# NCERT Solutions for Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations Exercise 5.1

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The CBSE Syllabus for 2022–2023 includes the chapter Complex Numbers and Quadratic Equations, which covers various important Mathematical theories and equations. Numerous practise problems are provided for each of these concepts in the NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 which will aid students in understanding more complex ideas in the future. Extramarks offers remedies to all of these issues along with adequate justifications. These Extramarks NCERT solutions assist students who want to do well in their examinations.

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*NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations (Ex 5.1) Exercise 5.1 *

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NCERT Solutions for Class 11 Mathematics –Chapter 1 – Sets

This chapter discusses the concept of sets and how they are representedEmpty set, Finite and Infinite sets, Equal sets, Subsets, Power set, and Universal set are among the topics covered in this chapter. The chapter also teaches students how to create Venn diagrams and introduces them to the ideas of set intersection and union. This chapter also discusses the properties of a set’s complement and the difference between sets. The chapter includes six exercises and one more, giving students enough problems to work through and, as a result, better understanding the idea.

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The chapter on the “Principle of Mathematical Induction” discusses a variety of subjects, including how to demonstrate an induction and why it is useful by using natural numbers as the least inductive subset of real numbers. The chapter’s exercise covers issues with the Principle of Mathematical Induction, as well as some of its straightforward applications.

NCERT Solutions for Class 11 Mathematics – Chapter 5 –Complex Numbers and Quadratic Equations

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As the name suggests, the chapter Linear Inequalities discusses the idea of linear inequality. This chapter also discusses the concepts of algebraic solutions of linear inequalities in one variable and their representation on the number line, graphical representations of linear inequalities in two variables, as well as the graphical method of solving a system of linear inequalities in two variables. This chapter has a total of 3 activities and a supplemental exercise that cover all the chapter’s subjects.

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### NCERT Solution Class 11 Maths Chapter 5 Exercise

(Include a Table for Exercises and No of Questions)

Extramarks’ subject specialists have put up the NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 that are available on the Extramarks mobile application and the Extramarks website. The solutions have mostly been created to make all the Class 11 Mathematics problems that are presented in the textbooks recommended by the CBSE syllabus simpler. All 16 chapters on the syllabus are covered by these NCERT Solutions for Class 11 Maths along with NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 which are arranged methodically. Therefore, the NCERT Class 11 Maths solutions provide comprehensive learning and aid in the improvement of students’ arithmetic abilities as well as their logical and reasoning abilities. To make learning much more comfortable, the complex problems have been clarified and made simpler in these solutions.

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### Complex Numbers and Quadratic Equations Exercise 5.1 – An Overview

The inability to resolve some quadratic equations is used as a motivation for the necessity of complex numbers, particularly the number 1. In this chapter, students can also learn about the polar representation of complex numbers, the Argand plane, and the algebraic features of complex numbers. The fundamental theorem of algebra is stated, as are the square root of a complex number and the solutions to quadratic equations (with real coefficients) in the complex number system. This chapter has three exercises that cover these subjects, as well as a random exercise. Complex numbers and quadratic equations are covered in NCERT’s fifth chapter. Complex numbers and quadratic equations are covered in detail in the NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1. The key formulas from NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 are also included in the revision notes. NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 provide thorough explanations of concepts using examples and exercises. The topics covered in the NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 range from complex number algebra through quadratic equations. In addition to covering NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1, NCERT notes also cover chapter 5 notes.

### Class 11 Maths Chapter 5 Exercise 5.1 – All Questions

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

$\begin{array}{l}1.\text{\hspace{0.33em}}\left(5\text{i}\right)\left(-\frac{3}{5}\text{i}\right)\\ \text{2.\hspace{0.33em}}{\mathrm{i}}^{9}+{\mathrm{i}}^{19}\\ \text{3.\hspace{0.33em}}{\mathrm{i}}^{-39}\\ \text{4.\hspace{0.33em}}3(7+\mathrm{i}7)+\mathrm{i}(7+\mathrm{i}7)\\ \text{5.\hspace{0.33em}}(1-\mathrm{i})-(-1+\mathrm{i}6)\\ \text{6.\hspace{0.33em}}\left(\frac{1}{5}+\mathrm{i}\frac{2}{5}\right)-\left(4+\mathrm{i}\frac{5}{2}\right)\\ \text{7.\hspace{0.33em}}\left[\left(\frac{1}{3}+\mathrm{i}\frac{7}{3}\right)+\left(4+\mathrm{i}\frac{1}{3}\right)\right]-\left(-\frac{4}{3}+\mathrm{i}\right)\\ \text{8.\hspace{0.33em}}{(1-\mathrm{i})}^{4}\\ 9.\text{\hspace{0.33em}}{\left(\frac{1}{3}+3\text{i}\right)}^{3}\\ 10.\text{\hspace{0.33em}}{\left(-2-\frac{1}{3}\text{i}\right)}^{3}\end{array}$

**Ans.**

1.

$\begin{array}{l}\left(5\mathrm{i}\right)(-\frac{3}{5}\mathrm{i})=5\times -\frac{3}{5}{\mathrm{i}}^{2}\\ =-3(-1)[\because \text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{i}}^{2}=-1]\\ =3+0\text{\hspace{0.17em}}\mathrm{i},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}which is in the form of a}+\text{ib.}\end{array}$

2.

$\begin{array}{l}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i}}^{\text{9}}+{\text{i}}^{\text{19}}={\mathrm{i}}^{4\times 2+1}+{\mathrm{i}}^{4\times 4+3}\\ ={\left({\mathrm{i}}^{4}\right)}^{2}\mathrm{i}+{\left({\mathrm{i}}^{4}\right)}^{4}{\mathrm{i}}^{3}\\ ={\left(1\right)}^{2}\times \mathrm{i}+{\left(1\right)}^{4}(-\mathrm{i})[\because {\mathrm{i}}^{4}=1{\text{and i}}^{\text{3}}=-\text{\hspace{0.17em}i}]\\ =\mathrm{i}-\mathrm{i}\\ =0+0\mathrm{i},\text{which is in the form of a}+\text{ib.}\end{array}$

3.

${\text{\hspace{0.17em}\hspace{0.17em}i}}^{\u2013\text{39}}=\frac{1}{{\mathrm{i}}^{39}}$ \begin{array}{l}\text{}\end{array} $\begin{array}{l}=\frac{1}{{\mathrm{i}}^{4\times 9+3}}\\ =\frac{1}{{\left({\mathrm{i}}^{4}\right)}^{9}.{\mathrm{i}}^{3}}\\ =\frac{1}{{\left(1\right)}^{9}.(-\mathrm{i})}\\ =-\frac{1}{\mathrm{i}}\times \frac{\mathrm{i}}{\mathrm{i}}\\ =-\frac{\mathrm{i}}{{\mathrm{i}}^{2}}\\ =-\frac{\mathrm{i}}{-1}\\ =0+\mathrm{i},\text{which is in the form of a}+\text{ib.}\end{array}$

4.

3(7 + i 7)+ i(7 + i 7) = 21 + 21i + 7i + 7i^{2}

= 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.

6.

$\begin{array}{l}\text{\hspace{0.17em}}(\frac{1}{5}+\mathrm{i}\frac{2}{5})-(4+\mathrm{i}\frac{5}{2})=(\frac{1}{5}-4)+\mathrm{i}(\frac{2}{5}-\frac{5}{2})\\ =-\frac{19}{5}+\mathrm{i}\left(\frac{4-25}{10}\right)\\ =-\frac{19}{5}-\mathrm{i}\frac{21}{10},\\ \mathrm{which}\text{is in the form of a}+\text{ib.}\end{array}$

7.

$\begin{array}{l}[(\frac{1}{3}+\mathrm{i}\frac{7}{3})+(4+\mathrm{i}\frac{1}{3})]-(-\frac{4}{3}+\mathrm{i})=(\frac{1}{3}+4+\frac{4}{3})+\mathrm{i}(\frac{7}{3}+\frac{1}{3}-1)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left(\frac{1+12+4}{3}\right)+\mathrm{i}\left(\frac{7+1-3}{3}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\left(\frac{17}{3}\right)+\mathrm{i}\left(\frac{5}{3}\right),\\ \mathrm{Which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}\mathrm{.}\end{array}$

8.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{(1\u2013\mathrm{i})}^{4}={\left\{{(1\u2013\mathrm{i})}^{2}\right\}}^{2}\\ ={(1-2\mathrm{i}+{\mathrm{i}}^{2})}^{2}\end{array}$ \begin{array}{l}\text{}\text{}\end{array} $\begin{array}{l}={(1-2\mathrm{i}-1)}^{2}\\ ={(-2\mathrm{i})}^{2}\\ =4{\mathrm{i}}^{2}\\ =4(-1)\\ =-4+\mathrm{i}0,\\ \mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}.\end{array}$

9.

$\begin{array}{l}{(\frac{1}{3}+3\mathrm{i})}^{3}={\left(\frac{1}{3}\right)}^{3}+3{\left(\frac{1}{3}\right)}^{2}\left(3\mathrm{i}\right)+3\left(\frac{1}{3}\right){\left(3\mathrm{i}\right)}^{2}+{\left(3\mathrm{i}\right)}^{3}\\ [\because {(\mathrm{a}+\mathrm{b})}^{3}={\mathrm{a}}^{3}+3{\mathrm{a}}^{2}\mathrm{b}+3{\mathrm{ab}}^{2}+{\mathrm{b}}^{3}]\\ =\frac{1}{27}+\text{\hspace{0.17em}}\mathrm{i}+{\mathrm{i}}^{2}\text{\hspace{0.17em}}9\text{\hspace{0.17em}}+{\mathrm{i}}^{3}\text{\hspace{0.17em}}27\text{\hspace{0.17em}}\\ =\frac{1}{27}+\mathrm{i}\text{\hspace{0.17em}}-9-\mathrm{i}\text{\hspace{0.17em}}27\\ =(\frac{1}{27}-9)+\mathrm{i}(1-27)\\ =\left(\frac{1-243}{27}\right)+\mathrm{i}(-26)\\ =\left(\frac{-242}{27}\right)-\mathrm{i}\text{\hspace{0.17em}}26\\ \mathrm{which}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}+\mathrm{ib}.\end{array}$

10.

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

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

11. 4 – 3i

12.

$\sqrt{5}+3\mathrm{i}$

13. – i

**Ans.**

$\begin{array}{l}\mathrm{Multiplicative}\text{inverse of}(\text{4}-\text{3i})=\frac{1}{(\text{4}-\text{3i})}\times \frac{(\text{4}+\text{3i})}{(\text{4}+\text{3i})}\\ =\frac{(\text{4}+\text{3i})}{({\text{4}}^{\text{2}}-{\text{9i}}^{\text{2}})}\end{array}$

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

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

$\begin{array}{l}\mathrm{Multiplicative}\text{inverse of}(\u2013\text{i})=\frac{1}{(\u2013\text{i})}\times \frac{\mathrm{i}}{\mathrm{i}}\\ =\frac{\mathrm{i}}{-{\mathrm{i}}^{2}}\\ =\frac{\mathrm{i}}{-(-1)}\\ =\frac{\mathrm{i}}{1}\\ =\mathrm{i},\\ \mathrm{Thus},\text{the multiplicative inverse of}-\text{i is i.}\end{array}$

**Q.3 **

$\begin{array}{l}\text{Express the following expression in the form of a+ib:}\\ \frac{\left(\text{3+i}\sqrt{\text{5}}\right)\left(\text{3-i}\sqrt{\text{5}}\right)}{\left(\sqrt{\text{3}}\text{+}\sqrt{\text{2}}\text{i}\right)\text{\u2013}\left(\sqrt{\text{3}}\text{\u2013}\sqrt{\text{2}}\text{i}\right)}\end{array}$

**Ans.**

$\begin{array}{l}\frac{(3+\mathrm{i}\sqrt{5})(3-\mathrm{i}\sqrt{5})}{(\sqrt{3}+\sqrt{2}\mathrm{i})-(\sqrt{3}-\sqrt{2}\mathrm{i})}=\frac{{3}^{2}-{\left(\mathrm{i}\sqrt{5}\right)}^{2}}{\sqrt{3}+\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}-\sqrt{3}+\sqrt{2\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\mathrm{i}}\text{\hspace{0.17em}}\left[\begin{array}{l}\because (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{(9-{\mathrm{i}}^{2}5)}{2\sqrt{2}\mathrm{i}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\{9-(-1)5\}}{2\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}=\frac{14}{2\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7}{\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}\times \frac{\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}{\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}{{\mathrm{i}}^{2}\left(2\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}{(-1)2}[\because {\mathrm{i}}^{2}=-1]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0-\frac{7\sqrt{2}\text{\hspace{0.17em}}\mathrm{i}}{2},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{which}\text{in the form of a + ib.}\end{array}$

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