NCERT Solutions Class 11 Maths Chapter 5

The students must not underestimate the importance of revision. They must plan their study schedule in advance so that they can cover all the concepts essential for the examination. For students studying Mathematics, it is advised to revise all the important questions, theorems, derivation, and formulae and practice several problems related to each concept regularly. The NCERT Solutions Class 11 Mathematics Chapter 5 is an ultimate set of study material which covers all the necessary information and formulas.

The NCERT Solutions Class 11 Mathematics Chapter 5 talks about various concepts related to Complex numbers and Quadratic equations. To gain expertise in this chapter, students must refer to the NCERT Solutions Class 8, NCERT Solutions Class 9 and NCERT Solutions Class 10 to recall the concepts related to linear equations and quadratic equations.

The NCERT Solutions Class 11 Mathematics Chapter 5 help students gain conceptual understanding and attain high scores. With the help of these CBSE revision notes, students can comprehend all important concepts and theorems from this chapter. Students must consider using several academic notes provided by Extramarks to enhance their preparation.

Key Topics Covered In NCERT Solutions Class 11 Mathematics Chapter 5

In Extramarks NCERT Solutions for Class 11 Mathematics Chapter 5, students can be assured that all the concepts are defined and explained in detail.

The subject matter experts at Extramarks have covered the following key topics in the ch 5 Mathematics Class 11 notes:

Exercise	Topic
5.1	Introduction
5.2	Complex numbers
5.3	Algebra of Complex Numbers
5.4	Modulus and the Conjugate
5.5	Argand Plane and Polar Representation
5.6	Quadratic Equations

5.1 Introduction

This chapter helps students to understand the types of numbers: Imaginary and Complex. The chapter includes the basic introduction to the algebra of complex numbers, the power of i, modulus, conjugate and polar representation of complex numbers. Using the NCERT Solutions Class 11 Mathematics Chapter 5, students can easily ace their exams. Students will learn about the argand plane and solution of quadratic equations so that they can prepare for ch 5 Mathematics Class 11 thoroughly.

5.2 Complex numbers

In this section, the definition of complex numbers and their representation in the general form are given. Also, the concepts of the Real part (Re z) and the Imaginary part (Img z) are explained here. Solved examples have also been given for better understanding of the concept.

Refer to the NCERT Solutions Class 11 Mathematics Chapter 5 to understand each concept thoroughly.

5.3 Algebra of Complex Numbers

This section is the most important in the entire chapter. Students will learn to develop algebraic operations for complex numbers. Firstly, the addition of two numbers is explained. The properties under addition are given below: Closure, the commutative law, the associative law, the existence of additive identity and the existence of an additive inverse.

This section in the NCERT Solutions Class 11 Mathematics Chapter 5 further explains the concept of the difference between two complex numbers and the properties it holds. Furthermore, students will gain information about the multiplication of given complex numbers. The properties of multiplication are explained in detail using proofs. The properties of multiplication are the closure law, the commutative law, the associative law, the existence of a multiplicative identity, the existence of multiplicative inverse and the distributive law. This section in the NCERT Solutions Class 11 Mathematics Chapter 5 further explains the division of two complex numbers.

Students will also learn about the Power and Values of i. In this section, the square roots of negative real numbers are only mentioned.

Several solved examples are used to help students understand the special identities of complex numbers along. These identities are proved in a stepwise manner.

5.4 Modulus and the Conjugate

The modulus and conjugate of the complex number z = a + ib are explained in this topic. The modulus is denoted by |z|, whereas the conjugate is z. Also, the multiplicative inverse of the complex number z is mentioned in this section.

The properties of modulus and conjugate are given below:

|z1.z2|= |z1| .|z2|
z1z2=z1z2, where |z2| 0
z1.z2 = z1 . z2
z1z2 = z1 z2
(z1z2) = z1z2, where z2 0

Unlimited problems are included in the NCERT Solutions Class 11 Mathematics Chapter 5 for students to practice.

5.5 Argand Plane and Polar Representation

In this section, the Argand plane is defined. This section geometrically represents the complex

number x + iy corresponding to the ordered pair (x, y). The x and y-axes in the Argand plane are known as the real axis and the imaginary axis, respectively. The mirror image of a point is also explained in this section. In the Class 11 Mathematics chapter 5 notes, students will learn how to represent a complex number in the polar form. Concepts such as the principal arguments and amplitude of a complex number are explained in the NCERT Solutions. Students have to answer several questions based on this topic.

5.6 Quadratic Equations

In this section, students will learn about quadratic or polynomial equations with degree n having n roots. Several solved examples are included in the NCERT Solutions Class 11 Mathematics Chapter 5. The Fundamental Theorem of Algebra is also mentioned in this topic.

List of NCERT Solutions Class 11 Mathematics Chapter 5 Exercise & Solutions

Students can access the NCERT Solutions Class 11 Mathematics Chapter 5 Complex numbers and Quadratic equations on the Extramarks mobile application and web portal. The notes include a detailed explanation of all the concepts, several practice problems and solved examples for students to understand each topic clearly. Chapter 5 Class 11 Mathematics notes enable the students to quickly revise and glance through all important formulas and derivations of the chapter. Using these NCERT Solutions Class 11 Mathematics Chapter 5, students will be able to tackle all questions and solve them on their own.

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NCERT Solutions Class 1
NCERT Solutions Class 2
NCERT Solutions Class 3
NCERT Solutions Class 4
NCERT Solutions Class 5
NCERT Solutions Class 6
NCERT Solutions Class 7

NCERT Exemplar Class 11 Mathematics

The NCERT Exemplar Mathematics notes are one of the most important reference materials. Extramarks have provided the NCERT Exemplar for students to gain an in-depth understanding of any particular topic. It includes several tricky and challenging questions which enhance the thinking and problem-solving ability of students. Along with the NCERT Exemplar, all students can also refer to the NCERT Solutions Class 11 Mathematics Chapter 5 to prepare effectively for both board exams and entrance examinations.

Extramarks aims to provide the best learning experience to every student. Therefore, we provide guidance and support through our academic notes and all-inclusive NCERT Solutions for Class 11 Mathematics chapter 5. Students are also advised to solve several question papers and sample papers for extra practice.

Key Features of NCERT Solutions Class 11 Mathematics Chapter 5

In the NCERT Solutions Class 11 Mathematics Chapter 5, every problem has been solved using reasoning and logic.
The information in the notes strictly adheres to the CBSE syllabus, marking system, and weightage allotted for each concept.
Studying with the help of the NCERT Solutions Class 11 Mathematics Chapter 5 aids students to boost confidence so that they can establish a strong command of the subject.
Students will also learn tips, tricks and shortcut techniques essential for solving Mathematics problems without making any silly mistakes.
Students can effectively prepare for their annual examinations using the NCERT Solutions Class 11 Mathematics Chapter 5.
These notes prepare them to tackle any question in the exam easily.

Q.1 Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

$\begin{array}{l} 1 . „ (5 i) (- \frac{3}{5} i) \\ 2.„ i^{9} + i^{19} \\ 3.„ i^{- 39} \\ 4.„ 3 (7 + i 7) + i (7 + i 7) \\ 5.„ (1 - i) - (- 1 + i 6) \\ 6.„ (\frac{1}{5} + i \frac{2}{5}) - (4 + i \frac{5}{2}) \\ 7.„ [(\frac{1}{3} + i \frac{7}{3}) + (4 + i \frac{1}{3})] - (- \frac{4}{3} + i) \\ 8.„ {(1 - i)}^{4} \\ 9 . „ {(\frac{1}{3} + 3 i)}^{3} \\ 10 . „ {(- 2 - \frac{1}{3} i)}^{3} \end{array}$

Ans

$\begin{array}{l} (5 i) (- \frac{3}{5} i) = 5 \times - \frac{3}{5} i^{2} \\ = - 3 (- 1) [âˆµ i^{2} = - 1] \\ = 3 + 0 i, which is in the form of a + ib. \end{array}$

$\begin{array}{l} i^{9} + i^{19} = i^{4 \times 2 + 1} + i^{4 \times 4 + 3} \\ = {(i^{4})}^{2} i + {(i^{4})}^{4} i^{3} \\ = {(1)}^{2} \times i + {(1)}^{4} (- i) [âˆµ i^{4} = 1 {and i}^{3} = - i] \\ = i - i \\ = 0 + 0 i, which is in the form of a + ib. \end{array}$

$i^{- 39} = \frac{1}{i^{39}}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \frac{1}{i^{4 \times 9 + 3}} \\ = \frac{1}{{(i^{4})}^{9} . i^{3}} \\ = \frac{1}{{(1)}^{9} . (- i)} \\ = - \frac{1}{i} \times \frac{i}{i} \\ = - \frac{i}{i^{2}} \\ = - \frac{i}{- 1} \\ = 0 + i, which is in the form of a + ib. \end{array}$

4.
3(7 + i 7)+ i(7 + i 7) = 21 + 21i + 7i + 7i²
= 21 + 28i + 7(– 1)
= 21 + 28i – 7
= 14 + 28i, which is in the form of a + ib.

5.
(1 – i) – (–1 + i 6) = 1 – i + 1 – i 6
= 2 – 7i, which is in the form of a + ib.

$\begin{array}{l} (\frac{1}{5} + i \frac{2}{5}) - (4 + i \frac{5}{2}) = (\frac{1}{5} - 4) + i (\frac{2}{5} - \frac{5}{2}) \\ = - \frac{19}{5} + i (\frac{4 - 25}{10}) \\ = - \frac{19}{5} - i \frac{21}{10}, \\ which is in the form of a + ib. \end{array}$

$\begin{array}{l} [(\frac{1}{3} + i \frac{7}{3}) + (4 + i \frac{1}{3})] - (- \frac{4}{3} + i) = (\frac{1}{3} + 4 + \frac{4}{3}) + i (\frac{7}{3} + \frac{1}{3} - 1) \\ = (\frac{1 + 12 + 4}{3}) + i (\frac{7 + 1 - 3}{3}) \\ = (\frac{17}{3}) + i (\frac{5}{3}), \\ Which is in the form of a + ib . \end{array}$

$\begin{array}{l} {(1 - i)}^{4} = {{(1 - i)}^{2}}^{2} \\ = {(1 - 2 i + i^{2})}^{2} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = {(1 - 2 i - 1)}^{2} \\ = {(- 2 i)}^{2} \\ = 4 i^{2} \\ = 4 (- 1) \\ = - 4 + i 0, \\ which is in the form of a + ib . \end{array}$

$\begin{array}{l} {(\frac{1}{3} + 3 i)}^{3} = {(\frac{1}{3})}^{3} + 3 {(\frac{1}{3})}^{2} (3 i) + 3 (\frac{1}{3}) {(3 i)}^{2} + {(3 i)}^{3} \\ [âˆµ {(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 {ab}^{2} + b^{3}] \\ = \frac{1}{27} + i + i^{2} 9 + i^{3} 27 \\ = \frac{1}{27} + i - 9 - i 27 \\ = (\frac{1}{27} - 9) + i (1 - 27) \\ = (\frac{1 - 243}{27}) + i (- 26) \\ = (\frac{- 242}{27}) - i 26 \\ which is in the form of a + ib . \end{array}$

10.

$\begin{array}{l} {(- 2 - \frac{1}{3} i)}^{3} = {(- 2)}^{3} + 3 {(- 2)}^{2} (- \frac{1}{3} i) + 3 (- 2) {(- \frac{1}{3} i)}^{2} + {(- \frac{1}{3} i)}^{3} \\ [âˆµ {(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 {ab}^{2} + b^{3}] \\ = - 8 - 4 i - \frac{2}{3} i^{2} - \frac{1}{27} i^{3} \\ = - 8 - 4 i - \frac{2}{3} (- 1) - \frac{1}{27} (- i) [âˆµ i^{2} = - 1 {and i}^{3} = - i] \\ = - 8 + \frac{2}{3} - 4 i + \frac{1}{27} i \\ = - \frac{22}{3} - (\frac{108 - 1}{27}) i \\ = - \frac{22}{3} - (\frac{107}{27}) i, \\ which is in the form of a + ib . \end{array}$

Q.2 Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. 4 – 3i
12.

$\sqrt{5} + 3 i$

13. – i

Ans

11.

$\begin{array}{l} Multiplicative inverse of (4 - 3i) = \frac{1}{(4 - 3i)} \times \frac{(4 + 3i)}{(4 + 3i)} \\ = \frac{(4 + 3i)}{(4^{2} - {9i}^{2})} \end{array}$ $\begin{array}{l} [âˆµ (a - b) (a + b) = a^{2} - b^{2}] \\ = \frac{(4 + 3i)}{(16 + 9)} \\ = \frac{4}{25} + \frac{3}{25} i \\ Thus, the multiplicative inverse of (4 - 3 i) is (\frac{4}{25} + \frac{3}{25} i) . \end{array}$

12.

$\begin{array}{l} Multiplicative inverse of (\sqrt{5} + 3 i) = \frac{1}{(\sqrt{5} + 3 i)} \times \frac{(\sqrt{5} - 3 i)}{(\sqrt{5} - 3 i)} \\ = \frac{(\sqrt{5} - 3 i)}{{{(\sqrt{5})}^{2} - {(3 i)}^{2}}} \\ = \frac{(\sqrt{5} - 3 i)}{{5 - 9 i^{2}}} \\ = \frac{(\sqrt{5} - 3 i)}{(5 + 9)} \\ = \frac{\sqrt{5}}{14} - i \frac{3}{14} \\ Thus, the multiplicative inverse of (\sqrt{5} + 3 i) is (\frac{\sqrt{5}}{14} - i \frac{3}{14}) . \end{array}$

13.

$\begin{array}{l} Multiplicative inverse of (- i) = \frac{1}{(- i)} \times \frac{i}{i} \\ = \frac{i}{- i^{2}} \\ = \frac{i}{- (- 1)} \\ = \frac{i}{1} \\ = i, \\ Thus, the multiplicative inverse of - i is i. \end{array}$

Q.3

$\begin{array}{l} Express the following expression in the form of a+ib: \\ \frac{(3+i \sqrt{5}) (3-i \sqrt{5})}{(\sqrt{3} + \sqrt{2} i) – (\sqrt{3} – \sqrt{2} i)} \end{array}$

Ans

$\begin{array}{l} \frac{(3 + i \sqrt{5}) (3 - i \sqrt{5})}{(\sqrt{3} + \sqrt{2} i) - (\sqrt{3} - \sqrt{2} i)} = \frac{3^{2} - {(i \sqrt{5})}^{2}}{\sqrt{3} + \sqrt{2} i - \sqrt{3} + \sqrt{2} i} [\begin{array}{l} âˆµ (a - b) (a + b) \\ = a^{2} - b^{2} \end{array}] \\ = \frac{(9 - i^{2} 5)}{2 \sqrt{2} i} \\ = \frac{{9 - (- 1) 5}}{2 \sqrt{2} i} \end{array}$ $\begin{array}{l} = \frac{14}{2 \sqrt{2} i} \\ = \frac{7}{\sqrt{2} i} \times \frac{\sqrt{2} i}{\sqrt{2} i} \\ = \frac{7 \sqrt{2} i}{i^{2} (2)} \\ = \frac{7 \sqrt{2} i}{(- 1) 2} [âˆµ i^{2} = - 1] \\ = 0 - \frac{7 \sqrt{2} i}{2}, which in the form of a + ib. \end{array}$

Q.4 Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

$\begin{array}{l} 1 . „ z = - 1 - i \sqrt{3} \\ 2 . „z = - \sqrt{3} + i \end{array}$

Ans

$\begin{array}{l} z = - 1 - i \sqrt{3} \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - 1 and rsin θ = - \sqrt{3} \\ Squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- 1)}^{2} + {(- \sqrt{3})}^{2} \\ r^{2} = 1 + 3 \\ \Rightarrow r = 2 \\ So, ‹ modulus = 2 \end{array}$

$\begin{array}{l} ∴ cosθ = \frac{- 1}{2} and sin θ = - \frac{\sqrt{3}}{2} \\ \Rightarrow cosθ = - \cos \frac{π}{3} and sin θ = - \sin \frac{π}{3} \\ Since, sin θ and cos θ are negative in III quadrant. \\ So, θ = - (π - \frac{π}{3}) = - \frac{2 π}{3} \\ Thus, modulus and argument of (- 1 - i \sqrt{3}) is 2 and - \frac{2 π}{3} \\ respectively. \end{array}$

$\begin{array}{l} z = - \sqrt{3} + i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - \sqrt{3} and rsin θ = 1 \\ Squarring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- \sqrt{3})}^{2} + {(1)}^{2} \\ r^{2} = 3 + 1 = 4 \\ \Rightarrow r = 2 \\ So, ‹ modulus = 2 \\ ∴ cosθ = \frac{- \sqrt{3}}{2} and sin θ = \frac{1}{2} \\ \Rightarrow cosθ = - \cos \frac{π}{6} and sin θ = \sin \frac{π}{6} \\ Since, sin θ is positive and cos θ is negative in II quadrant. \end{array}$

$\begin{array}{l} So, θ = (π - \frac{π}{6}) = \frac{5 π}{6} \\ Thus, modulus and argument of (- \sqrt{3} + i) is 2 and \frac{5 π}{6} \\ respectively. \end{array}$

Q.5 Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i
4. – 1 + i
5. – 1 – i
6. – 3
7.

$\sqrt{3} + i$

8. i

Ans

$\begin{array}{l} Let z = 1 - i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = 1 and rsin θ = - 1 \\ Squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(1)}^{2} + {(- 1)}^{2} \\ r^{2} = 1 + 1 \\ \Rightarrow r = \sqrt{2} \\ ∴ cosθ = \frac{1}{\sqrt{2}} and sin θ = \frac{- 1}{\sqrt{2}} \\ \Rightarrow cosθ = \cos \frac{π}{4} and sin θ = - \sin \frac{π}{4} \\ Since, cos θ is positive and sin θ is negative in IV quadrant. \\ So, θ = - \frac{π}{4} [âˆµ - π < θ \leq π] \\ Thus, polar form of z = \sqrt{2} {\cos (- \frac{π}{4}) + isin (- \frac{π}{4})} . \end{array}$

$\begin{array}{l} Let z = - 1 + i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - 1 and rsin θ = 1 \\ Squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- 1)}^{2} + {(1)}^{2} \end{array}$ $\begin{array}{l} r^{2} = 1 + 1 \\ \Rightarrow r = \sqrt{2} \\ ∴ cosθ = \frac{- 1}{\sqrt{2}} and sin θ = \frac{1}{\sqrt{2}} \\ \Rightarrow cosθ = - \cos \frac{π}{4} and sin θ = \sin \frac{π}{4} \\ Since, sin θ is positive and cos θ is negative in II quadrant. \\ So, θ = π - \frac{π}{4} = \frac{3 π}{4} \\ Thus, polar form of z = \sqrt{2} {\cos (\frac{3 π}{4}) + isin (\frac{3 π}{4})} . \end{array}$

$\begin{array}{l} Let z = - 1 - i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - 1 and rsin θ = - 1 \\ Squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- 1)}^{2} + {(- 1)}^{2} \\ r^{2} = 1 + 1 \\ \Rightarrow r = \sqrt{2} \\ ∴ cosθ = \frac{- 1}{\sqrt{2}} and sin θ = \frac{- 1}{\sqrt{2}} \\ \Rightarrow cosθ = - \cos \frac{π}{4} and sin θ = - \sin \frac{π}{4} \\ Since, sin θ and cos θ is negative in III quadrant. \end{array}$ $\begin{array}{l} So, θ = - (π - \frac{π}{4}) = - \frac{3 π}{4} [âˆµ - π < θ \leq π] \\ Thus, polar form of z = \sqrt{2} {\cos (- \frac{3 π}{4}) + isin (- \frac{3 π}{4})} . \end{array}$

$\begin{array}{l} Let z = - 3 + 0 i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - 3 and rsin θ = 0 \\ Squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- 3)}^{2} + {(0)}^{2} \\ r^{2} = 9 \\ \Rightarrow r = 3 \\ ∴ cosθ = \frac{- 3}{3} = - 1 and sin θ = 0 \\ \Rightarrow cosθ = - \cos 0 and sin θ = \sin 0 \\ Since, sin θ is positive and cos θ is negative in II quadrant. \\ So, θ = (π - 0) = π \\ Thus, polar form of z = 3 {cosπ + isinπ} . \end{array}$

$\begin{array}{l} Let z = \sqrt{3} + i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = \sqrt{3} and rsin θ = 1 \\ Squaring and adding, we get \end{array}$ $\begin{array}{l} r^{2} (\cos^{2} θ + \sin^{2} θ) = {(\sqrt{3})}^{2} + {(1)}^{2} \\ r^{2} = 4 \\ \Rightarrow r = 2 \\ ∴ cosθ = \frac{\sqrt{3}}{2} and sin θ = \frac{1}{2} \\ \Rightarrow cosθ = \cos \frac{π}{6} and sin θ = \sin \frac{π}{6} \\ So, θ = \frac{π}{6} \\ Thus, polar form of z = 2 {\cos \frac{π}{6} + isin \frac{π}{6}} . \end{array}$

$\begin{array}{l} Let z = 0 + i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = 0 and rsin θ = 1 \\ Squaring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(0)}^{2} + {(1)}^{2} \\ r^{2} = 1 \\ \Rightarrow r = 1 \\ ∴ cosθ = 0 and sin θ = 1 \\ \Rightarrow cosθ = \cos \frac{π}{2} and sin θ = \sin \frac{π}{2} \\ So, θ = \frac{π}{2} \end{array}$ $Thus, polar form of z = 1 {\cos \frac{π}{2} + isin \frac{π}{2}} .$

Q.6 Solve the following equation:
x² + 3 = 0

Ans

$\begin{array}{l} x^{2} + 3 = 0 \\ c o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = 1, b = 0 and c = 3 \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- 0 \pm \sqrt{0^{2} - 4 (1) (3)}}{2 (1)} \\ = \pm \frac{\sqrt{- 12}}{2} \\ = \pm \frac{2 i \sqrt{3}}{2} [âˆµ \sqrt{- 1} = i] \\ x = \pm i \sqrt{3} \end{array}$

Q.7 Solve the following equation:
2x² + x + 1 = 0

Ans

$\begin{array}{l} {2x}^{2} + x + 1 = 0 \\ c o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = 2, b = 1 and c = 1 \end{array}$ $\begin{array}{l} B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- 1 \pm \sqrt{1^{2} - 4 (2) (1)}}{2 (2)} \\ = \frac{- 1 \pm \sqrt{1 - 8}}{4} \\ = \frac{- 1 \pm \sqrt{- 7}}{4} \\ x = \frac{- 1 \pm i \sqrt{7}}{4} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.8 Solve the following equation:
x² + 3x + 9 = 0

Ans

$\begin{array}{l} x^{2} + 3x + 9 = 0 \\ c o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = 1, b = 3 and c = 9 \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- 3 \pm \sqrt{3^{2} - 4 (1) (9)}}{2 (1)} \\ = \frac{- 3 \pm \sqrt{9 - 36}}{2} = \frac{- 3 \pm \sqrt{- 27}}{4} \end{array}$ $x = \frac{- 3 \pm 3 \sqrt{3} i}{2} [âˆµ \sqrt{- 1} = i]$

Q.9 Solve the following equation:
– x² + x – 2 = 0

Ans

$\begin{array}{l} - x^{2} + x - 2 = 0 \\ c o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = - 1, b = 1 and c = - 2 \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- 1 \pm \sqrt{1^{2} - 4 (- 1) (- 2)}}{2 (- 1)} \\ = \frac{- 1 \pm \sqrt{1 - 8}}{- 2} = \frac{- 1 \pm \sqrt{- 7}}{- 2} \\ x = \frac{- 1 \pm i \sqrt{7}}{- 2} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.10 Solve the following equation:
x² + 3x + 5 = 0

Ans

$\begin{array}{l} x^{2} + 3x + 5 = 0 \\ c o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = 1, b = 3 and c = 5 \\ B y using quadratic formula, \end{array}$ $\begin{array}{l} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- 3 \pm \sqrt{3^{2} - 4 (1) (5)}}{2 (1)} \\ = \frac{- 3 \pm \sqrt{9 - 20}}{2} = \frac{- 1 \pm \sqrt{- 11}}{2} \\ x = \frac{- 3 \pm i \sqrt{11}}{2} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.11 Solve the following equation:
x² – x + 2 = 0

Ans

$\begin{array}{l} x^{2} - x + 2 = 0 \\ C o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = 1, b = - 1 and c = 2 \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (2)}}{2 (1)} \\ = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm \sqrt{- 7}}{2} \\ x = \frac{1 \pm \sqrt{7} i}{2} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.12 Solve the following equation:

$\sqrt{2} x^{2} + x + \sqrt{2} = 0$

Ans

$\begin{array}{l} \sqrt{2} x^{2} + x + \sqrt{2} = 0 \\ C o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = \sqrt{2}, b = 1 and c = \sqrt{2} \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- 1 \pm \sqrt{1^{2} - 4 (\sqrt{2}) (\sqrt{2})}}{2 (\sqrt{2})} \\ = \frac{- 1 \pm \sqrt{1 - 8}}{2 \sqrt{2}} = \frac{- 1 \pm \sqrt{- 7}}{2 \sqrt{2}} \\ x = \frac{- 1 \pm i \sqrt{7}}{2 \sqrt{2}} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.13 Solve the following equation:

$\sqrt{3} x^{2} - \sqrt{2} x + 3 \sqrt{3} = 0$

Ans

$\begin{array}{l} \sqrt{3} x^{2} - \sqrt{2} x + 3 \sqrt{3} = 0 \\ C o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = \sqrt{3}, b = - \sqrt{2} and c = 3 \sqrt{3} \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{array}$ $\begin{array}{l} = \frac{- (- \sqrt{2}) \pm \sqrt{{(- \sqrt{2})}^{2} - 4 (\sqrt{3}) (3 \sqrt{3})}}{2 (\sqrt{3})} \\ = \frac{\sqrt{2} \pm \sqrt{2 - 36}}{2 \sqrt{3}} = \frac{\sqrt{2} \pm \sqrt{- 34}}{2 \sqrt{3}} \\ x = \frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.14 Solve the following equation:

$x^{2} + x + \frac{1}{\sqrt{2}} = 0$

Ans

$\begin{array}{l} x^{2} + x + \frac{1}{\sqrt{2}} = 0 \\ C o m p a r i n g {with ax}^{2} + b x + c = 0, we get \\ a = 1, b = 1 and c = \frac{1}{\sqrt{2}} \\ B y using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- (1) \pm \sqrt{{(1)}^{2} - 4 (1) (\frac{1}{\sqrt{2}})}}{2 (1)} \\ = \frac{- 1 \pm \sqrt{1 - 2 \sqrt{2}}}{2} = \frac{- 1 \pm \sqrt{1 - 2 \sqrt{2}}}{2} \end{array}$ $x = \frac{- 1 \pm i \sqrt{2 \sqrt{2} - 1}}{2} [âˆµ \sqrt{- 1} = i]$

Q.15 Solve the following equation:

$x^{2} + \frac{x}{\sqrt{2}} + 1 = 0$

Ans

$\begin{array}{l} x^{2} + \frac{x}{\sqrt{2}} + 1 = 0 \\ Comparing {with ax}^{2} + bx + c = 0, we get \\ a = 1, b = \frac{1}{\sqrt{2}} and c = 1 \\ By using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- (\frac{1}{\sqrt{2}}) \pm \sqrt{{(\frac{1}{\sqrt{2}})}^{2} - 4 (1) (1)}}{2 (1)} \\ = \frac{- \frac{1}{\sqrt{2}} \pm \sqrt{\frac{1}{2} - 4}}{2} \\ = \frac{- \frac{1}{\sqrt{2}} \pm \sqrt{- \frac{7}{2}}}{2} \\ x = \frac{- 1 \pm i \sqrt{7}}{2 \sqrt{2}} [âˆµ \sqrt{- 1} = i] \end{array}$

Q.16 Find the square root of the following: – 15 – 8i

Ans

$\begin{array}{l} Let ‹ x + iy = \sqrt{- 15 - 8i} \\ Squarring both sides, we get \\ {(x + iy)}^{2} = - 15 - 8i \\ x^{2} + 2 ixy + {(iy)}^{2} = - 15 - 8i \\ x^{2} + 2 ixy - y^{2} = - 15 - 8i \\ x^{2} - y^{2} + 2 ixy = - 15 - 8i \\ Comparing real and imaginary parts from both sides, we get \\ x^{2} - y^{2} = - 15 … (i) \\ and xy = - 4 \\ {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 {(xy)}^{2} \\ = {(- 15)}^{2} + 4 {(- 4)}^{2} \\ = 225 + 64 \\ = 289 \\ \Rightarrow x^{2} + y^{2} = 17 … (ii) \\ Adding equation (i) and equation (ii) \\ 2 x^{2} = 2 \Rightarrow x = \pm 1 \\ and putting x = 1 in xy = - 4, we get \end{array}$ $\begin{array}{l} \Rightarrow (1) y = - 4 \\ \Rightarrow y = - 4 \\ putting x = - 1 in xy = - 4, we get \\ \Rightarrow (- 1) y = - 4 \\ \Rightarrow y = 4 \\ ∴ x + iy = 1 - 4 i or - 1 + 4 i \\ \Rightarrow \sqrt{- 15 - 8i} = 1 - 4 i or - 1 + 4 i \end{array}$

Q.17 Find the square root of the following: – 8 – 6i.

Ans

$\begin{array}{l} Let ‹ x + iy = \sqrt{- 8 - 6i} \\ Squaring both sides, we get \\ {(x + iy)}^{2} = - 8 - 6i \\ x^{2} + 2 ixy + {(iy)}^{2} = - 8 - 6i \\ x^{2} + 2 ixy - y^{2} = - 8 - 6i \\ x^{2} - y^{2} + 2 ixy = - 8 - 6i \\ Comparing real and imaginary parts from both sides, we get \\ x^{2} - y^{2} = - 8 . .. (i) \\ and xy = - 3 \\ {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 {(xy)}^{2} \\ = {(- 8)}^{2} + 4 {(- 3)}^{2} \\ = 64 + 36 \\ = 100 \\ \Rightarrow x^{2} + y^{2} = 10 . .. (ii) \\ Adding equation (i) and equation (ii) \end{array}$ $\begin{array}{l} 2 x^{2} = 2 \Rightarrow x = \pm 1 \\ and putting x = 1 in xy = - 6, we get \\ \Rightarrow (1) y = - 3 \\ \Rightarrow y = - 3 \\ putting x = - 1 in xy = - 3, we get \\ \Rightarrow (- 1) y = - 3 \\ \Rightarrow y = 3 \\ ∴ x + iy = 1 - 3 i or - 1 + 3 i \\ \Rightarrow \sqrt{- 8 - 6i} = 1 - 3 i or - 1 + 3 i \end{array}$

Q.18 Find the square root of the following: 1 – i

Ans

$\begin{array}{l} Let ‹ x + iy = \sqrt{1 - i} \\ Squarring both sides, we get \\ {(x + iy)}^{2} = 1 - i \\ x^{2} + 2 ixy + {(iy)}^{2} = 1 - i \\ x^{2} + 2 ixy - y^{2} = 1 - i \\ x^{2} - y^{2} + 2 ixy = 1 - i \\ Comparing real and imaginary parts from both sides, we get \\ x^{2} - y^{2} = 1 … (i) \\ and xy = - \frac{1}{2} \\ {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 {(xy)}^{2} \\ = {(1)}^{2} + 4 {(- \frac{1}{2})}^{2} \end{array}$ $\begin{array}{l} = 1 + 1 \\ = 2 \\ \Rightarrow x^{2} + y^{2} = \sqrt{2} … (ii) \\ Adding equation (i) and equation (ii) \\ 2 x^{2} = 1 + \sqrt{2} \Rightarrow x = \pm \sqrt{\frac{1 + \sqrt{2}}{2}} \\ and putting x = \sqrt{\frac{1 + \sqrt{2}}{2}} in xy = - \frac{1}{2}, we get \\ \Rightarrow (\sqrt{\frac{1 + \sqrt{2}}{2}}) y = - \frac{1}{2} \\ \Rightarrow y = - \frac{1}{2} \times \sqrt{\frac{2}{\sqrt{2} + 1}} \\ \Rightarrow y = - \sqrt{\frac{\sqrt{2} + 1}{2 (\sqrt{2} + 1)} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}} \\ \Rightarrow y = - \sqrt{\frac{\sqrt{2} + 1}{2}} \\ Putting x = - \sqrt{\frac{1 + \sqrt{2}}{2}} in xy = - \frac{1}{2}, we get \\ \Rightarrow (- \sqrt{\frac{1 + \sqrt{2}}{2}}) y = - \frac{1}{2} \\ \Rightarrow y = \frac{1}{2} \times \sqrt{\frac{2}{\sqrt{2} + 1}} \end{array}$ $\begin{array}{l} \Rightarrow y = \sqrt{\frac{\sqrt{2} + 1}{2 (\sqrt{2} + 1)} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}} \\ \Rightarrow y = \sqrt{\frac{\sqrt{2} + 1}{2}} \\ ∴ x + iy = \sqrt{\frac{1 + \sqrt{2}}{2}} - \sqrt{\frac{\sqrt{2} + 1}{2}} i or - \sqrt{\frac{1 + \sqrt{2}}{2}} + \sqrt{\frac{\sqrt{2} + 1}{2}} i \\ \Rightarrow \sqrt{- 8 - 6i} = \sqrt{\frac{1 + \sqrt{2}}{2}} - \sqrt{\frac{\sqrt{2} + 1}{2}} i or - \sqrt{\frac{1 + \sqrt{2}}{2}} + \sqrt{\frac{\sqrt{2} + 1}{2}} \\ \Rightarrow \sqrt{- 8 - 6i} = \pm \sqrt{\frac{1 + \sqrt{2}}{2}} âˆ“ \sqrt{\frac{\sqrt{2} + 1}{2}} i \end{array}$

Q.19 Find the square root of the following: –i

Ans

$\begin{array}{l} Let x + iy = \sqrt{- i} \\ Squaring both sides, we get \\ {(x + iy)}^{2} = 0 - i \\ x^{2} + 2 ixy + {(iy)}^{2} = 0 - i \\ x^{2} + 2 ixy - y^{2} = 0 - i \\ x^{2} - y^{2} + 2 ixy = 0 - i \\ Comparing real and imaginary parts from both sides, we get \\ x^{2} - y^{2} = 0 . . . (i) \\ and xy = - \frac{1}{2} \\ {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 {(xy)}^{2} \end{array}$ $\begin{array}{l} \Rightarrow ‹‹‹‹‹‹‹‹‹‹„„„„„„„‹y = \sqrt{\frac{\sqrt{2} + 1}{2 (\sqrt{2} + 1)} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}} \\ \Rightarrow ‹‹‹‹‹‹‹‹‹‹‹„„„„„„„y = \sqrt{\frac{\sqrt{2} + 1}{2}} \\ ∴ „„„x + iy = \sqrt{\frac{1 + \sqrt{2}}{2}} - \sqrt{\frac{\sqrt{2} + 1}{2}} i„ or - \sqrt{\frac{1 + \sqrt{2}}{2}} + \sqrt{\frac{\sqrt{2} + 1}{2}} i \\ \Rightarrow \sqrt{1 - i} = \sqrt{\frac{1 + \sqrt{2}}{2}} - \sqrt{\frac{\sqrt{2} + 1}{2}} i„ or - \sqrt{\frac{1 + \sqrt{2}}{2}} + \sqrt{\frac{\sqrt{2} + 1}{2}} i \\ \Rightarrow \sqrt{1 - i} = \pm \sqrt{\frac{1 + \sqrt{2}}{2}} âˆ“ \sqrt{\frac{\sqrt{2} + 1}{2}} i \end{array}$

Q.20 Find the square root of the following: i

Ans

$\begin{array}{l} Let ‹ x + iy = \sqrt{i} \\ Squaring both sides, we get \\ {(x + iy)}^{2} = 0 + i \\ x^{2} + 2 ixy + {(iy)}^{2} = 0 + i \\ x^{2} + 2 ixy - y^{2} = 0 + i \\ x^{2} - y^{2} + 2 ixy = 0 + i \\ Comparing real and imaginary parts from both sides, we get \\ x^{2} - y^{2} = 0 … (i) \\ and xy = \frac{1}{2} \\ {(x^{2} + y^{2})}^{2} = {(0)}^{2} + 4 {(\frac{1}{2})}^{2} \\ = 1 \\ \Rightarrow x^{2} + y^{2} = 1 … (ii) \\ Adding equation (i) and equation (ii) \\ 2 x^{2} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}} \\ and putting x = \frac{1}{\sqrt{2}} in xy = \frac{1}{2}, we get \\ \Rightarrow \frac{1}{\sqrt{2}} y = \frac{1}{2} \\ \Rightarrow y = \frac{1}{\sqrt{2}} \end{array}$ $\begin{array}{l} Putting x = - \frac{1}{\sqrt{2}} in xy = \frac{1}{2}, we get \\ \Rightarrow (- \frac{1}{\sqrt{2}}) y = \frac{1}{2} \\ \Rightarrow y = - \frac{1}{\sqrt{2}} \\ ∴ x + iy = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i or - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \\ \Rightarrow \sqrt{i} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i or - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i \\ \Rightarrow \sqrt{i} = \pm \frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} i \end{array}$

Q.21 Find the square root of the following: 1 + i

Ans

$\begin{array}{l} Let ‹ x + iy = \sqrt{1 + i} \\ Squaring both sides, we get \\ {(x + iy)}^{2} = 1 + i \\ x^{2} + 2 ixy + {(iy)}^{2} = 1 + i \\ x^{2} + 2 ixy - y^{2} = 1 + i \\ x^{2} - y^{2} + 2 ixy = 1 + i \\ Comparing real and imaginary parts from both sides, we get \\ x^{2} - y^{2} = 1 … (i) \\ and xy = \frac{1}{2} \\ {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 {(xy)}^{2} \end{array}$ $\begin{array}{l} {(x^{2} + y^{2})}^{2} = {(1)}^{2} + 4 {(\frac{1}{2})}^{2} \\ = 1 + 1 \\ x^{2} + y^{2} = \sqrt{2} . . . (ii) \\ Adding equation (i) and equation (ii), we get \\ {2x}^{2} = 1 + \sqrt{2} \\ \Rightarrow x^{2} = \frac{1 + \sqrt{2}}{2} \\ \Rightarrow x = \pm \sqrt{\frac{\sqrt{2} + 1}{2}} \\ Putting value of x in xy = \frac{1}{2}, we get \\ y = \frac{1}{2 x} \\ = \frac{1}{2} \times \pm \sqrt{\frac{2}{\sqrt{2} + 1}} \\ = \pm \sqrt{\frac{1}{2 \sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}} \\ = \pm \sqrt{\frac{\sqrt{2} - 1}{2 (2 - 1)}} \\ y = \pm \sqrt{\frac{\sqrt{2} - 1}{2}} \\ ∴ x + iy = \pm \sqrt{\frac{\sqrt{2} + 1}{2}} \pm \sqrt{\frac{\sqrt{2} - 1}{2}} i \\ \Rightarrow \sqrt{1 + i} = \pm \sqrt{\frac{\sqrt{2} + 1}{2}} \pm \sqrt{\frac{\sqrt{2} - 1}{2}} i \end{array}$

Q.22

$Evaluate : {[i^{18} + {(\frac{1}{i})}^{25}]}^{3}$

Ans

$\begin{array}{l} {[i^{18} + {(\frac{1}{i})}^{25}]}^{3} = {[i^{4 \times 4 + 2} + (\frac{1}{i^{4 \times 6 + 1}})]}^{3} \\ = {[{(i^{4})}^{4} i^{2} + {(\frac{1}{i^{4}})}^{6} \frac{1}{i}]}^{3} \\ = {[{(1)}^{4} (- 1) + {(\frac{1}{1})}^{6} \times \frac{1}{i}]}^{3} [âˆµ i^{4} = 1] \\ = {[- 1 + \frac{1}{i} \times \frac{i}{i}]}^{3} \\ = {[- 1 + \frac{i}{i^{2}}]}^{3} \\ = {[- 1 - i]}^{3} [âˆµ i^{2} = - 1] \\ = {(- 1)}^{3} + 3 {(- 1)}^{2} (- i) + 3 (- 1) {(- i)}^{2} + {(- i)}^{3} \\ [âˆµ {(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 {ab}^{2} + b^{3}] \end{array}$ $\begin{array}{l} = - 1 - 3 i + 3 + i [âˆµ {(- i)}^{2} = - 1, {(- i)}^{3} = i] \\ = 2 - 2 i \end{array}$

Q.23 For any two complex numbers, z₁ and z₂, prove that
Re (z₁ z₂) = Re z₁ Re z₂ – Imz₁ Imz₂.

Ans

$\begin{array}{l} Let {‹ z}_{1} = a + ib {and z}_{2} = c + id \\ Then, z_{1} . z_{2} = (a + ib) (c + id) \\ = ac + iad + ibc + i^{2} bd \\ = ac + i (ad + bc) - bd \\ = (ac - bd) + i (ad + bc) \\ Re (z_{1} z_{2}) = ac - bd \\ = (a) (c) - (b) (d) \\ = Re (z_{1}) Re (z_{2}) - Im (z_{1}) Im (z_{2}) [\begin{array}{l} âˆµ Im (z_{1}) = b, \\ Im (z_{2}) = d \end{array}] \\ Thus, Re (z_{1} z_{2}) = Re (z_{1}) Re (z_{2}) - Im (z_{1}) Im (z_{2}) \end{array}$

Q.24

$Reduce (\frac{1}{1 – 4i} – \frac{2}{1 + i}) (\frac{3 – 4i}{5 + i}) to the standard form.$

Ans

$\begin{array}{l} (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) = (\frac{1}{1 - 4 i} \times \frac{1 + 4 i}{1 + 4 i} - \frac{2}{1 + i} \times \frac{1 - i}{1 - i}) (\frac{3 - 4 i}{5 + i} \times \frac{5 - i}{5 - i}) \\ = (\frac{1 + 4 i}{1 - {(4 i)}^{2}} - \frac{2 (1 - i)}{1^{2} - i^{2}}) (\frac{(3 - 4 i) (5 - i)}{5^{2} - i^{2}}) \end{array}$ $\begin{array}{l} = (\frac{1 + 4 i}{1 - 16 i^{2}} - \frac{2 (1 - i)}{1^{2} - (- 1)}) (\frac{(15 - 20 i - 3 i + 4 i^{2})}{5^{2} - (- 1)}) \\ = {\frac{1 + 4 i}{1 - 16 (- 1)} - \frac{2 (1 - i)}{1 + 1}} [\frac{{15 - 20 i - 3 i + 4 (- 1)}}{25 + 1}] \\ = {\frac{1 + 4 i}{1 + 16} - \frac{2 (1 - i)}{2}} [\frac{(11 - 23 i)}{26}] \\ = {\frac{1 + 4 i}{17} - (1 - i)} [\frac{(11 - 23 i)}{26}] \\ = \frac{1}{17 \times 26} {1 + 4 i - 17 + 17 i} (11 - 23 i) \\ = \frac{1}{17 \times 26} (- 16 + 21 i) (11 - 23 i) \\ = \frac{1}{442} (- 176 + 368 i + 231 i - 483 i^{2}) \\ = \frac{1}{442} (- 176 + 599 i + 483) \\ (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) = \frac{1}{442} (307 + 599 i) \end{array}$

Q.25

$If x - iy = \sqrt{\frac{a - ib}{c - id}} prove that {(x^{2} + y^{2})}^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}} .$

Ans

$\begin{array}{l} Given : ‹‹x - iy = \sqrt{\frac{a - ib}{c - id}} \\ Squaring both sides, we get \\ {(x - iy)}^{2} = {(\sqrt{\frac{a - ib}{c - id}})}^{2} \\ x^{2} - 2 i‹„ xy + i^{2} y^{2} = \frac{a - ib}{c - id} \times \frac{c + id}{c + id} \\ {„„x}^{2} - 2 i‹„ xy - y^{2} = \frac{(a - ib) (c + id)}{c^{2} - i^{2} d^{2}} \\ {„„x}^{2} - y^{2} - 2 i‹„ xy = \frac{(ac - ibc + iad - i^{2} bd)}{c^{2} - (- 1) d^{2}} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹„„„„„„„„„„„„„„„„„„„„„„ = \frac{(ac - ibc + iad + bd)}{c^{2} + d^{2}} „‹‹‹‹ [{Z‹i}^{2} = - 1] \\ „„„„„„„„„„„„„„„„„„„„„„‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{ac + bd}{c^{2} + d^{2}} - i (\frac{bc - ad}{c^{2} + d^{2}}) \\ Comparing „ real „ part „ of „ both „ sides, „ we „ get \\ x^{2} - y^{2} = \frac{ac + bd}{c^{2} + d^{2}} \\ We know that {(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + {(2 xy)}^{2} \\ So {(x^{2} + y^{2})}^{2} = {(\frac{ac + bd}{c^{2} + d^{2}})}^{2} + {(\frac{bc - ad}{c^{2} + d^{2}})}^{2} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{a^{2} c^{2} + b^{2} d^{2} + 2 abcd}{{(c^{2} + d^{2})}^{2}} + \frac{b^{2} c^{2} + a^{2} d^{2} - 2 abcd}{{(c^{2} + d^{2})}^{2}} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{a^{2} c^{2} + b^{2} d^{2} + 2 abcd + b^{2} c^{2} + a^{2} d^{2} - 2 abcd}{{(c^{2} + d^{2})}^{2}} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{a^{2} c^{2} + b^{2} d^{2} + b^{2} c^{2} + a^{2} d^{2}}{{(c^{2} + d^{2})}^{2}} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{a^{2} (c^{2} + d^{2}) + b^{2} (c^{2} + d^{2})}{{(c^{2} + d^{2})}^{2}} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{(a^{2} + b^{2}) (c^{2} + d^{2})}{{(c^{2} + d^{2})}^{2}} \\ ‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹ = \frac{a^{2} + b^{2}}{c^{2} + d^{2}} \\ ∴ {(x^{2} + y^{2})}^{2} „ = \frac{a^{2} + b^{2}}{c^{2} + d^{2}} \\ Hence proved . \end{array}$

Q.26

$\begin{array}{l} Convert the following in the polar form: \\ (i) „ \frac{1+7i}{{(2 - i)}^{2}} \\ (ii) „ \frac{1+3i}{1 - 2i} \end{array}$

Ans

$\begin{array}{l} (i) \frac{1 + 7 i}{{(2 - i)}^{2}} = \frac{1 + 7 i}{(4 - 4 i + i^{2})} \\ = \frac{1 + 7 i}{(4 - 4 i - 1)} \end{array}$ $\begin{array}{l} = \frac{1 + 7 i}{(3 - 4 i)} \times \frac{(3 + 4 i)}{(3 + 4 i)} \\ = \frac{3 + 4 i + 21 i + 28 i^{2}}{9 - {(4 i)}^{2}} \\ = \frac{3 + 25 i + 28 (- 1)}{9 - {(4)}^{2} i^{2}} \\ = \frac{- 25 + 25 i}{9 - 16 (- 1)} \\ = \frac{- 25 + 25 i}{9 + 16} \\ = \frac{- 25}{25} + \frac{25}{25} i \\ = - 1 + i \\ Let z = - 1 + i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - 1 and rsin θ = 1 \\ Squarring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- 1)}^{2} + {(1)}^{2} \\ r^{2} = 1 + 1 \\ \Rightarrow r = \sqrt{2} \\ ∴ cosθ = \frac{- 1}{\sqrt{2}} and sin θ = \frac{1}{\sqrt{2}} \\ \Rightarrow cosθ = - \cos \frac{π}{4} and sin θ = \sin \frac{π}{4} \end{array}$ $\begin{array}{l} Since, sin θ is positive and cos θ is negative in II quadrant. \\ So, θ = π - \frac{π}{4} = \frac{3 π}{4} \\ Thus, polar form of z = \sqrt{2} {\cos (\frac{3 π}{4}) + isin (\frac{3 π}{4})} . \\ (ii) \frac{1 + 3 i}{1 - 2 i} = = \frac{1 + 3 i}{1 - 2 i} \times \frac{1 + 2 i}{1 + 2 i} \\ = \frac{1 + 2 i + 3 i + 6 i^{2}}{1 - 4 i^{2}} \\ = \frac{1 + 2 i + 3 i + 6 (- 1)}{1 - 4 (- 1)} \\ = \frac{1 + 5 i - 6}{1 + 4} \\ = \frac{- 5}{5} + \frac{5 i}{5} \\ = - 1 + i \\ Let z = - 1 + i \\ Comparing with z = r (cosθ + isinθ), we get \\ rcos θ = - 1 and rsin θ = 1 \\ Squarring and adding, we get \\ r^{2} (\cos^{2} θ + \sin^{2} θ) = {(- 1)}^{2} + {(1)}^{2} \\ r^{2} = 1 + 1 \\ \Rightarrow r = \sqrt{2} \\ ∴ cosθ = \frac{- 1}{\sqrt{2}} and sin θ = \frac{1}{\sqrt{2}} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow cosθ = - \cos \frac{π}{4} and sin θ = \sin \frac{π}{4} \\ Since, sin θ is positive and cos θ is negative in II quadrant. \\ So, θ = π - \frac{π}{4} = \frac{3 π}{4} \\ Thus, polar form of z = \sqrt{2} {\cos (\frac{3 π}{4}) + isin (\frac{3 π}{4})} . \end{array}$

Q.27 Solve each of the equation in Exercises 6 to 9.

$\begin{array}{l} 6 . „ 3 x^{2} - 4 x + \frac{20}{3} = 0 \\ 7 . {„x}^{2} - 2 x + \frac{3}{2} = 0 \\ 8.„ 27 x^{2} - 10 x + 1 = 0 \\ 9.„ 21 x^{2} - 28 x + 10 = 0 \end{array}$

Ans

$\begin{array}{l} 3 x^{2} - 4 x + \frac{20}{3} = 0 \\ comparing {with ax}^{2} + bx + c = 0, we get \\ a = 3, b = - 4 and c = \frac{20}{3} \\ By using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (3) (\frac{20}{3})}}{2 (3)} \\ = \frac{4 \pm \sqrt{16 - 80}}{6} = \frac{4 \pm \sqrt{- 64}}{6} \end{array}$ $\begin{array}{l} x = \frac{4 \pm 8 i}{6} [âˆµ \sqrt{- 1} = i] \\ = \frac{2}{3} \pm \frac{4}{3} i \end{array}$

$\begin{array}{l} We have, \\ x^{2} - 2 x + \frac{3}{2} = 0 \\ comparing {with ax}^{2} + bx + c = 0, we get \\ a = 1, b = - 2 and c = \frac{3}{2} \\ By using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (1) (\frac{3}{2})}}{2 (1)} \\ = \frac{2 \pm \sqrt{4 - 6}}{2} = \frac{2 \pm \sqrt{- 2}}{2} \\ x = \frac{2 \pm \sqrt{2} i}{2} [âˆµ \sqrt{- 1} = i] \\ = 1 \pm \frac{\sqrt{2}}{2} i \end{array}$

$We have, {27x}^{2} - 10 x + 1 = 0$ $\begin{array}{l} Comparing {with ax}^{2} + bx + c = 0, we get \\ a = 27, b = - 10 and c = 1 \\ By using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = \frac{- (- 10) \pm \sqrt{{(- 10)}^{2} - 4 (27) 1}}{2 (27)} \\ = \frac{10 \pm \sqrt{100 - 108}}{54} = \frac{10 \pm \sqrt{- 8}}{54} \\ x = \frac{10 \pm 2 \sqrt{2} i}{54} [âˆµ \sqrt{- 1} = i] \\ = \frac{5 \pm \sqrt{2} i}{27} \\ x = \frac{5}{\sqrt{27}} \pm \frac{\sqrt{2}}{27} i \end{array}$

$\begin{array}{l} We have, \\ {21x}^{2} - 28x + 10 = 0 \\ Comparing {with ax}^{2} + bx + c = 0, we get \\ a = 21, b = - 28 and c = 10 \\ By using quadratic formula, \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \end{array}$ $\begin{array}{l} = \frac{- (- 28) \pm \sqrt{{(- 28)}^{2} - 4 (21) (10)}}{2 (21)} \\ = \frac{28 \pm \sqrt{784 - 840}}{42} = \frac{28 \pm \sqrt{- 56}}{42} \\ x = \frac{28 \pm 2 \sqrt{14} i}{42} [âˆµ \sqrt{- 1} = i] \\ = \frac{14 \pm \sqrt{14} i}{21} \\ = \frac{14}{21} \pm \frac{\sqrt{14}}{21} i \\ x = \frac{2}{3} \pm \frac{\sqrt{14}}{21} i \end{array}$

Q.28

$If z_{1} = 2 - i, z_{2} = 1 + i, find | \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} | .$

Ans

$\begin{array}{l} Given : z_{1} = 2 - i, z_{2} = 1 + i \\ \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} = \frac{2 - i + 1 + i + 1}{2 - i - 1 - i + 1} \\ = \frac{4}{2 - 2 i} \\ = \frac{2}{1 - i} \times \frac{1 + i}{1 + i} \\ = \frac{2 + 2 i}{1 - i^{2}} \\ = \frac{2 + 2 i}{1 - (- 1)} \\ = \frac{2 + 2 i}{2} \\ = 1 + i \\ | \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} | = | 1 + i | \\ = \sqrt{{(1)}^{2} + {(1)}^{2}} \\ = \sqrt{1 + 1} \\ = \sqrt{2} \end{array}$

Q.29

$If a + ib = \frac{{(x + iy)}^{2}}{{2x}^{2} +1} {, prove that a}^{2} {+b}^{2} = \frac{{(x^{2} +1)}^{2}}{{({2x}^{2} +1)}^{2}} .$

Ans

$\begin{array}{l} Given : a + ib = \frac{{(x + iy)}^{2}}{2 x^{2} + 1} \\ = \frac{x^{2} + 2 xy i + i^{2} y^{2}}{2 x^{2} + 1} \\ = \frac{x^{2} + 2 xy i - y^{2}}{2 x^{2} + 1} \\ = \frac{x^{2} - y^{2}}{2 x^{2} + 1} + \frac{2 xy}{2 x^{2} + 1} i \\ Taking modulus of both sides, we get \\ | a + ib | = | \frac{x^{2} - y^{2}}{2 x^{2} + 1} + \frac{2 xy}{2 x^{2} + 1} i | \end{array}$ $\begin{array}{l} \sqrt{a^{2} + b^{2}} = \sqrt{{(\frac{x^{2} - y^{2}}{2 x^{2} + 1})}^{2} + {(\frac{2 xy}{2 x^{2} + 1})}^{2}} \\ Squaring both sides, we get \\ a^{2} + b^{2} = {(\frac{x^{2} - y^{2}}{2 x^{2} + 1})}^{2} + {(\frac{2 xy}{2 x^{2} + 1})}^{2} \\ = \frac{{(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2}}{{(2 x^{2} + 1)}^{2}} \\ = \frac{x^{4} - 2 x^{2} y^{2} + y^{4} + 4 x^{2} y^{2}}{{(2 x^{2} + 1)}^{2}} \\ = \frac{x^{4} + 2 x^{2} y^{2} + y^{4}}{{(2 x^{2} + 1)}^{2}} \\ = \frac{{(x^{2} + y^{2})}^{2}}{{(2 x^{2} + 1)}^{2}} \\ Thus, \\ a^{2} + b^{2} = \frac{{(x^{2} + y^{2})}^{2}}{{(2 x^{2} + 1)}^{2}} . \end{array}$

Q.30

$\begin{array}{l} {Let z}_{1} {= 2 – i, z}_{2} = -2 + i. Find \\ (i) Re (\frac{z_{1} z_{2}}{{\bar{z}}_{1}}), (ii) Im (\frac{1}{z_{1} {\bar{z}}_{1}}) \end{array}$

Ans

$\begin{array}{l} We „have, \\ {„„„„„„z}_{1} = 2 - i, „ \bar{z_{1}} = 2 + {i and z}_{2} = - 2 + i \\ ∴ {„z}_{1} . z_{2} = (2 - i) (- 2 + i) \\ „„„„„„„„„„ = - 4 + 2 i + 2 i - i^{2} \\ „„„„„„„„„„ = - 4 + 4 i - (- 1) \\ „„„„„„„„„„ = - 3 + 4 i \\ Now, „ \frac{z_{1} z_{2}}{_{1}} = \frac{- 3 + 4 i}{2 + i} \\ „„„„„„„„„„„„„„„ = \frac{- 3 + 4 i}{2 + i} \times \frac{2 - i}{2 - i} \\ „„„„„„„„„„„„„„„ = \frac{(- 3 + 4 i) (2 - i)}{2^{2} - i^{2}} \\ „„„„„„„„„„„„„„„ = \frac{- 6 + 3 i + 8 i - 4 i^{2}}{4 + 1} \\ „„„„„„„„„„„„„„„ = \frac{- 6 + 3 i + 8 i - 4 (- 1)}{5} \\ „„„„„„„„„„„„„„„ = \frac{- 2 + 11 i}{5} \\ „„„„„„„„„„„„„„„ = \frac{- 2}{5} + \frac{11}{5} i \\ „ Re (\frac{z_{1} z_{2}}{_{1}}) = \frac{- 2}{5} \\ {„„„„„„„„„z}_{1}_{1} = (2 - i) (2 + i) \\ „„„„„„„„„„„„„„„ = 4 - i^{2} \\ „„„„„„„„„„„„„„„ = 4 - (- 1) \\ „„„„„„„„„„„„„„„ = 4 + 1 \\ „„„„„„„„„„„„„„„ = 5 \\ „„„„„„„„„ \frac{1}{{z_{1}}_{1}} = \frac{1}{5} + 0 i \\ „ Im „ (\frac{1}{{z_{1}}_{1}}) = 0 \end{array}$

Q.31

$Find the modulus and argument of the complex number \frac{1+2i}{1-3i} .$

Ans

$\begin{array}{l} We „have, \\ \frac{1 + 2 i}{1 - 3 i} = \frac{1 + 2 i}{1 - 3 i} \times \frac{1 + 3 i}{1 + 3 i} \\ „„„„„„„„„ = \frac{1 + 3 i + 2 i + 6 i^{2}}{1 - {(3 i)}^{2}} \\ „„„„„„„„„ = \frac{1 + 5 i + 6 (- 1)}{1 - 9 (- 1)} \\ „„„„„„„„„ = \frac{1 + 5 i - 6}{1 + 9} \\ „„„„„„„„„ = \frac{- 5 + 5 i}{10} \\ „„„„„„„„„ = - \frac{5}{10} + \frac{5}{10} i \\ „„„„„„„„„ = - \frac{1}{2} + \frac{1}{2} i \\ Modulus (- \frac{1}{2} + \frac{1}{2} i) \\ „„„„„„„„„„ = \sqrt{{(- \frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}} \\ „„„„„„„„„„ = \sqrt{\frac{1}{4} + \frac{1}{4}} \\ „„„„„„„„„„ = \sqrt{\frac{2}{4}} \\ „„„„„„„„„„ = \frac{\sqrt{2}}{2} \\ argument (θ) = \tan^{- 1} (\frac{(\frac{1}{2})}{(- \frac{1}{2})}) [\begin{array}{l} If z = a + ib \\ θ = \tan^{- 1} (\frac{b}{a}) \end{array}] \\ „„„„„„„„„„„„„„„„„„ = \tan^{- 1} (\frac{(\frac{1}{2})}{(- \frac{1}{2})}) \\ „„„„„„„„„„„„„„„„„„ = \tan^{- 1} (- 1) = \frac{3 π}{4} \end{array}$

Q.32 Find the real numbers x and y if
(x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Ans

$\begin{array}{l} (x - i y) (3 + 5i) = 3 x + 5 xi - 3 y i - 5 {yi}^{2} \\ = 3 x + 5 xi - 3 y i + 5 y \\ = (3 x + 5 y) + (5 x - 3 y) i \\ Conjugate of (3 x + 5 y) + (5 x - 3 y) i \\ = (3 x + 5 y) - (5 x - 3 y) i . . . (i) \\ Comparing equation (i) with - 6 - 24i, we get \\ 3 x + 5 y = - 6 and - (5 x - 3 y) = - 24 \\ 3 x + 5 y = - 6 . . . (ii) \\ 5 x - 3 y = 24 . . . (iii) \\ Multiplying equation (ii) by 3 and equation (iii) by 5, we get \\ 9 x + 15 y = - 18 . . . (iv) \\ 25 x - 15 y = 120 . . . (v) \\ Adding equation (iv) and equation (v), we get \\ 34 x = 102 \Rightarrow x = \frac{102}{34} = 3 \\ Putting value of x in equation (ii), we get \\ 3 (3) + 5 y = - 6 \\ \Rightarrow 5 y = - 6 - 9 \\ \Rightarrow y = - \frac{15}{5} = - 3 \\ Thus, the values of x and y are 3 and - 3 respectively. \end{array}$

Q.33

$Find the modulus of \frac{1 + i}{1 – i} - \frac{1 - i}{1 + i} .$

Ans

$\begin{array}{l} \frac{1 + i}{1 - i} - \frac{1 - i}{1 + i} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} - \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \\ = \frac{1 + 2 i + i^{2}}{1^{2} - i^{2}} - \frac{1 - 2 i + i^{2}}{1^{2} - i^{2}} \\ = \frac{1 + 2 i - 1}{1 - (- 1)} - \frac{1 - 2 i - 1}{1 - (- 1)} \\ = \frac{2 i}{2} - \frac{- 2 i}{2} \\ = i + i \\ = 2 i \\ ∴ | \frac{1 + i}{1 - i} - \frac{1 - i}{1 + i} | \\ = | 0 + 2 i | \\ = \sqrt{0 + 4} \\ = 2 \end{array}$

Q.34

${If (x + iy)}^{3} = u + iv, then show that \frac{u}{x} + \frac{v}{y} = 4 (x^{2} {– y}^{2}) .$

Ans

$\begin{array}{l} Given : \\ {(x + iy)}^{3} = u + iv \\ x^{3} + 3 x^{2} (iy) + 3 x {(iy)}^{2} + {(iy)}^{3} = u + iv \\ x^{3} + 3 {ix}^{2} y - 3 {xy}^{2} - {iy}^{3} = u + iv \\ (x^{3} - 3 {xy}^{2}) + (3 x^{2} y - y^{3}) i = u + iv \end{array}$ $\begin{array}{l} Comparing real and imaginary parts of both sides,we get \\ u = x^{3} - 3 {xy}^{2} and v = 3 x^{2} y - y^{3} \\ \Rightarrow \frac{u}{x} = x^{2} - 3 y^{2} and \frac{v}{y} = 3 x^{2} - y^{2} \\ L . H . S . = \frac{u}{x} + \frac{v}{y} \\ = x^{2} - 3 y^{2} + 3 x^{2} - y^{2} \\ = 4 x^{2} - 4 y^{2} \\ = 4 (x^{2} - y^{2}) = R . H . S . \\ Hence proved. \end{array}$

Q.35

$If α and β are different complex numbers with |β| =1, then find |\frac{β-α}{1- \bar{α} β}| .$

Ans

$\begin{array}{l} Since, {| z |}^{2} = z . \bar{z} \\ ∴ {| \frac{β - α}{1 - \bar{α} β} |}^{2} = (\frac{β - α}{1 - \bar{α} β}) \bar{(\frac{β - α}{1 - \bar{α} β})} \\ = (\frac{β - α}{1 - \bar{α} β}) (\frac{\bar{β} - \bar{α}}{1 - \bar{\bar{α}} \bar{β}}) \\ = (\frac{β - α}{1 - \bar{α} β}) (\frac{\bar{β} - \bar{α}}{1 - α \bar{β}}) [âˆµ \bar{\bar{z}} = z] \\ = \frac{β \bar{β} - β \bar{α} - α \bar{β} + α \bar{α}}{1 - α \bar{β} - \bar{α} β + \bar{α} βα \bar{β}} \end{array}$ $\begin{array}{l} = \frac{{| β |}^{2} - β \bar{α} - α \bar{β} + {| α |}^{2}}{1 - α \bar{β} - \bar{α} β + (α \bar{α}) (β \bar{β})} [âˆµ z \bar{z} = {| z |}^{2}] \\ = \frac{{| β |}^{2} - β \bar{α} - α \bar{β} + {| α |}^{2}}{1 - α \bar{β} - \bar{α} β + {| α |}^{2} {| β |}^{2}} \\ = \frac{(1^{2}) - β \bar{α} - α \bar{β} + {| α |}^{2}}{1 - α \bar{β} - \bar{α} β + {| α |}^{2} (1^{2})} \\ = \frac{1 - β \bar{α} - α \bar{β} + {| α |}^{2}}{1 - α \bar{β} - \bar{α} β + {| α |}^{2} . 1} \\ = \frac{1 - β \bar{α} - α \bar{β} + {| α |}^{2}}{1 - α \bar{β} - \bar{α} β + {| α |}^{2}} \\ = 1 \\ \Rightarrow | \frac{β - α}{1 - \bar{α} β} | = \sqrt{1} = 1 \end{array}$

Q.36

$\begin{array}{l} Find the number of non-zero integral solutions of the \\ equation {|1 – i|}^{x} {= 2}^{x} . \end{array}$

Ans

$\begin{array}{l} „„„„„„„„„„„„„„„ {|1 - i|}^{x} = 2^{x} \\ \Rightarrow {\{\sqrt{1^{2} + {(- 1)}^{2}}\}}^{x} = 2^{x} \\ \Rightarrow „„„„„„„„„„„„„ {\{\sqrt{2}\}}^{x} = 2^{x} \end{array}$ $\begin{array}{l} Squaring both sides, we get \\ „„„„ {(\sqrt{2})}^{2 x} = 2^{2 x} \\ \Rightarrow „„„„„„„ 2^{x} = 2^{2 x} \\ \Rightarrow „„„„„„„„x = 2 x \\ \Rightarrow „„„„„„ - x = 0 \\ \Rightarrow „„„„„„„„„x = 0 \\ Thus, x = 0. \end{array}$

Q.37 If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²
Ans

$\begin{array}{l} Given: \\ (a + ib) (c + id) (e + if) (g + ih) = A + iB \\ Taking modulus of both sides, we get \\ | (a + ib) (c + id) (e + if) (g + ih) | = | A + iB | \\ \Rightarrow | (a + ib) | \times | (c + id) | \times | (e + if) | \times | (g + ih) | = | A + iB | \\ \Rightarrow \sqrt{(a^{2} + b^{2})} \sqrt{(c^{2} + d^{2})} \sqrt{(e^{2} + f^{2})} \sqrt{(g^{2} + h^{2})} = \sqrt{A^{2} + B^{2}} \\ Squaring both sides, we get \\ (a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2} \\ Hence proved. \end{array}$

Q.38

$If {(\frac{1 + i}{1 – i})}^{m} =1, then find the least positive integral„value of m.$

Ans

$We have,$ $\begin{array}{l} {(\frac{1 + i}{1 - i})}^{m} = 1 \\ \Rightarrow {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{m} = 1 \\ \Rightarrow {(\frac{1 + 2 i + i^{2}}{1 - i^{2}})}^{m} = 1 \\ \Rightarrow {(\frac{1 + 2 i - 1}{1 - (- 1)})}^{m} = 1 \\ \Rightarrow {(\frac{2 i}{2})}^{m} = 1 \\ \Rightarrow i^{m} = 1 \\ \Rightarrow i^{m} = i^{4 k} \\ \Rightarrow m = 4 k \\ For least positive integral value of m, k = 1. \\ So, m = 4. \end{array}$

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FAQs (Frequently Asked Questions)

1. Which study material should I refer to for preparing for NCERT Solutions Class 11 Mathematics Chapter 5?

Students are advised to use the best academic notes provided by the Extramarks platform. Along with NCERT Solutions Class 11 Mathematics Chapter 5, students are advised to use the NCERT Exemplar notes and several reference books. They can also appear for several test series and solve CBSE sample papers and various CBSE past years’ question papers to assess themselves based on their preparation level. Below we have listed some important reference books.

Mathematics: for Class 11 by RS Agarwal
Mathematics for Class 11 by RD Sharma
Mathematics by S. Chand publications
Problems in single variable calculus by I. A. Maron
Higher Algebra by Hall & Knight

2. Which chapters are included in the NCERT Exemplar Class 11 Mathematics?

Extramarks provide the best academic notes. The NCERT Solutions covers all the essential topics included in the textbooks in a well-structured and detailed manner. The solutions of every chapter are very easy to comprehend. The chapters in the NCERT Solutions Class 11 Mathematics are as follows:

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 4 Principle of Mathematical Induction

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight Lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability

NCERT Solutions Class 11 Maths Chapter 5

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Key Topics Covered In NCERT Solutions Class 11 Mathematics Chapter 5

List of NCERT Solutions Class 11 Mathematics Chapter 5 Exercise & Solutions

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Key Features of NCERT Solutions Class 11 Mathematics Chapter 5

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NCERT Solutions Class 11 Maths Chapter 5

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