JEE Mathematics Complex Numbers MCQs Set 19

Answer: (a) Circle of radius \( \sqrt5/2 \)

Explanation:
1. With \( z=x+iy \), \( \frac{x+i(y+1)}{(x+2)+iy} \) has real part \( \frac{x(x+2)+y(y+1)}{(x+2)^2+y^2} \).
2. Setting it to zero: \( x^2+y^2+2x+y=0 \).
3. Completing squares: \( (x+1)^2+\left(y+\tfrac12\right)^2=\tfrac54 \), a circle of radius \( \tfrac{\sqrt5}2 \).

Teacher's Note:
1. Purely imaginary means the real part is zero.
2. The circle is on the segment joining −2 and −i as diameter.

Answer: (c) Interior and boundary of an ellipse

Explanation:
1. The sum of distances from \( \pm1 \) is at most 4, with \( 4>2 \).
2. Equality gives the ellipse itself (foci \( \pm1 \), major axis 4) and the inequality gives the points inside it.
3. So the region is the interior together with the boundary.

Teacher's Note:
1. The inequality uses ≤, so the boundary is included.
2. Option (a) omits the boundary.

Answer: (c) The exterior of the circle

Explanation:
1. Let \( t=|z-1| \). The logarithm is defined only if \( \frac{t+4}{3t-2}>0 \), i.e. \( t>\tfrac23 \).
2. With base \( \tfrac12<1 \) the inequality reverses: \( \frac{t+4}{3t-2}<\tfrac12 \), so \( 2t+8<3t-2 \) and \( t>10 \).
3. So \( |z-1|>10 \): the exterior of the circle of radius 10 centred at 1.

Teacher's Note:
1. The argument of the log must be positive.
2. A base less than 1 reverses the inequality.

Answer: (b) A hyperbola

Explanation:
1. Let the moving circle have centre z and radius r. External contact gives \( |z-z_1|=a+r \) and \( |z-z_2|=b+r \).
2. Subtracting: \( |z-z_1|-|z-z_2|=a-b \), a constant.
3. A constant difference of distances from two fixed points is a branch of a hyperbola.

Teacher's Note:
1. Eliminate the unknown radius r by subtraction.
2. Constant difference (smaller than the distance between foci) means hyperbola.

Answer: (a) \( \dfrac{\sqrt3+i}2 \)

Explanation:
1. \( i=e^{i\pi/2} \), so its principal cube root is \( e^{i\pi/6}=\cos\tfrac\pi6+i\sin\tfrac\pi6=\frac{\sqrt3+i}2 \).

Teacher's Note:
1. Principal root: divide the argument by 3.
2. Check by cubing: (e^(iπ/6))³ = e^(iπ/2) = i.

Answer: (c) \( 2^{53}\left(\tfrac{\sqrt3}2+\tfrac12i\right) \)

Explanation:
1. \( -\sqrt3+i=2e^{5\pi i/6} \).
2. \( (-\sqrt3+i)^{53}=2^{53}e^{265\pi i/6} \) and \( \tfrac{265\pi}6=44\pi+\tfrac\pi6 \).
3. So the value is \( 2^{53}\left(\cos\tfrac\pi6+i\sin\tfrac\pi6\right)=2^{53}\left(\tfrac{\sqrt3}2+\tfrac12i\right) \).

Teacher's Note:
1. Reduce 265π/6 modulo 2π.
2. −√3 + i lies in the second quadrant with argument 5π/6.

Answer: (a) \( -i \)

Explanation:
1. \( z=e^{i\pi/6} \), so \( z^{69}=e^{69i\pi/6}=e^{23i\pi/2} \).
2. \( \tfrac{23\pi}2=10\pi+\tfrac{3\pi}2 \), so \( z^{69}=\cos\tfrac{3\pi}2+i\sin\tfrac{3\pi}2=-i \).

Teacher's Note:
1. z is a 12th root of unity, so z^69 = z^(69 mod 12) = z^9.
2. z^9 = e^(i3π/2) = −i.

Answer: (c) \( \cos n\left(\tfrac\pi2-\theta\right)+i\sin n\left(\tfrac\pi2-\theta\right) \)

Explanation:
1. Write \( \sin\theta+i\cos\theta=\cos\left(\tfrac\pi2-\theta\right)+i\sin\left(\tfrac\pi2-\theta\right) \).
2. By De Moivre's theorem the nth power is \( \cos n\left(\tfrac\pi2-\theta\right)+i\sin n\left(\tfrac\pi2-\theta\right) \).

Teacher's Note:
1. De Moivre's theorem needs the cos + i sin form with the same angle.
2. Never apply it directly to sin + i cos.

Answer: (b) 1

Explanation:
1. The four values are \( e^{i(\pi/4+3k\pi/2)} \) for \( k=0,1,2,3 \), i.e. \( e^{i\pi/4}\{1,-i,-1,i\} \).
2. Product: \( e^{i\pi}\cdot(1)(-i)(-1)(i)=(-1)\cdot(-1)=1 \).

Teacher's Note:
1. There are exactly four fourth-root values.
2. Their product is the constant term (with sign) of the equation w⁴ = z³, here e^(iπ) · (−1).

Answer: (d) \( \cos8\theta+i\sin8\theta \)

Explanation:
1. Write \( \sin\theta+i\cos\theta=i(\cos\theta-i\sin\theta)=ie^{-i\theta} \).
2. The ratio is \( \frac{e^{i\theta}}{ie^{-i\theta}}=-ie^{2i\theta}=e^{i(2\theta-\pi/2)} \).
3. Fourth power: \( e^{i(8\theta-2\pi)}=e^{8i\theta}=\cos8\theta+i\sin8\theta \).

Teacher's Note:
1. Factor i out of sinθ + i cosθ to bring it to polar form.
2. Then apply De Moivre's theorem.

Answer: (d) \( \sin9\theta-i\cos9\theta \)

Explanation:
1. \( \sin\theta+i\cos\theta=ie^{-i\theta} \), so its fifth power is \( i^5e^{-5i\theta}=ie^{-5i\theta} \).
2. The quotient is \( \frac{e^{4i\theta}}{ie^{-5i\theta}}=-ie^{9i\theta} \).
3. \( -ie^{9i\theta}=-i\cos9\theta+\sin9\theta=\sin9\theta-i\cos9\theta \).

Teacher's Note:
1. Convert sinθ + i cosθ to i e^(−iθ).
2. 1/i = −i.

Answer: (d) 4

Explanation:
1. The numerator is \( z=1+\cos\theta+i\sin\theta=2\cos\tfrac\theta2\,e^{i\theta/2} \).
2. The denominator is \( i(1+\cos\theta-i\sin\theta)=i\bar z \).
3. So the ratio is \( \frac{z}{i\bar z}=\frac{e^{i\theta}}{i}=e^{i(\theta-\pi/2)} \).
4. Fourth power: \( e^{i(4\theta-2\pi)}=e^{4i\theta} \), so \( n=4 \).

Teacher's Note:
1. The denominator is i times the conjugate of the numerator.
2. z / z̄ = e^(2i arg z).

Answer: (b) -1

Explanation:
1. With \( \phi=\tfrac\pi{20} \): numerator \( =2\sin\phi(\sin\phi+i\cos\phi)=2\sin\phi\cdot ie^{-i\phi} \).
2. Denominator \( =2\sin\phi(\sin\phi-i\cos\phi)=2\sin\phi\cdot(-i)e^{i\phi} \).
3. The ratio is \( -e^{-2i\phi}=-e^{-i\pi/10} \), and its tenth power is \( (-1)^{10}e^{-i\pi}=-1 \).

Teacher's Note:
1. Use half-angle formulas to factor out 2 sin(π/20).
2. (−1)^10 = 1 and e^(−iπ) = −1.

Answer: (c) \( i(\cos25\theta-i\sin25\theta) \)

Explanation:
1. Using exponentials: \( (e^{2i\theta})^{-5}=e^{-10i\theta} \) and \( (e^{-3i\theta})^6=e^{-18i\theta} \).
2. Also \( \sin\theta-i\cos\theta=-ie^{i\theta} \), so \( (-i)^3e^{3i\theta}=ie^{3i\theta} \).
3. Product: \( ie^{-25i\theta}=i(\cos25\theta-i\sin25\theta) \).

Teacher's Note:
1. Convert sinθ − i cosθ to −i e^(iθ).
2. (−i)³ = i.

Answer: (d) \( i \)

Explanation:
1. \( \sin\phi-i\cos\phi=-i(\cos\phi+i\sin\phi)=-ie^{i\phi} \).
2. The sum becomes \( -i\sum_{k=1}^6e^{2\pi ik/7} \).
3. The sum of the seven 7th roots of unity is 0, so \( \sum_{k=1}^{6}e^{2\pi ik/7}=-1 \). The total is \( -i(-1)=i \).

Teacher's Note:
1. The sum of all nth roots of unity is zero.
2. Omitting the root 1 leaves −1.

Answer: (a) \( \dfrac{1+i\sqrt3}2 \)

Explanation:
1. The total angle is \( \pi\sum_{n\ge1}4^{-n}=\pi\cdot\frac{1/4}{1-1/4}=\tfrac\pi3 \).
2. So the product is \( \cos\tfrac\pi3+i\sin\tfrac\pi3=\frac{1+i\sqrt3}2 \).

Teacher's Note:
1. The infinite geometric series of angles sums to π/3.
2. Product of unit complex numbers = e^(i · sum of angles).

Answer: (b) \( \cos2\theta+i\sin2\theta \)

Explanation:
1. The angles add to \( \theta\left(1+\tfrac12+\tfrac14+\cdots\right)=2\theta \).
2. Hence the product is \( \cos2\theta+i\sin2\theta \).

Teacher's Note:
1. Sum the geometric series 1 + 1/2 + 1/4 + ... = 2.
2. De Moivre's theorem for products.

Answer: (a) -1

Explanation:
1. \( z_1z_2z_3z_4=e^{i\pi(1+2+3+4)/10}=e^{i\pi}=-1 \).

Teacher's Note:
1. Add the angles 1 + 2 + 3 + 4 = 10 in units of π/10.
2. e^(iπ) = −1.

Answer: (c) 1

Explanation:
1. \( \sin\tfrac\pi8+i\cos\tfrac\pi8=ie^{-i\pi/8} \) and \( \sin\tfrac\pi8-i\cos\tfrac\pi8=-ie^{i\pi/8} \).
2. The ratio is \( -e^{-i\pi/4} \).
3. Eighth power: \( (-1)^8e^{-2\pi i}=1 \).

Teacher's Note:
1. Convert both parts to i e^(−iθ) form.
2. (−1)^8 = 1 and e^(−2πi) = 1.

Answer: (c) \( 3\cos(\alpha+\beta+\gamma) \)

Explanation:
1. Let \( a=e^{i\alpha},\ b=e^{i\beta},\ c=e^{i\gamma} \). The hypothesis says \( a+b+c=0 \).
2. Then \( a^3+b^3+c^3=3abc \) (since \( a^3+b^3+c^3-3abc=(a+b+c)(\dots) \)).
3. Taking real parts: \( \cos3\alpha+\cos3\beta+\cos3\gamma=3\cos(\alpha+\beta+\gamma) \).

Teacher's Note:
1. If a + b + c = 0 then a³ + b³ + c³ = 3abc.
2. Real part of abc = cos(α + β + γ).

Answer: (c) 0

Explanation:
1. With \( a,b,c \) as before, \( a+b+c=0 \) and, since \( \bar a=1/a \) etc., \( ab+bc+ca=abc(\bar a+\bar b+\bar c)=0 \).
2. So \( a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0 \).
3. The real part gives \( \cos2\alpha+\cos2\beta+\cos2\gamma=0 \).

Teacher's Note:
1. For unit numbers, the reciprocals are the conjugates.
2. a² + b² + c² = (a + b + c)² − 2(ab + bc + ca).

Answer: (d) \( x^2+x+2=0 \)

Explanation:
1. Since \( a \) is a primitive 7th root of unity, \( a+a^2+\cdots+a^6=-1 \), so \( \alpha+\beta=-1 \).
2. Multiplying out, \( \alpha\beta \) has nine terms: three equal to \( a^7=1 \) and the other six are the six distinct powers \( a,a^2,\dots,a^6 \) once each, so \( \alpha\beta=3+(a+a^2+\cdots+a^6)=3-1=2 \).
3. So the equation is \( x^2-(\alpha+\beta)x+\alpha\beta=x^2+x+2=0 \).

Teacher's Note:
1. Use a⁷ = 1 and 1 + a + ... + a⁶ = 0.
2. A numerical check with a = e^(2πi/7) gives α + β = −1 and αβ = 2.

Answer: (d) For any two positive integers \( n_1,n_2 \)

Explanation:
1. \( i^3=-i,\ i^5=i,\ i^7=-i \), so the expression is \( (1+i)^{n_1}+(1-i)^{n_1}+(1+i)^{n_2}+(1-i)^{n_2} \).
2. Each pair \( (1+i)^n+(1-i)^n \) is a number plus its conjugate, which is real.
3. So the sum is real for all positive integers \( n_1,n_2 \).

Teacher's Note:
1. Reduce the powers of i first.
2. A number plus its conjugate is always real.

Answer: (d) 1

Explanation:
1. \( \frac bc=e^{i(\beta-\gamma)},\ \frac ca=e^{i(\gamma-\alpha)},\ \frac ab=e^{i(\alpha-\beta)} \).
2. The real part of the given sum is the required expression.
3. The sum equals 1, whose real part is 1.

Teacher's Note:
1. The real part of e^(iφ) is cos φ.
2. Take real parts on both sides.