RD Sharma Solutions for Class 11 Mathematics Chapter 13 Complex Numbers

RD Sharma Solutions for Class 11 Mathematics Chapter 13 “Complex Numbers” are available here, solved by expert teachers. These solutions follow the latest CBSE Board syllabus and NCERT Book guidelines. Students will find many complex numbers problems with their answers on this page. This will help students to learn about imaginary numbers like 'i'. It will also help you practice complex number operations like adding, subtracting, multiplying, and dividing. You can revise your full syllabus and score more marks easily in your exams.

In the RD Sharma Solutions for Class 11 Maths Chapter 13, find detailed, step-by-step solutions to all textbook exercises on Complex Numbers. Focus is given to essential topics like the imaginary unit 'i', algebra of complex numbers (addition, subtraction, multiplication, division), and properties such as modulus and conjugate. These foundational concepts are crucial for solving a variety of problems in both school level examinations and competitive entrance tests like JEE. The chapter builds upon previously learned number systems and equips students with the tools to solve quadratic equations with no real roots, represent numbers on the Argand plane, and understand their polar form efficiently.

Download RD Sharma Class 11 Chapter 13 Complex Numbers Solutions PDF Free

Get comprehensive RD Sharma Class 11 Maths solutions for Complex Numbers, including step-by-step answers, solved examples, and extra practice questions to help master complex number properties and their applications. Learn about the real part and imaginary part of a complex number, and how to find the square roots of a complex number. Download a free PDF to enhance Class 11 Maths preparation with trusted and exam oriented RD Sharma Solutions.

RD Sharma Class 11 Chapter 13: Complex Numbers - Step-by-Step Solutions

Question 1: Evaluate i⁴⁵ + 1/i⁶⁷

Solution:

Step 1: Find i⁴⁵

Since i⁴ = 1, we divide 45 by 4: 45 = 4 × 11 + 1

Therefore, i⁴⁵ = i^4×11+1 = (i⁴)¹¹ × i¹ = 1¹¹ × i = i

Step 2: Find 1/i⁶⁷

67 = 4 × 16 + 3, so i⁶⁷ = i³ = -i

Therefore, 1/i⁶⁷ = 1/(-i) = -1/i = -i/i² = -i/(-1) = i

Step 3: Final answer

i⁴⁵ + 1/i⁶⁷ = i + i = 2i

Question 2: Simplify (i⁴¹ + 1/i²⁵⁷)

Solution:

Step 1: Find i⁴¹

41 = 4 × 10 + 1, so i⁴¹ = i¹ = i

Step 2: Find 1/i²⁵⁷

257 = 4 × 64 + 1, so i²⁵⁷ = i¹ = i

Therefore, 1/i²⁵⁷ = 1/i = -i

Step 3: Final answer

i⁴¹ + 1/i²⁵⁷ = i + (-i) = 0

Question 3: Find the value of i³⁰ + i⁴⁰ + i⁶⁰

Solution:

Step 1: Find each power

30 = 4 × 7 + 2, so i³⁰ = i² = -1

40 = 4 × 10 + 0, so i⁴⁰ = i⁰ = 1

60 = 4 × 15 + 0, so i⁶⁰ = i⁰ = 1

Step 2: Final answer

i³⁰ + i⁴⁰ + i⁶⁰ = -1 + 1 + 1 = 1

Question 4: Show that 1 + i¹⁰ + i²⁰ + i³⁰ is a real number

Solution:

Step 1: Find each power

10 = 4 × 2 + 2, so i¹⁰ = i² = -1

20 = 4 × 5 + 0, so i²⁰ = i⁰ = 1

30 = 4 × 7 + 2, so i³⁰ = i² = -1

Step 2: Calculate the sum

1 + i¹⁰ + i²⁰ + i³⁰ = 1 + (-1) + 1 + (-1) = 0

Step 3: Conclusion

Since the result is 0, which is a real number, the expression is proven to be real.

Question 5: Express (1 + i)(1 + 2i) in the standard form a + ib

Solution:

Step 1: Multiply using FOIL method

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

= 1 + 2i + i + 2i²

= 1 + 3i + 2(-1)

= 1 + 3i - 2

= -1 + 3i

Question 6: Simplify (3 + 2i)/(-2 + i) into the form a + ib

Solution:

Step 1: Multiply numerator and denominator by conjugate of denominator

Conjugate of (-2 + i) is (-2 - i)

Step 2: Calculate

(3 + 2i)/(-2 + i) × (-2 - i)/(-2 - i)

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

Step 3: Expand numerator

(3 + 2i)(-2 - i) = -6 - 3i - 4i - 2i² = -6 - 7i + 2 = -4 - 7i

Step 4: Expand denominator

(-2 + i)(-2 - i) = 4 - i² = 4 + 1 = 5

Step 5: Final answer

(3 + 2i)/(-2 + i) = (-4 - 7i)/5 = -4/5 - 7i/5

Question 7: Find 1/(2 + i)² in the form a + ib

Solution:

Step 1: Calculate (2 + i)²

(2 + i)² = 4 + 4i + i² = 4 + 4i - 1 = 3 + 4i

Step 2: Find 1/(3 + 4i)

Multiply by conjugate: (3 - 4i)/(3 - 4i)

= (3 - 4i)/[(3 + 4i)(3 - 4i)]

= (3 - 4i)/(9 + 16)

= (3 - 4i)/25

= 3/25 - 4i/25

Question 8: Express (1-i)/(1+i) in the form a + ib

Solution:

Step 1: Multiply by conjugate of denominator

(1-i)/(1+i) × (1-i)/(1-i)

= (1-i)²/[(1+i)(1-i)]

Step 2: Calculate numerator

(1-i)² = 1 - 2i + i² = 1 - 2i - 1 = -2i

Step 3: Calculate denominator

(1+i)(1-i) = 1 - i² = 1 + 1 = 2

Step 4: Final answer

(1-i)/(1+i) = -2i/2 = -i = 0 - i

Question 9: Simplify (2 + i)³/(2 + 3i) to the form a + ib

Solution:

Step 1: Calculate (2 + i)³

First find (2 + i)² = 4 + 4i + i² = 3 + 4i

Then (2 + i)³ = (2 + i)(3 + 4i) = 6 + 8i + 3i + 4i² = 6 + 11i - 4 = 2 + 11i

Step 2: Divide by (2 + 3i)

(2 + 11i)/(2 + 3i) × (2 - 3i)/(2 - 3i)

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

Step 3: Calculate numerator

(2 + 11i)(2 - 3i) = 4 - 6i + 22i - 33i² = 4 + 16i + 33 = 37 + 16i

Step 4: Calculate denominator

(2 + 3i)(2 - 3i) = 4 + 9 = 13

Step 5: Final answer

(2 + i)³/(2 + 3i) = (37 + 16i)/13 = 37/13 + 16i/13

Question 10: Express [(1 + i)(1 + √3i)]/(1 - i) in standard form

Solution:

Step 1: Calculate (1 + i)(1 + √3i)

= 1 + √3i + i + √3i² = 1 + √3i + i - √3 = (1 - √3) + (1 + √3)i

Step 2: Divide by (1 - i)

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

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

Step 3: Calculate numerator

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

= (1 - √3) + (1 - √3)i + (1 + √3)i - (1 + √3)

= -2√3 + 2i

Step 4: Calculate denominator

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

Step 5: Final answer

= (-2√3 + 2i)/2 = -√3 + i

Question 11: Find the conjugate and modulus of 4 - 5i

Solution:

Step 1: Find conjugate

Conjugate of (4 - 5i) = 4 + 5i

Step 2: Find modulus

|4 - 5i| = √(4² + (-5)²) = √(16 + 25) = √41

Question 12: Find the modulus and argument of 3 + 4i

Solution:

Step 1: Find modulus

|3 + 4i| = √(3² + 4²) = √(9 + 16) = √25 = 5

Step 2: Find argument

arg(3 + 4i) = tan^-1(4/3)

Since both real and imaginary parts are positive, the complex number is in the first quadrant.

Therefore, arg(3 + 4i) = tan^-1(4/3)

Question 13: If z₁ = 2 + 3i and z₂ = 1 - 4i, find z₁ + z₂, z₁ - z₂, and z₁ · z₂

Solution:

Step 1: Find z₁ + z₂

z₁ + z₂ = (2 + 3i) + (1 - 4i) = 3 - i

Step 2: Find z₁ - z₂

z₁ - z₂ = (2 + 3i) - (1 - 4i) = 2 + 3i - 1 + 4i = 1 + 7i

Step 3: Find z₁ · z₂

z₁ · z₂ = (2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i² = 2 - 5i + 12 = 14 - 5i

Question 14: Find the square root of -5 + 12i

Solution:

Let √(-5 + 12i) = a + bi, where a and b are real

Then (a + bi)² = -5 + 12i

a² + 2abi - b² = -5 + 12i

(a² - b²) + 2abi = -5 + 12i

Comparing real and imaginary parts:

a² - b² = -5 ... (1)

2ab = 12, so ab = 6 ... (2)

From (2): b = 6/a

Substituting in (1): a² - 36/a² = -5

a⁴ + 5a² - 36 = 0

Let u = a²: u² + 5u - 36 = 0

(u + 9)(u - 4) = 0

u = 4 (taking positive value), so a² = 4, a = ±2

If a = 2, then b = 6/2 = 3

If a = -2, then b = 6/(-2) = -3

Therefore, √(-5 + 12i) = ±(2 + 3i)

Question 15: Find the reciprocal of 2 + 3i

Solution:

Step 1: Find 1/(2 + 3i)

Multiply by conjugate: (2 - 3i)/(2 - 3i)

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

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

= (2 - 3i)/13

= 2/13 - 3i/13

Question 16: Represent 1 + i geometrically in the Argand plane and write its modulus and argument

Solution:

Step 1: Geometric representation

In the Argand plane, 1 + i is represented by the point (1, 1)

The point is in the first quadrant, 1 unit right and 1 unit up from the origin.

Step 2: Find modulus

|1 + i| = √(1² + 1²) = √2

Step 3: Find argument

arg(1 + i) = tan^-1(1/1) = tan^-1(1) = π/4 or 45°

Question 17: Express -1 + √3i in polar form

Solution:

Step 1: Find modulus

r = |-1 + √3i| = √((-1)² + (√3)²) = √(1 + 3) = 2

Step 2: Find argument

Since real part is negative and imaginary part is positive, the point is in the second quadrant.

θ = π - tan^-1(√3/1) = π - π/3 = 2π/3

Step 3: Polar form

-1 + √3i = 2(cos(2π/3) + i sin(2π/3))

Question 18: If (1 + i)(x + iy) = 2 - 5i, find the real values of x and y

Solution:

Step 1: Expand left side

(1 + i)(x + iy) = x + ixy + ix + i²y = x + i(xy + x) - y = (x - y) + i(x + xy)

Step 2: Equate with right side

(x - y) + i(x + xy) = 2 - 5i

Step 3: Compare real and imaginary parts

Real part: x - y = 2 ... (1)

Imaginary part: x + xy = -5, so x(1 + y) = -5 ... (2)

Step 4: Solve the system

From (1): x = 2 + y

Substituting in (2): (2 + y)(1 + y) = -5

2 + 2y + y + y² = -5

y² + 3y + 7 = 0

Using quadratic formula: y = (-3 ± √(9 - 28))/2 = (-3 ± √(-19))/2

Since the discriminant is negative, let's recheck our work.

Actually, let's solve directly: (1 + i)(x + iy) = 2 - 5i

So x + iy = (2 - 5i)/(1 + i)

= (2 - 5i)(1 - i)/[(1 + i)(1 - i)]

= (2 - 2i - 5i + 5i²)/(1 + 1)

= (2 - 7i - 5)/2

= (-3 - 7i)/2

= -3/2 - 7i/2

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

Question 19: If (1 + i)²/(2 - i) = x + iy, find the value of x + y

Solution:

Step 1: Calculate (1 + i)²

(1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i

Step 2: Calculate 2i/(2 - i)

2i/(2 - i) × (2 + i)/(2 + i)

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

= (4i + 2i²)/(4 + 1)

= (4i - 2)/5

= -2/5 + 4i/5

Step 3: Find x + y

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

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

Question 20: Prove that the product of a complex number and its conjugate is always a non-negative real number

Solution:

Step 1: Let z be any complex number

Let z = a + bi, where a and b are real numbers

Step 2: Find the conjugate

z̄ = a - bi

Step 3: Calculate the product z · z̄

z · z̄ = (a + bi)(a - bi)

= a² - abi + abi - b²i²

= a² - b²(-1)

= a² + b²

Step 4: Conclusion

Since a and b are real numbers, a² ≥ 0 and b² ≥ 0

Therefore, a² + b² ≥ 0, which means z · z̄ is always non-negative.

Also, a² + b² is clearly a real number since it contains no imaginary part.

Hence, the product of a complex number and its conjugate is always a non-negative real number.