JEE Mathematics Complex Numbers MCQs Set 13

Answer: (b) \( -\sqrt6 \)

Explanation:
1. Write each root with iota: \( \sqrt{-2}=i\sqrt2,\ \sqrt{-3}=i\sqrt3 \).
2. Multiply: \( i^2\sqrt6=-\sqrt6 \).

Teacher's Note:
1. The rule \( \sqrt a\sqrt b=\sqrt{ab} \) is valid only when at least one of a, b is non-negative.
2. Always convert negative radicands to multiples of i before multiplying.

Answer: (d) 32

Explanation:
1. Combine the factors: \( (1+i)^5(1-i)^5=[(1+i)(1-i)]^5 \).
2. \( (1+i)(1-i)=1-i^2=2 \), so the product is \( 2^5=32 \).

Teacher's Note:
1. The product of a complex number and its conjugate is the square of its modulus.
2. Group equal powers before expanding.

Answer: (c) 32

Explanation:
1. \( (1+i)^2=2i\Rightarrow(1+i)^8=(2i)^4=16 \).
2. \( (1-i)^8=(-2i)^4=16 \).
3. Sum: \( 32 \).

Teacher's Note:
1. Even powers of 2i are real.
2. Use (1 ± i)² = ± 2i repeatedly.

Answer: (a) 0

Explanation:
1. \( (1+i)^6=(2i)^3=-8i \).
2. \( (1-i)^6=(-2i)^3=8i \).
3. Sum: \( 0 \).

Teacher's Note:
1. The two terms are conjugates, and odd powers of 2i are purely imaginary.
2. Conjugate pairs with purely imaginary values cancel.

Answer: (a) \( 2-i \)

Explanation:
1. Use \( i^2=-1,\ i^3=-i,\ i^6=-1,\ i^8=1 \).
2. Sum: \( 1-1-i+1+1=2-i \).

Teacher's Note:
1. Reduce each power of i by its remainder when the exponent is divided by 4.
2. Write the four basic values i, -1, -i, 1 before starting.

Answer: (b) -2

Explanation:
1. Numerator: exponents leave remainders 0, 2, 0, 2, 0 mod 4, giving \( 1-1+1-1+1=1 \).
2. Denominator: remainders 2, 0, 2, 0, 2 give \( -1+1-1+1-1=-1 \).
3. The expression equals \( \frac{1}{-1}-1=-2 \).

Teacher's Note:
1. Only the remainder of the exponent mod 4 matters.
2. Five alternating terms of the form i^(even) add to a single term.

Answer: (b) 0

Explanation:
1. Take moduli: \( |1-i|^n=(\sqrt2)^n=2^n \).
2. This forces \( 2^{n/2}=2^n \), so \( n=0 \).

Teacher's Note:
1. Comparing moduli is the fastest test for equations of this kind.
2. Check the answer by substituting n = 0: both sides equal 1.

Answer: (b) 4

Explanation:
1. \( \dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{2}=i \).
2. \( i^m=1 \) first happens at \( m=4 \).

Teacher's Note:
1. The ratio (1 + i)/(1 − i) simplifies to i.
2. i^m = 1 exactly when m is a multiple of 4.

Answer: (d) -1

Explanation:
1. The terms are \( -1,+1,-1,+1,\dots \), a geometric series with ratio \( -1 \).
2. An odd number of terms leaves one unpaired \( -1 \).

Teacher's Note:
1. Pair consecutive terms: each pair adds to zero.
2. An odd count always leaves the first term.

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

Explanation:
1. \( \sum_{n=1}^{13}i^n=i \) (12 terms vanish in blocks of four, leaving \( i^{13}=i \)).
2. \( \sum_{n=1}^{13}i^{n+1}=i^2+\dots+i^{14}=i^{14}=-1 \) (the first 12 terms cancel).
3. Total: \( i-1 \).

Teacher's Note:
1. Split the sum into two separate sums.
2. Full blocks of four consecutive powers always cancel.

Answer: (a) 0

Explanation:
1. \( i^{57}=i \) (57 leaves remainder 1) and \( i^{125}=i \).
2. \( \frac1i=-i \), so the sum is \( i-i=0 \).

Teacher's Note:
1. Use 1/i = −i.
2. Reduce both exponents modulo 4.

Answer: (d) -2

Explanation:
1. \( (1-i)^3=(1-i)(-2i)=-2i-2 \).
2. \( 1-i^3=1+i \).
3. Ratio: \( \frac{-2(1+i)}{1+i}=-2 \).

Teacher's Note:
1. Replace i³ by −i in the denominator.
2. Factor the numerator to see the cancellation.

Answer: (a) \( \beta+i\alpha \)

Explanation:
1. Write \( b+ia=i(a-ib) \).
2. Then \( (b+ia)^5=i^5(a-ib)^5=i\,\overline{(a+ib)^5}=i(\alpha-i\beta) \).
3. This equals \( \beta+i\alpha \).

Teacher's Note:
1. Taking the conjugate of a power conjugates the base.
2. b + ia is i times the conjugate of a + ib.

Answer: (c) \( 2^n \)

Explanation:
1. \( 1-\frac1i=1+i \).
2. So the expression is \( [(1-i)(1+i)]^n=2^n \).

Teacher's Note:
1. 1/i = −i, which turns 1 − 1/i into 1 + i.
2. Combine equal powers into one power of the product.

Answer: (c) 8

Explanation:
1. \( \frac{2i}{1+i}=\frac{2i(1-i)}{2}=1+i \).
2. \( (1+i)^2=2i,\ (1+i)^4=-4,\ (1+i)^8=16 \).
3. The first positive value occurs at \( n=8 \).

Teacher's Note:
1. Track (1 + i)^n through n = 2, 4, 8.
2. (1 + i)^4 is negative, so it does not qualify.

Answer: (b) Second

Explanation:
1. \( \frac{(1+2i)(1+i)}{2}=\frac{1+3i-2}{2}=\frac{-1+3i}{2} \).
2. Real part negative, imaginary part positive: second quadrant.

Teacher's Note:
1. Rationalise the denominator first.
2. Quadrant depends on the signs of real and imaginary parts.

Answer: (c) \( n\pi\pm\dfrac\pi3 \)

Explanation:
1. Multiply by the conjugate: real part of the numerator is \( \mathrm{Re}[(3+2is)(1+2is)]=3-4s^2 \), where \( s=\sin\theta \).
2. Set \( 3-4s^2=0 \) so \( \sin^2\theta=\tfrac34 \).
3. Hence \( \theta=n\pi\pm\frac\pi3 \).

Teacher's Note:
1. Purely imaginary means the real part is zero (and the number is not zero).
2. sin²θ = 3/4 gives θ = ±π/3 modulo π.

Answer: (a) \( \mathrm{Re}(z)=0\Rightarrow\mathrm{Im}(z^2)=0 \)

Explanation:
1. If \( \mathrm{Re}(z)=0 \), then \( z=iy \) and \( z^2=-y^2 \) is real, so \( \mathrm{Im}(z^2)=0 \): option (a) holds.
2. Option (b) fails for \( z=1+i \): \( z^2=2i \) has real part 0 and imaginary part 2.
3. Option (c) fails because \( \mathrm{Re}(z^2)=-y^2\ne0 \).

Teacher's Note:
1. Test each implication with a simple counter-example such as z = 1 + i.
2. A purely imaginary number squares to a real number.

Answer: (b) -1

Explanation:
1. \( z=\frac1z\Rightarrow z^2=1\Rightarrow z=\pm1 \).
2. Among the options only \( -1 \) qualifies.

Teacher's Note:
1. Self-inverse numbers satisfy z² = 1.
2. i has inverse −i, so it is not self-inverse.

Answer: (c) \( -\dfrac i2 \)

Explanation:
1. \( z^2=2i \).
2. \( \frac{1}{2i}=\frac{-i}{2} \).

Teacher's Note:
1. Square first, then invert.
2. 1/i = −i.

Answer: (a) -2

Explanation:
1. Cube: \( z=(p+iq)^3=(p^3-3pq^2)+i(3p^2q-q^3) \).
2. So \( x=p^3-3pq^2 \) and \( -y=3p^2q-q^3 \), giving \( y=q^3-3p^2q \).
3. \( \frac xp=p^2-3q^2,\ \frac yq=q^2-3p^2 \), sum \( -2(p^2+q^2) \).
4. The ratio is \( -2 \).

Teacher's Note:
1. Compare real and imaginary parts after cubing.
2. Note z = x − iy, so y picks up a sign.

Answer: (b) \( 4(a^2-b^2) \)

Explanation:
1. Cube: \( x+iy=(a^3-3ab^2)+i(3a^2b-b^3) \).
2. \( \frac xa=a^2-3b^2,\ \frac yb=3a^2-b^2 \).
3. Sum: \( 4a^2-4b^2=4(a^2-b^2) \).

Teacher's Note:
1. Expand (a + ib)³ using the binomial theorem.
2. Divide each part by a and b respectively.

Answer: (c) \( b+ia \)

Explanation:
1. Multiply numerator and denominator by \( 1+b+ia \): the denominator becomes \( (1+b)^2+a^2=2+2b \) (using \( a^2+b^2=1 \)).
2. The numerator is \( (1+b+ia)^2=(1+b)^2-a^2+2ia(1+b)=2b(1+b)+2ia(1+b) \).
3. Dividing gives \( b+ia \).

Teacher's Note:
1. Use a² + b² = 1 to simplify (1 + b)² − a².
2. The result has unit modulus, as it must for a ratio of conjugates.

Answer: (c) 1

Explanation:
1. Modulus of a product or quotient is the product or quotient of moduli.
2. \( \frac{\sqrt2\cdot\sqrt5}{\sqrt{10}}=1 \).

Teacher's Note:
1. A modulus can never be negative, which rules out options (a) and (d).
2. Use |z1 z2| = |z1||z2| and |z1/z2| = |z1|/|z2|.