Free PDF Download Available on Vedantu for RD Sharma Solutions Class 11
FAQs on RD Sharma Class 11 Solutions Chapter 13 - Complex Numbers (Ex 13.2) Exercise 13.2 - Free PDF
1. How should I use the Vedantu solutions for RD Sharma Class 11 Maths Exercise 13.2?
It is best to first attempt the problems from Exercise 13.2 on your own. Afterwards, you can use the Vedantu solutions to verify your final answer and, more importantly, to check the step-by-step methodology. If you are stuck on a problem, these solutions provide a clear, logical path to the correct answer, helping you understand the underlying concepts of complex number operations as per the 2025-26 CBSE syllabus.
2. What is the correct method for solving problems on the equality of two complex numbers in Exercise 13.2?
To solve problems involving the equality of two complex numbers, such as a + ib = c + id, the solutions follow a standard procedure:
- First, ensure both complex numbers are in the standard form (real part + i × imaginary part).
- Next, equate the real parts on both sides of the equation (a = c).
- Then, equate the imaginary parts on both sides (b = d).
- Finally, solve the two resulting equations to find the values of any unknown variables. This principle is fundamental to many problems in Exercise 13.2.
3. What key properties of complex numbers are demonstrated in the solutions for Exercise 13.2?
The solutions for RD Sharma Exercise 13.2 frequently apply the fundamental algebraic properties of complex numbers, which are crucial for your exams. These include:
- Commutative and Associative Laws for both addition and multiplication.
- The Distributive Law, which is essential for correctly multiplying two complex numbers.
- The concepts of Additive Identity (0 + i0) and Multiplicative Identity (1 + i0).
- The use of the Additive Inverse (-z) and the Multiplicative Inverse (z⁻¹).
4. Why is it incorrect to just multiply the real parts and imaginary parts separately when finding the product of two complex numbers?
This is a common misconception. A complex number like (a + ib) must be treated as a single algebraic binomial, not as two separate components. The correct method uses the distributive property, similar to expanding (x+y)(u+v).
For (a + ib) × (c + id), the expansion is ac + i(ad) + i(bc) + i²(bd). The crucial step is remembering that i² = -1. This changes the last term, and the final product correctly groups into a new real part (ac - bd) and a new imaginary part i(ad + bc).
5. How do the solutions calculate the multiplicative inverse of a complex number z = a + ib?
The solutions find the multiplicative inverse (z⁻¹) using a standard formula derived from its properties. The step-by-step process is:
- First, find the conjugate of the complex number, which is z̄ = a - ib.
- Second, calculate the modulus squared of the complex number, which is |z|² = a² + b².
- Finally, apply the formula: z⁻¹ = z̄ / |z|². This simplifies to (a - ib) / (a² + b²), a method that ensures the denominator is always a real number.
6. What is the main role of the complex conjugate when solving problems in Exercise 13.2?
The conjugate of a complex number is a powerful tool with two primary applications demonstrated in the solutions:
- Division of Complex Numbers: To divide one complex number by another, we multiply both the numerator and the denominator by the conjugate of the denominator. This process, known as rationalisation, is essential because it converts the denominator into a real number, making the expression simpler to solve.
- Finding the Multiplicative Inverse: The formula for the multiplicative inverse (z⁻¹) directly uses the conjugate in the numerator, highlighting its fundamental role in finding the reciprocal of a complex number.
7. What is the most efficient first step to simplify expressions with large powers of iota (i), like i⁹⁹⁸?
The solutions demonstrate a systematic method based on the cyclic nature of powers of 'i' (where i⁴ = 1). The most efficient first step is to divide the power by 4 and identify the remainder.
- For i⁹⁹⁸, divide 998 by 4. The result is 249 with a remainder of 2.
- Rewrite the expression as i⁹⁹⁸ = i⁽⁴ × ²⁴⁹ ⁺ ²⁾ = (i⁴)²⁴⁹ × i².
- Since i⁴ = 1 and i² = -1, the expression simplifies to (1)²⁴⁹ × (-1), which equals -1. This method is faster and less error-prone than repeated multiplication.
