Solutions for Class 11 Maths Chapter 13 by RD Sharma- Free PDF Download
FAQs on Solutions for CBSE Class 11 Maths Chapter 13 - Complex Numbers (By RD Sharma)
1. What is the standard method for dividing two complex numbers as explained in RD Sharma solutions?
To divide one complex number by another, you must eliminate the imaginary part from the denominator. The step-by-step method shown in RD Sharma solutions is as follows:
First, find the conjugate of the complex number in the denominator. The conjugate is found by changing the sign of the imaginary part.
Multiply both the numerator and the denominator of the fraction by this conjugate.
Simplify the expression in the numerator by expanding the terms. The denominator will become a real number because a complex number multiplied by its conjugate, (a + ib)(a - ib), equals a² + b².
Finally, express the result in the standard form a + ib by separating the real and imaginary parts.
2. What are the key properties of complex numbers essential for solving problems in RD Sharma Class 11 Chapter 13?
To solve the exercises in RD Sharma Chapter 13, a thorough understanding of the following properties is essential:
Equality of Complex Numbers: Two complex numbers, p + qi and r + si, are equal if and only if their real parts are equal (p = r) and their imaginary parts are equal (q = s).
Additive Identity: The number 0 (or 0 + 0i) is the additive identity, as (a + ib) + 0 = a + ib.
Multiplicative Identity: The number 1 (or 1 + 0i) is the multiplicative identity, since (a + ib) * 1 = a + ib.
Conjugate Properties: The sum and product of a complex number and its conjugate are always purely real numbers.
Modulus Properties: The modulus of a product of two complex numbers is the product of their individual moduli, i.e., |z₁z₂| = |z₁||z₂|.
3. What are the step-by-step instructions to find the square root of a complex number using the methods from RD Sharma?
Finding the square root of a complex number (a + ib) is a common problem type in this chapter. The method involves assuming the square root is another complex number (x + iy) and then solving for x and y:
Let √(a + ib) = x + iy.
Square both sides to get a + ib = (x + iy)², which simplifies to a + ib = (x² - y²) + i(2xy).
Equate the real and imaginary parts: x² - y² = a and 2xy = b.
Use the identity (x² + y²)² = (x² - y²)² + (2xy)² to find the value of x² + y². This gives x² + y² = √(a² + b²).
Solve the two simultaneous equations for x² and y²: (x² - y² = a) and (x² + y² = √(a² + b²)).
Determine the values of x and y, ensuring their signs are consistent with the sign of b from the equation 2xy = b.
4. How do you find the multiplicative inverse of a complex number as shown in RD Sharma Chapter 13?
The multiplicative inverse of a non-zero complex number z = a + ib is 1/z. To express this in the standard form, you use its conjugate. The correct method is:
Identify the complex number, z = a + ib.
The multiplicative inverse is z⁻¹ = 1 / (a + ib).
Multiply the numerator and denominator by the conjugate of z, which is (a - ib).
This results in (a - ib) / ((a + ib)(a - ib)) = (a - ib) / (a² + b²).
Separate the real and imaginary parts to get the final answer: [a / (a² + b²)] - i[b / (a² + b²)].
5. How can RD Sharma solutions for Complex Numbers help in solving Multiple Choice Questions (MCQs) effectively?
RD Sharma solutions provide detailed, step-by-step derivations for a wide variety of problems. For MCQs, this builds a strong foundation in two ways:
Conceptual Clarity: By working through the detailed solutions, you gain a deep understanding of properties like conjugates, modulus, and arguments. This allows you to quickly identify the correct property to apply in an MCQ without performing full calculations.
Pattern Recognition: Practising the numerous problems in RD Sharma helps you recognise common problem formats and their solutions, enabling you to solve MCQs faster and with greater accuracy during exams.
6. Why is it important to understand that a real number is also a complex number for solving certain problems in this chapter?
This concept is fundamental. A real number, 'x', can be expressed as the complex number x + 0i. Understanding this is crucial for problems where you must equate complex expressions. For instance, if an equation simplifies to z = 5, you implicitly understand it as z = 5 + 0i. This allows you to correctly equate the real part to 5 and the imaginary part to 0, which is a common step required to find unknown variables in RD Sharma exercises.
7. Why is correctly identifying the quadrant of a complex number critical when finding its argument?
Identifying the correct quadrant is critical because the formula for the argument (θ) changes depending on the signs of the real (x) and imaginary (y) parts. While the basic angle α is often calculated as tan⁻¹|y/x|, the principal argument θ is determined as follows:
First Quadrant (x > 0, y > 0): θ = α
Second Quadrant (x < 0, y > 0): θ = π - α
Third Quadrant (x < 0, y < 0): θ = α - π
Fourth Quadrant (x > 0, y < 0): θ = -α
Failing to check the quadrant leads to an incorrect argument, which in turn makes the polar representation (r(cosθ + isinθ)) of the complex number incorrect. This is a common point of error in exams.
8. How are complex numbers applied to solve quadratic equations in RD Sharma problems?
Complex numbers provide a way to find solutions for quadratic equations of the form ax² + bx + c = 0 when the discriminant (D = b² - 4ac) is negative. In the real number system, such an equation has no solution. However, in the complex number system, the solution is found as follows:
Calculate the negative discriminant, D.
Rewrite the square root of the discriminant using the imaginary unit 'i'. For example, √(-D) can be written as i√D.
Substitute this into the standard quadratic formula: x = [-b ± i√|D|] / 2a.
This gives two complex conjugate roots, demonstrating a key application of complex numbers in algebra.
