JEE Mathematics Complex Numbers MCQs Set 17

Answer: (b) \( -\tfrac32i \)

Explanation:
1. \( (3+2i)^2=5+12i \), so with negative real part \( \sqrt{5+12i}=-(3+2i) \) and \( \sqrt{5-12i}=-(3-2i) \).
2. Numerator: \( -(3+2i)-(3-2i)=-6 \). Denominator: \( -(3+2i)+(3-2i)=-4i \).
3. So \( z=\frac{-6}{-4i}=\frac{3}{2i}=-\frac32i \).

Teacher's Note:
1. Spot the perfect square: 5 + 12i = (3 + 2i)².
2. The negative real parts select the sign of each root.

Answer: (b) \( 2\left(\cos\tfrac\pi6+i\sin\tfrac\pi6\right) \)

Explanation:
1. Modulus: \( \sqrt{3+1}=2 \).
2. Argument: \( \tan^{-1}\tfrac1{\sqrt3}=\tfrac\pi6 \), so the polar form is \( 2\left(\cos\tfrac\pi6+i\sin\tfrac\pi6\right) \).

Teacher's Note:
1. Polar form needs both modulus and argument.
2. Option (a) has modulus 1, not 2.

Answer: (c) 0

Explanation:
1. \( 1+i\sqrt3=2e^{i\pi/3} \).
2. \( (1+i\sqrt3)^9=2^9e^{3\pi i}=512(\cos3\pi+i\sin3\pi)=-512 \).
3. So \( a=-512 \) and \( b=0 \).

Teacher's Note:
1. Use De Moivre's theorem on the polar form.
2. sin 3π = 0.

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

Explanation:
1. The two terms are conjugates \( e^{\pm i\pi/6} \) raised to the 5th power.
2. So \( z=2\cos\tfrac{5\pi}6=-\sqrt3 \), which is real: \( \mathrm{Im}(z)=0 \).

Teacher's Note:
1. A number plus its conjugate is real.
2. Re(z) is negative here, so options (a), (c), (d) fail.

Answer: (b) \( e^{-r\sin\theta} \)

Explanation:
1. \( iz=ir\cos\theta-r\sin\theta \).
2. So \( |e^{iz}|=e^{\mathrm{Re}(iz)}=e^{-r\sin\theta} \).

Teacher's Note:
1. |e^w| = e^(Re w).
2. Re(iz) = −Im(z).

Answer: (c) \( (\sqrt2)^{n+2}\cos\tfrac{n\pi}4 \)

Explanation:
1. \( 1\pm i=\sqrt2\,e^{\pm i\pi/4} \).
2. So the sum is \( (\sqrt2)^n\cdot2\cos\tfrac{n\pi}4=(\sqrt2)^{n+2}\cos\tfrac{n\pi}4 \) since \( 2=(\sqrt2)^2 \).

Teacher's Note:
1. Conjugates give twice the real part.
2. Check n = 2: (1+i)² + (1−i)² = 0 and cos(π/2) = 0.

Answer: (b) \( -\sin\theta \)

Explanation:
1. \( e^{-i\theta}=\cos\theta-i\sin\theta \).
2. So \( e^{e^{-i\theta}}=e^{\cos\theta}\,e^{-i\sin\theta} \), whose amplitude is \( -\sin\theta \).

Teacher's Note:
1. Split the exponent into real and imaginary parts.
2. The real factor e^(cosθ) is positive, so it carries no argument.

Answer: (d) 2

Explanation:
1. \( \cos(iy)=\cosh y \), so \( \cos z=\cosh\log(2-\sqrt3)=\frac12\left[(2-\sqrt3)+\frac1{2-\sqrt3}\right] \).
2. \( \frac1{2-\sqrt3}=2+\sqrt3 \), so the value is \( \tfrac12(4)=2 \).

Teacher's Note:
1. cos(iy) = cosh y.
2. Rationalise 1/(2 − √3) to get 2 + √3.

Answer: (b) \( \dfrac{2ab}{a^2-b^2} \)

Explanation:
1. With \( \theta=\tan^{-1}\frac ba \), \( \frac{a-ib}{a+ib}=e^{-2i\theta} \), so \( \log\frac{a-ib}{a+ib}=-2i\theta \).
2. \( i\cdot(-2i\theta)=2\theta \), so the expression is \( \tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{2ab}{a^2-b^2} \).

Teacher's Note:
1. The ratio of a number to its conjugate has argument −2θ.
2. Finish with the double-angle formula for tan.

Answer: (a) -1, 0

Explanation:
1. Using principal values, \( i^i=e^{i\log i}=e^{i\cdot i\pi/2}=e^{-\pi/2} \).
2. Then \( \log i^i=-\tfrac\pi2 \) and \( \sin\left(-\tfrac\pi2\right)=-1 \).
3. So \( a=-1,\ b=0 \).

Teacher's Note:
1. i^i is a real number, a famous fact.
2. Use the principal logarithm of a positive real.

Answer: (b) \( \dfrac1{\log2}\left\{\log5+2\pi ni+\tfrac{i\pi}2\right\} \)

Explanation:
1. \( \log_2(5i)=\frac{\log(5i)}{\log2} \).
2. The general logarithm is \( \log(5i)=\log5+i\left(\tfrac\pi2+2n\pi\right) \).
3. So the general value is \( \frac1{\log2}\left\{\log5+2\pi ni+\tfrac{i\pi}2\right\} \).

Teacher's Note:
1. Change of base: log_2 w = log w / log 2.
2. The general argument of 5i is π/2 + 2nπ.

Answer: (b) z lies on the real axis

Explanation:
1. Note \( 1-iz=-i(z+i) \), so \( |w|=\frac{|z+i|}{|z-i|} \).
2. Setting this to 1 means z is equidistant from \( -i \) and \( i \), i.e. z lies on the real axis.

Teacher's Note:
1. Factor −i out of the numerator.
2. Equidistant from i and −i is the real axis.

Answer: (a) Circle

Explanation:
1. Let \( w=-1+5z \). Then \( |w+1|=5|z|=10 \).
2. This is a circle with centre \( -1 \) and radius 10.

Teacher's Note:
1. A scaling and translation maps a circle to a circle.
2. Express z in terms of w and use |z| = 2.

Answer: (b) \( |a|^2>b \)

Explanation:
1. Complete the square: \( (z+a)(\bar z+\bar a)=|a|^2-b \), i.e. \( |z+a|^2=|a|^2-b \).
2. This is a circle with centre \( -a \) and radius \( \sqrt{|a|^2-b} \) when \( |a|^2>b \).
3. If \( |a|^2=b \) it is a single point and if \( |a|^2<b \) there is no locus.

Teacher's Note:
1. Rewrite the equation as |z + a|² = constant.
2. The constant must be positive for a genuine circle.

Answer: (a) 0

Explanation:
1. The origin is equidistant from the three vertices, so it is the circumcentre.
2. In an equilateral triangle the circumcentre and centroid coincide, so \( \frac{z_1+z_2+z_3}{3}=0 \).

Teacher's Note:
1. In an equilateral triangle all the centres coincide.
2. The centroid is the average of the vertices.

Answer: (b) 2

Explanation:
1. \( z_2 \) lies on a circle of radius 5 centred at \( 3+4i \), which is at distance 5 from the origin.
2. So \( |z_2| \) ranges from 0 to 10.
3. With \( |z_1|=12 \), the least value of \( |z_1-z_2| \) is \( 12-10=2 \).

Teacher's Note:
1. The circle |z2 − 3 − 4i| = 5 passes through the origin.
2. Minimum distance between points of two circles about the origin is the difference of the largest |z2| and |z1|.

Answer: (c) -1

Explanation:
1. \( e^{iA}e^{iB}e^{iC}=e^{i(A+B+C)}=e^{i\pi}=-1 \).

Teacher's Note:
1. The angles of a triangle sum to π.
2. e^(iπ) = −1 (Euler's identity).

Answer: (c) A straight line

Explanation:
1. Let \( z=x+iy \). The real part is \( \frac{2x(x+4)+y(2y-1)}{4x^2+(2y-1)^2} \).
2. Setting it to \( \tfrac12 \): \( 4x^2+16x+4y^2-2y=4x^2+4y^2-4y+1 \).
3. The quadratic terms cancel, leaving \( 16x+2y=1 \), a straight line.

Teacher's Note:
1. Multiply numerator and denominator by the conjugate of the denominator.
2. The quadratic terms cancel, so the locus is a line.

Answer: (b) 3

Explanation:
1. For a circle on diameter \( z_1z_2 \), the point z satisfies \( |z-z_1|^2+|z-z_2|^2=|z_1-z_2|^2 \) (Pythagoras).
2. With \( z_{1,2}=\omega,\omega^2 \): \( |\omega-\omega^2|=|i\sqrt3|=\sqrt3 \).
3. Hence \( \lambda=3 \).

Teacher's Note:
1. The angle in a semicircle is a right angle.
2. The distance between ω and ω² is √3.

Answer: (d) \( \alpha\bar\beta+\bar\alpha\beta=0 \)

Explanation:
1. \( \alpha\bar z+\bar\alpha z=2\,\mathrm{Re}(\bar\alpha z) \), so the first line is \( \mathrm{Re}(\bar\alpha z)=-\tfrac12 \) with normal direction \( \alpha \).
2. Similarly the second line has normal direction \( \beta \).
3. The lines are perpendicular when the normals are: \( \mathrm{Re}(\alpha\bar\beta)=0 \), i.e. \( \alpha\bar\beta+\bar\alpha\beta=0 \).

Teacher's Note:
1. Re(conj(a) z) is the dot product of a and z.
2. Perpendicular normals have zero dot product.

Answer: (d) \( z_1+z_2+z_3 \)

Explanation:
1. With the circumcentre O at the origin the centroid is \( G=\frac{z_1+z_2+z_3}3 \).
2. Since O, G, H are collinear with \( OH=3\,OG \) (Euler line), \( H=z_1+z_2+z_3 \).

Teacher's Note:
1. The orthocentre, centroid and circumcentre lie on the Euler line.
2. H = 3G − 2O.

Answer: (b) \( 90^\circ \)

Explanation:
1. \( i=\cos90^\circ+i\sin90^\circ \), so multiplication adds \( 90^\circ \) to the argument.

Teacher's Note:
1. Multiplying by e^(iα) rotates by α anticlockwise.
2. The modulus is unchanged.

Answer: (b) \( 1+\sqrt2 \)

Explanation:
1. Let \( r=|\omega| \). Then \( 2\ge r-\frac1r \), so \( r^2-2r-1\le0 \).
2. Thus \( r\le1+\sqrt2 \), and the bound is attained when \( \omega \) and \( 1/\omega \) point in opposite directions.

Teacher's Note:
1. Use |a + b| ≥ |a| − |b|.
2. Solve the quadratic inequality for r.

Answer: (b) \( \pi-2\cos^{-1}\tfrac35 \)

Explanation:
1. The disc has centre \( 25i \) and radius 15. The tangents from the origin make angle \( \varphi \) with the imaginary axis where \( \sin\varphi=\tfrac{15}{25}=\tfrac35 \).
2. The difference of the extreme arguments is \( 2\varphi=2\sin^{-1}\tfrac35 \).
3. Since \( \sin^{-1}\tfrac35=\tfrac\pi2-\cos^{-1}\tfrac35 \), this equals \( \pi-2\cos^{-1}\tfrac35 \).

Teacher's Note:
1. The total angle between the two tangents is twice the half-angle φ.
2. Convert with sin⁻¹x + cos⁻¹x = π/2.