JEE Mathematics Complex Numbers MCQs Set 16

Answer: (b) \( \tfrac\pi4 \)

Explanation:
1. \( \frac{(13-5i)(4+9i)}{97}=\frac{52+117i-20i+45}{97}=\frac{97+97i}{97}=1+i \).
2. So the argument is \( \tfrac\pi4 \).

Teacher's Note:
1. Multiply by the conjugate of the denominator.
2. |4 − 9i|² = 97 and the numerator simplifies to 97(1 + i).

Answer: (c) \( 2\pi/3 \)

Explanation:
1. \( z=\frac{-2(1-\sqrt3i)}{4}=\frac{-1+\sqrt3i}2 \).
2. This lies in the second quadrant with reference angle \( \tfrac\pi3 \), so \( \arg z=\tfrac{2\pi}3 \).

Teacher's Note:
1. Rationalise before reading off the argument.
2. The negative real part and positive imaginary part place z in quadrant II.

Answer: (d) \( \tfrac\pi2-\tfrac\alpha2 \)

Explanation:
1. Use half-angles: \( 1-\cos\alpha=2\sin^2\tfrac\alpha2 \) and \( \sin\alpha=2\sin\tfrac\alpha2\cos\tfrac\alpha2 \).
2. So \( z=2\sin\tfrac\alpha2\left(\sin\tfrac\alpha2+i\cos\tfrac\alpha2\right)=2\sin\tfrac\alpha2\left[\cos\left(\tfrac\pi2-\tfrac\alpha2\right)+i\sin\left(\tfrac\pi2-\tfrac\alpha2\right)\right] \).
3. The modulus factor is positive, so \( \mathrm{amp}\,z=\tfrac\pi2-\tfrac\alpha2 \).

Teacher's Note:
1. Half-angle identities turn 1 − cosα and sinα into a common factor.
2. For 0 < α < π, 2sin(α/2) is positive, so it does not alter the argument.

Answer: (b) \( |z|=1,\ \arg z=\tfrac\pi6 \)

Explanation:
1. The number is already in polar form \( r(\cos\theta+i\sin\theta) \) with \( r=1 \) and \( \theta=\tfrac\pi6 \).

Teacher's Note:
1. Check that cos² + sin² = 1 to confirm modulus 1.
2. Polar form is read directly: no calculation is required.

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

Explanation:
1. Conjugation reflects z in the real axis, so the angle with the positive real axis changes sign.
2. Hence \( \arg\bar z=-\theta \).

Teacher's Note:
1. For θ = π the principal values coincide, which is the only exception.
2. Compare with z = 1 + i and its conjugate 1 − i.

Answer: (a) \( \pi \)

Explanation:
1. Since \( \arg z<0 \), \( \arg z\in(-\pi,0) \).
2. Multiplying by \( -1 \) adds \( \pi \) and the result \( \arg z+\pi\in(0,\pi) \) is principal, so the difference is \( \pi \).

Teacher's Note:
1. If arg z were positive, subtracting π would be needed instead.
2. Check the sign condition to decide which way to rotate.

Answer: (c) \( -\pi\text{ to }\pi \)

Explanation:
1. By definition the principal argument satisfies \( -\pi<\arg z\le\pi \).

Teacher's Note:
1. The principal range is one full turn starting just after −π.
2. The range −π/2 to π/2 is that of tan⁻¹, not of arg.

Answer: (c) \( \tfrac\pi2 \)

Explanation:
1. Squaring gives \( \mathrm{Re}(z_1\bar z_2)=0 \).
2. So \( z_1/z_2 \) is purely imaginary and the amplitudes differ by \( \tfrac\pi2 \) (up to sign).

Teacher's Note:
1. Equal diagonals mean a rectangle.
2. The two vectors are perpendicular.

Answer: (a) \( \pi \)

Explanation:
1. A negative real number lies on the negative real axis, at angle \( \pi \) from the positive real axis.

Teacher's Note:
1. The principal argument of a negative real number is π, not −π.
2. Positive reals have argument 0.

Answer: (d) \( -\tfrac\pi2 \)

Explanation:
1. A purely imaginary number with negative imaginary part lies on the negative imaginary axis, so \( \arg z=-\tfrac\pi2 \).

Teacher's Note:
1. Draw the point on the Argand plane.
2. Positive imaginary axis is π/2, negative is −π/2.

Answer: (c) \( \alpha+\beta-2\pi \)

Explanation:
1. \( \arg(z_1z_2)\equiv\alpha+\beta\pmod{2\pi} \).
2. Since \( \pi<\alpha+\beta\le2\pi \), subtracting \( 2\pi \) brings the value into \( (-\pi,0] \), the principal range.

Teacher's Note:
1. Principal values lie in (−π, π].
2. Use the nearest multiple of 2π that places the sum in range.

Answer: (a) \( \arg(z-1)=2\arg z \)

Explanation:
1. Write \( z-1=e^{i\varphi} \), so \( z=1+e^{i\varphi}=2\cos\tfrac\varphi2\,e^{i\varphi/2} \).
2. Therefore \( \arg z=\tfrac\varphi2 \) (the factor \( 2\cos\tfrac\varphi2 \) is positive for \( |\varphi|<\pi \)).
3. Hence \( \arg(z-1)=\varphi=2\arg z \).

Teacher's Note:
1. The circle |z − 1| = 1 passes through the origin, and the angle at the centre is twice the angle at the circumference.
2. This is the geometric meaning of the identity.

Answer: (b) 1 : 3

Explanation:
1. Equal amplitudes mean \( \frac{y}{x-1}=\frac{y+3}{x} \) (same slope).
2. Cross-multiply: \( xy=(x-1)(y+3)=xy+3x-y-3 \), so \( y=3(x-1) \).
3. Therefore \( (x-1):y=1:3 \).

Teacher's Note:
1. Equal arguments mean the points z − 1 and z + 3i lie on the same ray from the origin.
2. Compare tan of the arguments.

Answer: (d) \( x^2+y^2-2y=1 \)

Explanation:
1. \( \frac{z-1}{z+1}=\frac{(x^2+y^2-1)+2iy}{(x+1)^2+y^2} \).
2. An amplitude of \( \tfrac\pi4 \) requires real part equal to imaginary part, and both positive: \( x^2+y^2-1=2y \).
3. So \( x^2+y^2-2y=1 \) (with \( y>0 \)).

Teacher's Note:
1. For amplitude π/4, tan = 1, so Re = Im > 0.
2. The locus is an arc of a circle through 1 and −1.

Answer: (c) \( 3\sqrt2 \)

Explanation:
1. The points \( z_1,z_2 \) are 6 units apart, and they subtend the constant angle \( \tfrac\pi4 \) at z, so z lies on a circle through them.
2. Radius: \( R=\frac{6}{2\sin(\pi/4)}=3\sqrt2 \). The centre lies on the perpendicular bisector \( \mathrm{Re}(z)=7 \) at distance \( R\cos\tfrac\pi4=3 \) from the chord, on the side of z.
3. Since the arg is positive, z lies above the chord, so the centre is \( 7+9i \).
4. Therefore \( |z-(7+9i)|=3\sqrt2 \).

Teacher's Note:
1. The chord length and inscribed angle fix the radius through the sine rule.
2. The sign of the argument tells which side of the chord the centre lies on.

Answer: (a) 0

Explanation:
1. Write \( z_2=\bar z_1 \) and \( z_4=\bar z_3 \).
2. The sum of arguments is \( \arg\left(\frac{z_1}{\bar z_3}\cdot\frac{\bar z_1}{z_3}\right)=\arg\frac{|z_1|^2}{|z_3|^2} \).
3. A positive real number has argument 0.

Teacher's Note:
1. The argument of a product is the sum of the arguments.
2. z times conj(z) is a positive real number.

Answer: (d) All of the above

Explanation:
1. \( \mathrm{Re}(z_1\bar z_2)=ac+bd=0 \) and \( a^2+b^2=c^2+d^2=1 \) mean the rows of \( \begin{pmatrix}a&b\\c&d\end{pmatrix} \) are orthonormal.
2. Hence the matrix is orthogonal and its columns are also orthonormal: \( a^2+c^2=1 \), \( b^2+d^2=1 \) and \( ab+cd=0 \).
3. These say \( |w_1|=|w_2|=1 \) and \( \mathrm{Re}(w_1\bar w_2)=ab+cd=0 \).

Teacher's Note:
1. Orthonormal rows of a square matrix imply orthonormal columns.
2. Re(w1 conj w2) = ab + cd.

Answer: (b) \( z_0z+\bar z_0\bar z=12 \)

Explanation:
1. Amplitude 0 means \( \frac{z-2}{2z+3i} \) is a positive real number, so \( z-2=k(2z+3i) \) for real \( k \). This gives the line through \( 2 \) and \( -\tfrac32i \) with equation \( 3x-4y=6 \).
2. Now \( z_0z+\bar z_0\bar z=2\,\mathrm{Re}(z_0z)=2(3x-4y)=12 \).
3. This is option (b).

Teacher's Note:
1. Amplitude zero means a real ratio, so the three points 2, −3i/2 and z are collinear.
2. Re(z0 z) = 3x − 4y for z0 = 3 + 4i.

Answer: (b) \( -\tfrac{7\pi}{18},\ -2\cos\tfrac{11\pi}{18} \)

Explanation:
1. With \( \phi=\tfrac{11\pi}9 \): \( 1+\cos\phi=2\cos^2\tfrac\phi2 \) and \( \sin\phi=2\sin\tfrac\phi2\cos\tfrac\phi2 \), so \( z=2\cos\tfrac\phi2\left(\cos\tfrac\phi2+i\sin\tfrac\phi2\right) \).
2. Here \( \tfrac\phi2=\tfrac{11\pi}{18} \) (110 degrees) has a negative cosine, so the modulus is \( -2\cos\tfrac{11\pi}{18} \).
3. The argument is \( \tfrac{11\pi}{18}-\pi=-\tfrac{7\pi}{18} \).

Teacher's Note:
1. When the factor 2cos(φ/2) is negative, absorb the sign as an extra π in the argument.
2. Principal argument must lie in (−π, π].

Answer: (c) 2

Explanation:
1. With \( z=x+iy \): \( 2x=2\sqrt{(x-1)^2+y^2} \) gives \( y^2=2x-1 \), a parabola.
2. For \( z_1,z_2 \) on it: \( y_1^2-y_2^2=2(x_1-x_2) \), so \( y_1+y_2=\frac{2(x_1-x_2)}{y_1-y_2} \).
3. The slope of \( z_1-z_2 \) is \( \tan\tfrac\pi4=1 \), so \( x_1-x_2=y_1-y_2 \) and \( y_1+y_2=2 \).

Teacher's Note:
1. Subtract the parabola equations for the two points.
2. A direction with argument π/4 has slope 1.

Answer: (a) \( |z|=\tfrac12\sqrt{26},\ \arg z=-\pi+\tan^{-1}\tfrac{19}{17} \)

Explanation:
1. Numerator: \( z_1+\bar z_2=2+3i \), and \( (2+3i)(1+2i)=-4+7i \). Denominator: \( z_2\bar z_1=(1-i)(1-2i)=-1-3i \).
2. So \( z=\frac{-4+7i}{-1-3i}=\frac{(-4+7i)(-1+3i)}{10}=\frac{-17-19i}{10} \).
3. Modulus \( \frac{\sqrt{289+361}}{10}=\frac{\sqrt{26}}2 \); the point lies in the third quadrant, so \( \arg z=-\pi+\tan^{-1}\frac{19}{17} \).

Teacher's Note:
1. Third quadrant: principal argument = tan⁻¹(y/x) − π.
2. Conjugate in the denominator is the conjugate of z1, not of z2.

Answer: (b) \( \pm(1-3i) \)

Explanation:
1. Let \( (x+iy)^2=-8-6i \): \( x^2-y^2=-8 \), \( xy=-3 \), and \( x^2+y^2=10 \).
2. So \( x^2=1,\ y^2=9 \) with \( xy<0 \): \( (1,-3) \) or \( (-1,3) \).
3. Square roots: \( \pm(1-3i) \).

Teacher's Note:
1. Use the modulus relation x² + y² = |w| to shorten the algebra.
2. The sign of xy equals the sign of the imaginary part.

Answer: (b) 25

Explanation:
1. Square both sides and take moduli: \( x^2+y^2=|x-iy|^2=|-7-24i|=\sqrt{49+576}=25 \).

Teacher's Note:
1. For a square root w of N, |w|² = |N|.
2. 7² + 24² = 25².

Answer: (b) \( \sqrt2 \)

Explanation:
1. \( \sqrt i=\pm\frac{1+i}{\sqrt2} \) and \( \sqrt{-i}=\pm\frac{1-i}{\sqrt2} \).
2. Taking the principal roots, the sum is \( \frac{(1+i)+(1-i)}{\sqrt2}=\sqrt2 \).

Teacher's Note:
1. Use cos45° + i sin45° for √i.
2. Only the choice of signs (+,+) gives a value among the options.