omega+3|\Rightarrow |\omega-1|\lt\left|\omega+\frac35\right| \).
3. With \( \omega=x+iy \): \( (x-1)^2\lt\left(x+\frac35\right)^2\Rightarrow -2x+1\lt\frac65x+\frac9{25} \), i.e. \( x\gt\frac15 \).
4. So \( 5\,\mathrm{Re}(\omega)\gt1 \). Answer: (b) (Check: \( z=-0.9 \) gives \( 5\mathrm{Re}(\omega)\approx1.2\lt4 \), so (d) is not always true.)

Teacher's Note:
1. Express z in terms of \( \omega \) and apply the condition \( |z|\lt1 \).
2. Squaring a modulus inequality removes the square roots.

Answer: (b) \( \sin x=|\sin y| \)

Explanation:
1. \( \sin^2x-2\sin x+5=(\sin x-1)^2+4\ge4 \), so the exponent \( \sqrt{\cdot}\ge2 \).
2. The inequality is \( 2^{\sqrt{(\sin x-1)^2+4}}\le 2^{2\sin^2y} \), i.e. \( \sqrt{(\sin x-1)^2+4}\le2\sin^2y \).
3. The left side is at least 2 and the right side is at most 2, so both equal 2: \( \sin x=1 \) and \( \sin^2y=1 \).
4. Then \( \sin x=1=|\sin y| \). Answer: (b) The other options would need \( 2=\pm3 \) or \( 1=\pm2 \).

Teacher's Note:
1. Compare the minimum of one side with the maximum of the other.
2. When both extremes coincide, equality must hold in each part.

Answer: (b) the line passing through the origin with slope 1

Explanation:
1. \( |z-i|=|z-1| \) says z is equidistant from \( i=(0,1) \) and \( 1=(1,0) \).
2. With \( z=x+iy \): \( x^2+(y-1)^2=(x-1)^2+y^2\Rightarrow -2y+1=-2x+1\Rightarrow y=x \).
3. This is the line through the origin with slope 1. Answer: (b)

Teacher's Note:
1. Equal distances from two fixed points give a perpendicular bisector.
2. Square both sides and simplify.

Answer: (d) \( \frac{2\pi}{3} \)

Explanation:
1. Multiply by the conjugate: \( \frac{(3+2i\sin\theta)(1+2i\sin\theta)}{1+4\sin^2\theta}=\frac{3-4\sin^2\theta}{1+4\sin^2\theta}+i\frac{8\sin\theta}{1+4\sin^2\theta} \).
2. Purely imaginary needs the real part zero (and imaginary part non-zero): \( 3-4\sin^2\theta=0\Rightarrow\sin\theta=\pm\frac{\sqrt3}{2} \).
3. In \( \left(-\frac\pi2,\pi\right) \): \( \theta=-\frac\pi3,\frac\pi3,\frac{2\pi}{3} \). Sum \( =\frac{2\pi}{3} \). Answer: (d)

Teacher's Note:
1. Rationalise the denominator, then set the real part to zero.
2. List every solution inside the given interval before adding.

Answer: (c) \( \mathrm{Re}(z)=0 \)

Explanation:
1. \( \left|\frac{z_1}{z_2}\right|=\frac43 \), and the ratio is purely imaginary, so \( \frac{z_1}{z_2}=\pm\frac43i \).
2. For \( +\frac43i \): \( \frac{3z_1}{2z_2}=2i \) and \( \frac{2z_2}{3z_1}=\frac{1}{2i}=-\frac i2 \), so \( z=\frac{3i}{2} \).
3. For \( -\frac43i \) the result is \( -\frac{3i}2 \). In both cases \( z \) is purely imaginary, so \( \mathrm{Re}(z)=0 \), \( \mathrm{Im}(z)\ne0 \) and \( |z|=\frac32 \). Answer: (c)

Teacher's Note:
1. Write the ratio of the two numbers as a single quantity \( w \) and express z in terms of \( w \).
2. A purely imaginary \( w \) makes both terms purely imaginary.

Answer: (a) 91

Explanation:
1. \( -2-\frac i3=-\frac13(6+i) \), so the cube is \( -\frac1{27}(6+i)^3 \).
2. \( (6+i)^3=216+108i+18i^2+i^3=198+107i \).
3. So \( x+iy=-198-107i \), giving \( x=-198 \), \( y=-107 \) and \( y-x=91 \). Answer: (a)

Teacher's Note:
1. Factor out the common fraction before cubing.
2. Remember \( i^2=-1 \) and \( i^3=-i \).

Answer: (d) 2

Explanation:
1. A purely imaginary number satisfies \( w+\bar w=0 \): \( \frac{z-\alpha}{z+\alpha}+\frac{\bar z-\alpha}{\bar z+\alpha}=0 \) (as \( \alpha \) is real).
2. Cross-multiplying: \( (z-\alpha)(\bar z+\alpha)+(\bar z-\alpha)(z+\alpha)=0\Rightarrow2|z|^2-2\alpha^2=0 \).
3. So \( \alpha^2=|z|^2=4 \) and \( \alpha=\pm2 \). Answer: (d)

Teacher's Note:
1. Use the condition \( w+\bar w=0 \) for a purely imaginary w.
2. Replace \( z\bar z \) by \( |z|^2 \).

Answer: (c) 1

Explanation:
1. The roots of \( x^2-x+1=0 \) are \( -\omega \) and \( -\omega^2 \), where \( \omega \) is a non-real cube root of unity.
2. \( \alpha^{101}+\beta^{107}=(-\omega)^{101}+(-\omega^2)^{107}=-(\omega^{101}+\omega^{214})=-(\omega^2+\omega) \) (using \( \omega^3=1 \)).
3. Since \( 1+\omega+\omega^2=0 \), this equals \( -(-1)=1 \). Answer: (c)

Teacher's Note:
1. Identify the roots as negatives of cube roots of unity.
2. Reduce powers using \( \omega^3=1 \).

Answer: (d) \( \sin^{-1}\left(\frac1{\sqrt3}\right) \)

Explanation:
1. Rationalise: \( \frac{(2+3i\sin\theta)(1+2i\sin\theta)}{1+4\sin^2\theta}=\frac{2-6\sin^2\theta}{1+4\sin^2\theta}+i\frac{7\sin\theta}{1+4\sin^2\theta} \).
2. The real part is zero: \( 2-6\sin^2\theta=0\Rightarrow\sin^2\theta=\frac13 \), so \( \sin\theta=\pm\frac1{\sqrt3} \).
3. Thus \( \theta=\sin^{-1}\frac1{\sqrt3} \) is a value. Answer: (d)

Teacher's Note:
1. Multiply by the conjugate of the denominator.
2. Set only the real part to zero.

Answer: (b) lies in the interval \( (1,2) \)

Explanation:
1. \( |z|\ge2 \) is the region on or outside the circle of radius 2 centred at the origin.
2. \( \left|z+\frac12\right| \) is the distance of z from \( \left(-\frac12,0\right) \). The nearest points of the region lie on the circle, at \( z=-2 \).
3. Minimum distance \( =2-\frac12=\frac32 \), which lies in \( (1,2) \). Answer: (b) (It is not strictly greater than \( \frac32 \), so (d) fails.)

Teacher's Note:
1. Interpret the modulus as a distance from a fixed point.
2. The nearest point of the region is on the boundary circle along the line joining the centre and the point.

Answer: (a) either on the real axis or on a circle passing through the origin

Explanation:
1. Let \( z=x+iy \). \( \frac{z^2}{z-1}=\frac{[(x^2-y^2)+2ixy][(x-1)-iy]}{(x-1)^2+y^2} \).
2. The imaginary part of the numerator is \( 2xy(x-1)-y(x^2-y^2)=y(x^2+y^2-2x) \).
3. For a real value, \( y(x^2+y^2-2x)=0 \): either \( y=0 \) (real axis) or \( x^2+y^2-2x=0 \), a circle through the origin. Answer: (a)

Teacher's Note:
1. Split z into real and imaginary parts and set the imaginary part to zero.
2. Check whether the circle passes through \( (0,0) \).

Answer: (b) 1

Explanation:
1. \( z \) is equidistant from \( 1=(1,0) \), \( -1=(-1,0) \) and \( i=(0,1) \), i.e. \( z \) is the circumcentre of the triangle with these vertices.
2. A triangle has exactly one circumcentre. Here it is \( z=0 \), since the three points lie on the unit circle.
3. So there is exactly one such complex number. Answer: (b)

Teacher's Note:
1. Equal distances from three non-collinear points define the circumcentre.
2. Check that the three points are not on one line.

Answer: (d) \( -\frac1{i+1} \)

Explanation:
1. The complex number is the conjugate of \( \frac1{i-1} \).
2. \( \overline{\left(\frac1{i-1}\right)}=\frac1{-i-1}=-\frac1{i+1} \). Answer: (d)

Teacher's Note:
1. The conjugate of the conjugate is the number itself.
2. Replace \( i \) by \( -i \) everywhere.

Answer: (c) -i

Explanation:
1. \( \sin\theta+i\cos\theta=i(\cos\theta-i\sin\theta)=ie^{-i\theta} \), so the sum is \( i\sum_{k=1}^{10}e^{-2k\pi i/11} \).
2. The 11th roots of unity add to zero: \( \sum_{k=0}^{10}e^{-2k\pi i/11}=0 \), so \( \sum_{k=1}^{10}=-1 \).
3. The sum is \( i(-1)=-i \). Answer: (c)

Teacher's Note:
1. Factor out \( i \) to get the exponential form.
2. The sum of all n-th roots of unity is zero.

Answer: (b) \( -1,\ 1-2\omega,\ 1-2\omega^2 \)

Explanation:
1. \( (x-1)^3=-8=(-2)^3 \), so \( \left(\frac{x-1}{-2}\right)^3=1 \).
2. Hence \( \frac{x-1}{-2}\in\{1,\omega,\omega^2\} \), i.e. \( x-1=-2,-2\omega,-2\omega^2 \).
3. So \( x=-1,\ 1-2\omega,\ 1-2\omega^2 \). Answer: (b)

Teacher's Note:
1. Rewrite as a cube equal to 1 so that the cube roots of unity appear.
2. Multiply each cube root of unity by the constant.

Answer: (b) a straight line

Explanation:
1. \( |w|=1\Rightarrow|z|=\left|z-\frac i3\right| \).
2. This says \( z \) is equidistant from \( 0 \) and \( \frac i3=\left(0,\frac13\right) \).
3. The locus is the perpendicular bisector of these points, a straight line. Answer: (b)

Teacher's Note:
1. Equal distances from two fixed points give a line.
2. Squaring \( |z|=|z-\frac i3| \) gives a linear equation \( y=\frac16 \).

Answer: (d) -2

Explanation:
1. \( z=(p+iq)^3=(p^3-3pq^2)+i(3p^2q-q^3) \). Comparing with \( z=x-iy \): \( x=p^3-3pq^2 \), \( -y=3p^2q-q^3 \).
2. \( \frac xp=p^2-3q^2 \) and \( \frac yq=q^2-3p^2 \).
3. \( \frac xp+\frac yq=-2p^2-2q^2=-2(p^2+q^2) \), so the required value is \( -2 \). Answer: (d)

Teacher's Note:
1. Cube \( p+iq \) and compare real and imaginary parts.
2. Divide by \( p \) and \( q \) separately before adding.

Answer: (c) \( a^2=3b \)

Explanation:
1. For an equilateral triangle with vertices \( 0,z_1,z_2 \): \( z_1^2+z_2^2=z_1z_2 \), so \( (z_1+z_2)^2=3z_1z_2 \).
2. From the equation, \( z_1+z_2=-a \) and \( z_1z_2=b \).
3. So \( a^2=3b \). Answer: (c)

Teacher's Note:
1. Use the standard condition \( z_1^2+z_2^2=z_1z_2 \) for an equilateral triangle with a vertex at the origin.
2. Substitute the sum and product of roots.

Answer: (d) \( -128\omega^2 \)

Explanation:
1. \( 1+\omega=-\omega^2 \), so \( 1+\omega-\omega^2=-2\omega^2 \).
2. \( (-2\omega^2)^7=-128\,\omega^{14}=-128(\omega^3)^4\omega^2=-128\omega^2 \). Answer: (d)

Teacher's Note:
1. Replace \( 1+\omega \) by \( -\omega^2 \).
2. Reduce the power using \( \omega^3=1 \).

Answer: (d) \( x=0,\ y=0 \)

Explanation:
1. Apply \( R_1\to R_1+R_2 \): the first row becomes \( (6i+4,\ 0,\ 0) \).
2. Expand along the first row: \( (6i+4)\begin{vmatrix}3i&-1\\3&i\end{vmatrix}=(6i+4)(3i^2+3) \).
3. \( 3i^2+3=0 \), so the determinant is 0, giving \( x=0 \), \( y=0 \). Answer: (d)

Teacher's Note:
1. A row operation that creates zeros makes the expansion short.
2. Check whether a minor vanishes using \( i^2=-1 \).