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

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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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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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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)

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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.

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

Ans.
1.

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

2.

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

3.

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

4.
3(7 + i 7)+ i(7 + i 7) = 21 + 21i + 7i + 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.

6.

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

7.

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

8.

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

9.

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

10.

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

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.

11.

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

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

12.

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

13.

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

Q.3

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

Ans.

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

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