RD Sharma Class 11 Solutions Chapter 13 Exercise 13.2 (Updated for 2024)

1. Express the following complex numbers in the standard form a + ib.

(i) (1 + i) (1 + 2i)

(ii) (3 + 2i) / (-2 + i)

(iii) 1/(2 + i)²

(iv) (1 – i) / (1 + i)

(v) (2 + i)³ / (2 + 3i)

(vi) [(1 + i) (1 +√3i)] / (1 – i)

(vii) (2 + 3i) / (4 + 5i)

(viii) (1 – i)³ / (1 – i³)

(ix) (1 + 2i)^-3

(x) (3 – 4i) / [(4 – 2i) (1 + i)]

(xi)

(xii) (5 +√2i) / (1-√2i)

Solution:

(i) (1 + i) (1 + 2i)

Let us simplify and express in the standard form of (a + ib).

(1 + i) (1 + 2i) = (1+i)(1+2i)

= 1(1+2i)+i(1+2i)

= 1+2i+i+2i²

= 1+3i+2(-1) [since, i² = -1]

= 1+3i-2

= -1+3i

∴ The values of a and b are -1 and 3.

(ii) (3 + 2i) / (-2 + i)

Let us simplify and express in the standard form of (a + ib).

(3 + 2i) / (-2 + i) = [(3 + 2i) / (-2 + i)] × (-2-i) / (-2-i) [multiply and divide with (-2-i)]

= [3(-2-i) + 2i (-2-i)] / [(-2)² – (i)²]

= [-6 -3i – 4i -2i²] / (4-i²)

= [-6 -7i -2(-1)] / (4 – (-1)) [since, i² = -1]

= [-4 -7i] / 5

∴ The values of a and b are -4/5 and -7/5

(iii) 1/(2 + i)²

Let us simplify and express in the standard form of (a + ib).

1/(2 + i)² = 1/(2² + i² + 2(2) (i))

= 1/ (4 – 1 + 4i) [since, i² = -1]

= 1/(3 + 4i) [multiply and divide with (3 – 4i)]

= 1/(3 + 4i) × (3 – 4i)/ (3 – 4i)]

= (3-4i)/ (3² – (4i)²)

= (3-4i)/ (9 – 16i²)

= (3-4i)/ (9 – 16(-1)) [since, i² = -1]

= (3-4i)/25

∴ The values of a and b are 3/25 and -4/25

(iv) (1 – i) / (1 + i)

Let us simplify and express in the standard form of (a + ib).

(1 – i) / (1 + i) = (1 – i) / (1 + i) × (1-i)/(1-i) [multiply and divide with (1-i)]

= (1² + i² – 2(1)(i)) / (1² – i²)

= (1 + (-1) -2i) / (1 – (-1))

= -2i/2

= -i

∴ The values of a and b are 0 and -1

(v) (2 + i)³ / (2 + 3i)

Let us simplify and express in the standard form of (a + ib).

(2 + i)³ / (2 + 3i) = (2³ + i³ + 3(2)²(i) + 3(i)²(2)) / (2 + 3i)

= (8 + (i².i) + 3(4)(i) + 6i²) / (2 + 3i)

= (8 + (-1)i + 12i + 6(-1)) / (2 + 3i)

= (2 + 11i) / (2 + 3i)

= (2 + 11i)/(2 + 3i) × (2-3i)/(2-3i)

= [2(2-3i) + 11i(2-3i)] / (2² – (3i)²)

= (4 – 6i + 22i – 33i²) / (4 – 9i²)

= (4 + 16i – 33(-1)) / (4 – 9(-1)) [since, i² = -1]

= (37 + 16i) / 13

∴ The values of a and b are 37/13 and 16/13.

(vi) [(1 + i) (1 +√3i)] / (1 – i)

Let us simplify and express in the standard form of (a + ib).

= (1 + √3i + i + √3i²) / (1 – i)

= (1 + (√3+1)i + √3(-1)) / (1-i) [since, i² = -1]

= [(1-√3) + (1+√3)i] / (1-i)

= [(1-√3) + (1+√3)i] / (1-i) × (1+i)/(1+i)

= [(1-√3) (1+i) + (1+√3)i(1+i)] / (1² – i²)

= [1-√3+ (1-√3)i + (1+√3)i + (1+√3)i²] / (1-(-1)) [since, i² = -1]

= [(1-√3)+(1-√3+1+√3)i+(1+√3)(-1)] / 2

= (-2√3 + 2i) / 2

= -√3 + i

∴ The values of a and b are -√3 and 1.

(vii) (2 + 3i) / (4 + 5i)

Let us simplify and express in the standard form of (a + ib).

(2 + 3i) / (4 + 5i) = [multiply and divide with (4-5i)]

= (2 + 3i) / (4 + 5i) × (4-5i)/(4-5i)

= [2(4-5i) + 3i(4-5i)] / (4² – (5i)²)

= [8 – 10i + 12i – 15i²] / (16 – 25i²)

= [8+2i-15(-1)] / (16 – 25(-1)) [since, i² = -1]

= (23 + 2i) / 41

∴ The values of a and b are 23/41 and 2/41.

(viii) (1 – i)³ / (1 – i³)

Let us simplify and express in the standard form of (a + ib).

(1 – i)³ / (1 – i³) = [1³ – 3(1)²i + 3(1)(i)² – i³] / (1-i².i)

= [1 – 3i + 3(-1)-i².i] / (1 – (-1)i) [since, i² = -1]

= [-2 – 3i – (-1)i] / (1+i)

= [-2-4i] / (1+i)

= [-2-4i] / (1+i) × (1-i)/(1-i)

= [-2(1-i)-4i(1-i)] / (1² – i²)

= [-2+2i-4i+4i²] / (1 – (-1))

= [-2-2i+4(-1)] /2

= (-6-2i)/2

= -3 – i

∴ The values of a and b are -3 and -1.

(ix) (1 + 2i)^-3

Let us simplify and express in the standard form of (a + ib).

(1 + 2i)^-3 = 1/(1 + 2i)³

= 1/(1³+3(1)² (2i)+2(1)(2i)² + (2i)³)

= 1/(1+6i+4i²+8i³)

= 1/(1+6i+4(-1)+8i².i) [since, i² = -1]

= 1/(-3+6i+8(-1)i) [since, i² = -1]

= 1/(-3-2i)

= -1/(3+2i)

= -1/(3+2i) × (3-2i)/(3-2i)

= (-3+2i)/(3² – (2i)²)

= (-3+2i) / (9-4i²)

= (-3+2i) / (9-4(-1))

= (-3+2i) /13

∴ The values of a and b are -3/13 and 2/13.

(x) (3 – 4i) / [(4 – 2i) (1 + i)]

Let us simplify and express in the standard form of (a + ib).

(3 – 4i) / [(4 – 2i) (1 + i)] = (3-4i)/ [4(1+i)-2i(1+i)]

= (3-4i)/ [4+4i-2i-2i²]

= (3-4i)/ [4+2i-2(-1)] [since, i² = -1]

= (3-4i)/ (6+2i)

= (3-4i)/ (6+2i) × (6-2i)/(6-2i)

= [3(6-2i)-4i(6-2i)] / (6² – (2i)²)

= [18 – 6i – 24i + 8i²] / (36 – 4i²)

= [18 – 30i + 8 (-1)] / (36 – 4 (-1)) [since, i² = -1]

= [10-30i] / 40

= (1 – 3i) / 4

∴ The values of a and b are 1/4 and -3/4.

(xi)

(xii) (5 +√2i) / (1-√2i)

Let us simplify and express in the standard form of (a + ib).

(5 +√2i) / (1-√2i) = [Multiply and divide with (1+√2i)]

= (5 +√2i) / (1-√2i) × (1+√2i)/(1+√2i)

= [5(1+√2i) + √2i(1+√2i)] / (1² – (√2)²)

= [5+5√2i + √2i + 2i²] / (1 – 2i²)

= [5 + 6√2i + 2(-1)] / (1-2(-1)) [since, i² = -1]

= [3+6√2i]/3

= 1+ 2√2i

∴ The values of a and b are 1 and 2√2.

2. Find the real values of x and y, if

(i) (x + iy) (2 – 3i) = 4 + i

(ii) (3x – 2i y) (2 + i)² = 10(1 + i)

(iv) (1 + i) (x + iy) = 2 – 5i

Solution:

(i) (x + iy) (2 – 3i) = 4 + i

Given:

(x + iy) (2 – 3i) = 4 + i

Let us simplify the expression, and we get

x(2 – 3i) + iy(2 – 3i) = 4 + i

2x – 3xi + 2yi – 3yi² = 4 + i

2x + (-3x+2y)i – 3y (-1) = 4 + i [since, i² = -1]

2x + (-3x+2y)i + 3y = 4 + i [since, i² = -1]

(2x+3y) + i(-3x+2y) = 4 + i

Equating Real and Imaginary parts on both sides, we get

2x+3y = 4… (i)

And -3x+2y = 1… (ii)

Multiply (i) by 3 and (ii) by 2 and add

On solving, we get

6x – 6x – 9y + 4y = 12 + 2

13y = 14

y = 14/13

Substitute the value of y in (i) , and we get

2x+3y = 4

2x + 3(14/13) = 4

2x = 4 – (42/13)

= (52-42)/13

2x = 10/13

x = 5/13

x = 5/13, y = 14/13

∴ The real values of x and y are 5/13 and 14/13.

(ii) (3x – 2i y) (2 + i)² = 10(1 + i)

Given:

(3x – 2iy) (2+i)² = 10(1+i)

(3x – 2yi) (2²+i²+2(2)(i)) = 10+10i

(3x – 2yi) (4 + (-1)+4i) = 10+10i [since, i² = -1]

(3x – 2yi) (3+4i) = 10+10i

Let us divide with 3+4i into both sides, and we get

(3x – 2yi) = (10+10i)/(3+4i)

= Now multiply and divide with (3-4i)

= [10(3-4i) + 10i(3-4i)] / (3² – (4i)²)

= [30-40i+30i-40i²] / (9 – 16i²)

= [30-10i-40(-1)] / (9-16(-1))

= [70-10i]/25

Now, equating Real and Imaginary parts on both sides, we get

3x = 70/25 and -2y = -10/25

x = 70/75 and y = 1/5

x = 14/15 and y = 1/5

∴ The real values of x and y are 14/15 and 1/5.

(4+2i) x-3i-3 + (9-7i)y = 10i

(4x+9y-3) + i(2x-7y-3) = 10i

Now, equating Real and Imaginary parts on both sides, we get,

4x+9y-3 = 0 … (i)

And 2x-7y-3 = 10

2x-7y = 13 … (ii)

Multiply (i) by 7 and (ii) by 9 and add.

On solving these equations, we get

28x + 18x + 63y – 63y = 117 + 21

46x = 117 + 21

46x = 138

x = 138/46

= 3

Substitute the value of x in (i), and we get

4x+9y-3 = 0

9y = -9

y = -9/9

= -1

x = 3 and y = -1

∴ The real values of x and y are 3 and -1.

(iv) (1 + i) (x + iy) = 2 – 5i

Given:

(1 + i) (x + iy) = 2 – 5i

Divide with (1+i) into both the sides, and we get,

(x + iy) = (2 – 5i)/(1+i)

Multiply and divide by (1-i).

= (2 – 5i)/(1+i) × (1-i)/(1-i)

= [2(1-i) – 5i (1-i)] / (1² – i²)

= [2 – 7i + 5(-1)] / 2 [since, i² = -1]

= (-3-7i)/2

Now, equating Real and Imaginary parts on both sides, we get

x = -3/2 and y = -7/2

∴ The real values of x and y are -3/2 and -7/2.

3. Find the conjugates of the following complex numbers:

(i) 4 – 5i

(ii) 1 / (3 + 5i)

(iii) 1 / (1 + i)

(iv) (3 – i)² / (2 + i)

(v) [(1 + i) (2 + i)] / (3 + i)

(vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)]

Solution:

(i) 4 – 5i

Given:

4 – 5i

We know the conjugate of a complex number (a + ib) is (a – ib).

So,

∴ The conjugate of (4 – 5i) is (4 + 5i)

(ii) 1 / (3 + 5i)

Given:

1 / (3 + 5i)

Since the given complex number is not in the standard form of (a + ib),

Let us convert to standard form by multiplying and dividing with (3 – 5i).

We get,

We know the conjugate of a complex number (a + ib) is (a – ib).

So,

∴ The conjugate of (3 – 5i)/34 is (3 + 5i)/34

(iii) 1 / (1 + i)

Given:

1 / (1 + i)

Since the given complex number is not in the standard form of (a + ib),

Let us convert to standard form by multiplying and dividing with (1 – i).

We get,

We know the conjugate of a complex number (a + ib) is (a – ib).

So,

∴ The conjugate of (1-i)/2 is (1+i)/2

(iv) (3 – i)² / (2 + i)

Given:

(3 – i)² / (2 + i)

Since the given complex number is not in the standard form of (a + ib),

Let us convert to standard form.

We know the conjugate of a complex number (a + ib) is (a – ib).

So,

∴ The conjugate of (2 – 4i) is (2 + 4i)

(v) [(1 + i) (2 + i)] / (3 + i)

Given:

Since the given complex number is not in the standard form of (a + ib),

Let us convert to standard form.

We know the conjugate of a complex number (a + ib) is (a – ib).

So,

∴ The conjugate of (3 + 4i)/5 is (3 – 4i)/5

(vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)]

Given:

Since the given complex number is not in the standard form of (a + ib),

Let us convert to standard form.

We know the conjugate of a complex number (a + ib) is (a – ib).

So,

∴ The conjugate of (63 – 16i)/25 is (63 + 16i)/25

4. Find the multiplicative inverse of the following complex numbers.

(i) 1 – i

(ii) (1 + i √3)²

(iii) 4 – 3i

(iv) √5 + 3i

Solution:

(i) 1 – i

Given:

1 – i

We know the multiplicative inverse of a complex number (Z) is Z^-1 or 1/Z.

So,

∴ The multiplicative inverse of (1 – i) is (1 + i)/2

(ii) (1 + i √3)²

Given:

(1 + i √3)²

Z = (1 + i √3)²

= 1² + (i √3)² + 2 (1) (i√3)

= 1 + 3i² + 2 i√3

= 1 + 3(-1) + 2 i√3 [since, i² = -1]

= 1 – 3 + 2 i√3

= -2 + 2 i√3

We know the multiplicative inverse of a complex number (Z) is Z^-1 or 1/Z.

So,

Z = -2 + 2 i√3

∴ The multiplicative inverse of (1 + i√3)² is (-1-i√3)/8

(iii) 4 – 3i

Given:

4 – 3i

We know the multiplicative inverse of a complex number (Z) is Z^-1 or 1/Z.

So,

Z = 4 – 3i

∴ The multiplicative inverse of (4 – 3i) is (4 + 3i)/25

(iv) √5 + 3i

Given:

√5 + 3i

We know the multiplicative inverse of a complex number (Z) is Z^-1 or 1/Z.

So,

Z = √5 + 3i

∴ The multiplicative inverse of (√5 + 3i) is (√5 – 3i)/14

6. If z₁ = (2 – i), z₂ = (-2 + i), find

Solution:

Given:

z₁ = (2 – i) and z₂ = (-2 + i)

7. Find the modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)]

Solution:

Given:

So,

Z = [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)]

Let us simplify, and we get

= [(1+i) (1+i) – (1-i) (1-i)] / (1² – i²)

= [1² + i² + 2(1)(i) – (1² + i² – 2(1)(i))] / (1 – (-1)) [Since, i² = -1]

= 4i/2

= 2i

We know that for a complex number Z = (a+ib) it’s magnitude is given by |z| = √(a² + b²)

So,

|Z| = √(0² + 2²)

= 2

∴ The modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)] is 2.

8. If x + iy = (a+ib)/(a-ib), prove that x² + y² = 1

Solution:

Given:

x + iy = (a+ib)/(a-ib)

We know that for a complex number Z = (a+ib) it’s magnitude is given by |z| = √(a² + b²)

So,

|a/b| is |a| / |b|

Applying Modulus on both sides, we get,

9. Find the least positive integral value of n for which [(1+i)/(1-i)]ⁿ is real.

Solution:

Given:

Z = [(1+i)/(1-i)]ⁿ

Now, let us multiply and divide by (1+i), and we get

= i [which is not real]

For n = 2, we have

= -1 [which is real]

So, the smallest positive integral ‘n’ that can make [(1+i)/(1-i)]ⁿ real is 2.

∴ The smallest positive integral value of ‘n’ is 2.

10. Find the real values of θ for which the complex number (1 + i cos θ) / (1 – 2i cos θ) is purely real.

Solution:

Given:

(1 + i cos θ) / (1 – 2i cos θ)

Z = (1 + i cos θ) / (1 – 2i cos θ)

Let us multiply and divide by (1 + 2i cos θ)

For a complex number to be purely real, the imaginary part should be equal to zero.

So,

3cos θ = 0 (since, 1 + 4cos²θ ≥ 1)

cos θ = 0

cos θ = cos π/2

θ = [(2n+1)π] / 2, for n ∈ Z

= 2nπ ± π/2, for n ∈ Z

∴ The values of θ to get the complex number to be purely real is 2nπ ± π/2, for n ∈ Z

11. Find the smallest positive integer value of n for which (1+i) ⁿ / (1-i) ^n-2 is a real number.

Solution:

Given:

(1+i) ⁿ / (1-i) ^n-2

Z = (1+i) ⁿ / (1-i) ^n-2

Let us multiply and divide by (1 – i)²

For n = 1,

Z = -2i¹⁺¹

= -2i²

= 2, which is a real number.

∴ The smallest positive integer value of n is 1.

12. If [(1+i)/(1-i)]³ – [(1-i)/(1+i)]³ = x + iy, find (x, y)

Solution:

Given:

Let us rationalize the denominator, we get

i³–(-i)³ = x + iy

2i³ = x + iy

2i².i = x + iy

2(-1)I = x + iy

-2i = x + iy

Equating Real and Imaginary parts on both sides, we get

x = 0 and y = -2

∴ The values of x and y are 0 and -2.

13. If (1+i)² / (2-i) = x + iy, find x + y

Solution:

Given:

(1+i)² / (2-i) = x + iy

Upon expansion, we get

Let us equate real and imaginary parts on both sides, and we get

x = -2/5 and y = 4/5

so,

x + y = -2/5 + 4/5

= (-2+4)/5

= 2/5

∴ The value of (x + y) is 2/5