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# NCERT Solutions Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Last updated date: 13th Jul 2024
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## NCERT Solutions for Class 11 Maths Chapter 4 - FREE PDF Download

NCERT for Chapter 4 Complex Numbers Class 11 Solutions by Vedantu, introduces complex numbers and their applications in quadratic equations. Complex numbers, which involve real and imaginary parts, play a crucial role in mathematics and various scientific fields. Through these solutions, we aim to simplify complex concepts into easy-to-understand explanations, helping you grasp each topic effectively. From understanding the fundamentals of complex numbers to solving quadratic equations using complex roots, every aspect of this chapter is covered comprehensively. With Vedantu's NCERT Solutions,students find step-by-step explanations to all the exercises in your textbook, ensuring that you understand the concepts thoroughly.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 4 - FREE PDF Download
2. Glance on Maths Chapter 4 Class 11 - Complex Numbers and Quadratic Equations
3. Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 11
4. Exercises Under NCERT Solutions for Class 11 Maths Complex Numbers and Quadratic Equations
5. Access NCERT Solutions for Class 11 Maths Chapter 4 – Complex Numbers and Quadratic Equations
6. Overview of Deleted Syllabus for CBSE Class 11 Maths Complex Numbers and Quadratic Equations
7. Class 11 Maths Chapter 4: Exercises Breakdown
8. Other Study Material for CBSE Class  11 Maths Chapter 4
9. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

## Glance on Maths Chapter 4 Class 11 - Complex Numbers and Quadratic Equations

• NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations includes Complex numbers, algebra of complex numbers,The Modulus and the Conjugate of a Complex Number,Argand Plane and Polar Representation.

• Introduction to complex numbers, which consist of real and imaginary parts.

• complex numbers are represented in  the form a+ib where a is the real part and ib is the imaginary part.

• Operations that can be performed on complex numbers are Addition, subtraction, multiplication and division.

• Conjugate of a complex number keeps the same real part of the complex number but changes the sign of the imaginary part and it is denoted with a bar sign.

• Modulus of a complex number is the square root of the sum of the squares of the real part and the imaginary part of the complex.

• Multiplicative Inverse of a complex number is a number which when multiplied with the original number equals one.

• In Ch 4 Maths Class 11 there is one exercise(14 fully solved questions) and one Miscellaneous Exercise (14 fully solved questions) in class 11th maths chapter 4 Complex Numbers and Quadratic Equations.

## Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 11

 S.No. Current Syllabus Exercises of Class 11 Maths Chapter 4 1 NCERT Solutions of Class 11 Maths Complex Numbers and Quadratic Equations Exercise 4.1
Competitive Exams after 12th Science

## Exercises Under NCERT Solutions for Class 11 Maths Complex Numbers and Quadratic Equations

Exercise 4.1: This exercise consists of 14 questions and is focused on expressing the complex numbers in the form of a+ib and finding the multiplicative inverse of complex numbers.

Miscellaneous Exercise: Class 11 Complex Numbers Miscellaneous Solutions consists of 14 questions and covers the fundamental concepts such as operations with complex numbers, properties of conjugates, modulus, argument, and solving quadratic equations. These problems may involve finding roots of quadratic equations, determining geometric properties using complex numbers, or solving equations with complex coefficients.

## Access NCERT Solutions for Class 11 Maths Chapter 4 – Complex Numbers and Quadratic Equations

### Exercise 4.1

1. Express the given complex number in the form $\mathrm{a+ib:}\left( \mathrm{5i} \right)\left( \mathrm{-}\dfrac{\mathrm{3}}{\mathrm{5}}\mathrm{i} \right)$

And evaluate

Ans:

Evaluate the complex number

$\left( 5i \right)\left( -\dfrac{3}{5}i \right)=-5\times \dfrac{3}{5}\times i\times i$

$\left( 5i \right)\left( -\dfrac{3}{5}i \right)=-3{{i}^{2}}\cdots \left[ {{i}^{2}}=-1 \right]$

$\left( 5i \right)\left( -\dfrac{3}{5}i \right)=3$

2. Express the given complex number in the form ${\mathrm{a + ib}}:{{\mathrm{i}}^9}{\mathrm{ + }}{{\mathrm{i}}^{19}}$

And evaluate

Ans:

Evaluate the complex number

${{i}^{9}}+{{i}^{19}}={{i}^{4\times 2+1}}+{{i}^{4\times 4+3}}$

${{i}^{9}}+{{i}^{19}}={{\left( {{i}^{4}} \right)}^{2}}.i+{{\left( {{i}^{4}} \right)}^{4}}.{{i}^{3}}\cdots \left[ {{i}^{4}}=1,{{i}^{3}}=-1 \right]$

${{i}^{9}}+{{i}^{19}}=0$

3. Express the given complex number in the form $\mathrm{a+ib:}{{\mathrm{i}}^{\mathrm{-39}}}$

And evaluate

Ans:

Evaluate the complex number

${{i}^{-39}}={{i}^{4\times 9-3}}$

${{i}^{-39}}={{\left( {{i}^{4}} \right)}^{-9}}.{{i}^{-3}}$

${{i}^{-39}}=i\cdots \left[ i=-1 \right]$

${{i}^{-39}}=i$

4. Express the given complex number in the form $a+ib:3\left( 7+i7 \right)+i\left( 7+i7 \right)$

And evaluate

Ans:

Evaluate the complex number

$3\left( 7+i7 \right)+i\left( 7+i7 \right)=21+21i+7i+7{{i}^{2}}$

$3\left( 7+i7 \right)+i\left( 7+i7 \right)=21+28i+7{{i}^{2}}\cdots \left[ {{i}^{2}}=-1 \right]$

$3\left( 7+i7 \right)+i\left( 7+i7 \right)=14+28i$

5. Express the given complex number in the form

$\mathrm{a+ib:}\left( \mathrm{1-i} \right)\mathrm{-}\left( \mathrm{-1+6i} \right)$

And evaluate

Ans:

Evaluate the complex number

$\left( 1-i \right)-\left( -1+6i \right)=1-i+1-i6$

$\left( 1-i \right)-\left( -1+6i \right)=2-7i$

6. Express the given complex number in the form $\mathrm{a+ib:}\left( \dfrac{\mathrm{1}}{\mathrm{5}}\mathrm{+i}\dfrac{\mathrm{2}}{\mathrm{5}} \right)\mathrm{-}\left( \mathrm{4+i}\dfrac{\mathrm{5}}{\mathrm{2}} \right)$

And evaluate

Ans:

Evaluate the complex number $\left( \dfrac{\mathrm{1}}{\mathrm{5}}\mathrm{+i}\dfrac{\mathrm{2}}{\mathrm{5}} \right)\mathrm{-}\left( \mathrm{4+i}\dfrac{\mathrm{5}}{\mathrm{2}} \right)\mathrm{=}\dfrac{\mathrm{1}}{\mathrm{5}}\mathrm{+i}\dfrac{\mathrm{2}}{\mathrm{5}}\mathrm{-4-i}\dfrac{\mathrm{5}}{\mathrm{2}}$

$\left( \dfrac{1}{5}+i\dfrac{2}{5} \right)-\left( 4+i\dfrac{5}{2} \right)=\dfrac{-19}{5}+i\left[ \dfrac{-21}{10} \right]$

$\left( \dfrac{1}{5}+i\dfrac{2}{5} \right)-\left( 4+i\dfrac{5}{2} \right)=\dfrac{-19}{5}-\dfrac{21}{10}i$

7. Express the given complex number in the form $\mathrm{a+ib:}\left[ \left(\dfrac{\mathrm{1}}{\mathrm{3}}\mathrm{+i}\dfrac{\mathrm{7}}{\mathrm{3}} \right)\mathrm{+}\left( \mathrm{4+i}\dfrac{\mathrm{1}}{\mathrm{3}} \right)\mathrm{-}\left( \mathrm{-}\dfrac{\mathrm{4}}{\mathrm{3}}\mathrm{+i} \right) \right]$

And evaluate

Ans:

Evaluate the complex number

$\left[ \left( \dfrac{1}{3}+i\dfrac{7}{3} \right)+\left( 4+i\dfrac{1}{3} \right)-\left( -\dfrac{4}{3}+i \right) \right]=\dfrac{1}{3}+i\dfrac{7}{3}+4+i\dfrac{1}{3}+\dfrac{4}{3}-i$

$\left[ \left( \dfrac{1}{3}+i\dfrac{7}{3} \right)+\left( 4+i\dfrac{1}{3} \right)-\left( -\dfrac{4}{3}+i \right) \right]=\left( \dfrac{1}{3}+4+\dfrac{4}{3} \right)+i\left( \dfrac{7}{3}+\dfrac{1}{3}-1 \right)$

$\left[ \left( \dfrac{1}{3}+i\dfrac{7}{3} \right)+\left( 4+i\dfrac{1}{3} \right)-\left( -\dfrac{4}{3}+i \right) \right]=\dfrac{17}{3}+i\dfrac{5}{3}$

8. Express the given complex number in the form $\mathrm{a+ib:}{{\left( \mathrm{1-i} \right)}^{\mathrm{4}}}$

And evaluate

Ans:

Evaluate the complex number

${{\left( 1-i \right)}^{4}}={{\left[ 1+{{i}^{2}}-2i \right]}^{2}}$

${{\left( 1-i \right)}^{4}}={{\left[ 1-1-2i \right]}^{2}}$

${{\left( 1-i \right)}^{4}}=\left( -2i \right)\times \left( -2i \right)$

${{\left( 1-i \right)}^{4}}=-4$

9. Express the given complex number in the form $\mathrm{a+ib:}{{\left( \dfrac{\mathrm{1}}{\mathrm{3}}\mathrm{+3i} \right)}^{\mathrm{3}}}$

And evaluate

Ans:

Evaluate the complex number

${{\left( \dfrac{1}{3}+3i \right)}^{3}}={{\left( \dfrac{1}{3} \right)}^{3}}+{{\left( 3i \right)}^{3}}+\dfrac{3}{3}3i\left( \dfrac{1}{3}+3i \right)$

${{\left( \dfrac{1}{3}+3i \right)}^{3}}=\dfrac{1}{27}-\left( 27i \right)+3i\left( \dfrac{1}{3}+3i \right)$

${{\left( \dfrac{1}{3}+3i \right)}^{3}}=\dfrac{-242}{27}-26i$

10. Express the given complex number in the form $\mathrm{a+ib:}{{\left( \mathrm{-2-}\dfrac{\mathrm{1}}{\mathrm{3}}\mathrm{i} \right)}^{\mathrm{3}}}$

And evaluate

Ans:

Evaluate the complex number

${{\left( -2-\dfrac{1}{3}i \right)}^{3}}={{\left( -1 \right)}^{3}}{{\left( 2+\dfrac{1}{3}i \right)}^{3}}$

${{\left( -2-\dfrac{1}{3}i \right)}^{3}}=-\left( {{2}^{3}}+{{\left( \dfrac{i}{3} \right)}^{3}}+6\dfrac{i}{3}\left( 2+\dfrac{i}{3} \right) \right)$

${{\left( -2-\dfrac{1}{3}i \right)}^{3}}=-\dfrac{22}{3}-\dfrac{107}{27}i$

11. Find the multiplicative inverse of the complex number

$\mathrm{4-3i}$

And evaluate

Ans:

Let  $z=4-3i$

Then,

$\overline{z}=4+3i\And \left| \overline{z} \right|={{4}^{2}}+{{\left( -3 \right)}^{2}}=16+9=25$

Therefore, the multiplicative inverse of  $4-3i$  is given by

${{z}^{-1}}=\dfrac{\overline{z}}{{{\left| z \right|}^{2}}}=\dfrac{4+3i}{25}=\dfrac{4}{25}+\dfrac{3}{25}i$

12. Find the multiplicative inverse of the complex number $\sqrt{\mathrm{5}}\mathrm{+3i}$

And evaluate

Ans:

Let  $z=\sqrt{5}+3i$

Then,

$\bar{z}=\sqrt{5}-3i\text{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }|z{{|}^{2}}={{(\sqrt{5})}^{2}}+{{3}^{2}}=5+9=14$

Therefore, the multiplicative inverse of  $\sqrt{5}+3i$  is given by

${{z}^{-1}}=\dfrac{{\bar{z}}}{|z{{|}^{2}}}=\dfrac{\sqrt{5}-3i}{14}=\dfrac{\sqrt{5}}{14}-\dfrac{3i}{14}$

13. Find the multiplicative inverse of the complex number

$\mathrm{-i}$

And evaluate

Ans:

Let  $z=-i$

Then,

$\mathrm{\bar{z}=i }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\!\!|\!\!\text{ z}{{\mathrm{ }\!\!|\!\!\text{ }}^{\mathrm{2}}}\mathrm{=}{{\mathrm{1}}^{\mathrm{2}}}\mathrm{=1}$

Therefore, the multiplicative inverse of  $\mathrm{-i}$  is given by ${{\mathrm{z}}^{\mathrm{-1}}}\mathrm{=}\dfrac{{\mathrm{\bar{z}}}}{\mathrm{ }\!\!|\!\!\text{ z}{{\mathrm{ }\!\!|\!\!\text{ }}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{i}}{\mathrm{1}}\mathrm{=i}$

14. Express the following expression in the form of  $\mathrm{a+ib}$ $\dfrac{\left( \mathrm{3+i}\sqrt{\mathrm{5}} \right)\left( \mathrm{3-i}\sqrt{\mathrm{5}} \right)}{\left( \sqrt{\mathrm{3}}\mathrm{+i}\sqrt{\mathrm{2}} \right)\mathrm{-}\left( \sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}} \right)}$

Evaluate

Ans:

The following expression $\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{{{\mathrm{(3)}}^{\mathrm{2}}}\mathrm{-(i}\sqrt{\mathrm{5}}{{\mathrm{)}}^{\mathrm{2}}}}{\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i-}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i}}$

\begin{align} &\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{\mathrm{9-5}{{\mathrm{i}}^{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}} \\ &\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{\mathrm{9-5}\left( \mathrm{-1} \right)}{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}} \\ &\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{\mathrm{-7}\sqrt{\mathrm{2i}}}{\mathrm{2}} \\ \end{align}

Miscellaneous Exercise

1. Evaluate
${{\left[ {{\mathrm{i}}^{\mathrm{18}}}\mathrm{+}{{\left( \dfrac{\mathrm{1}}{\mathrm{i}} \right)}^{\mathrm{25}}} \right]}^{\mathrm{3}}}$

The expression

Ans:

Expression

${{\left[ {{\mathrm{i}}^{\mathrm{18}}}\mathrm{+}{{\left( \dfrac{\mathrm{1}}{\mathrm{i}} \right)}^{\mathrm{25}}} \right]}^{\mathrm{3}}}\mathrm{=}{{\left[ {{\mathrm{i}}^{\mathrm{4 }\!\!\times\!\!\text{ 4+2}}}\mathrm{+}\dfrac{\mathrm{1}}{{{\mathrm{i}}^{\mathrm{4 }\!\!\times\!\!\text{ 6+1}}}} \right]}^{\mathrm{3}}}$

\begin{align} & \mathrm{=}{{\left[ {{\left( {{\mathrm{i}}^{\mathrm{4}}} \right)}^{\mathrm{4}}}\mathrm{ }\!\!\times\!\!\text{ }{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+}\dfrac{\mathrm{1}}{{{\left( {{\mathrm{i}}^{\mathrm{4}}} \right)}^{\mathrm{6}}}\mathrm{ }\!\!\times\!\!\text{ i}} \right]}^{\mathrm{3}}} \\ & \mathrm{=}{{\left[ {{\mathrm{i}}^{\mathrm{2}}}\mathrm{+}\dfrac{\mathrm{1}}{\mathrm{i}} \right]}^{\mathrm{3}}}\quad \left[ {{\mathrm{i}}^{\mathrm{4}}}\mathrm{=1} \right] \\ & \mathrm{=}{{\left[ \mathrm{-1+}\dfrac{\mathrm{1}}{\mathrm{i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{i}}{\mathrm{i}} \right]}^{\mathrm{3}}}\quad \left[ {{\mathrm{i}}^{\mathrm{2}}}\mathrm{=-1} \right] \\ & \mathrm{=}{{\left[ \mathrm{-1+}\dfrac{\mathrm{i}}{{{\mathrm{i}}^{\mathrm{2}}}} \right]}^{\mathrm{3}}} \\ \end{align}

\begin{align} & \mathrm{= }\!\![\!\!\text{ -1-i}{{\mathrm{ }\!\!]\!\!\text{ }}^{\mathrm{3}}} \\ & \mathrm{=(-1}{{\mathrm{)}}^{\mathrm{3}}}{{\mathrm{ }\!\![\!\!\text{ 1+i }\!\!]\!\!\text{ }}^{\mathrm{3}}} \\ & \mathrm{=-}\left[ {{\mathrm{1}}^{\mathrm{3}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{3}}}\mathrm{+3 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ i(1+i)} \right] \\ & \mathrm{=-}\left[ \mathrm{1+}{{\mathrm{i}}^{\mathrm{3}}}\mathrm{+3i+3}{{\mathrm{i}}^{\mathrm{2}}} \right] \\ & \mathrm{=- }\!\![\!\!\text{ 1-i+3i-3 }\!\!]\!\!\text{ } \\ & \mathrm{=- }\!\![\!\!\text{ -2+2i }\!\!]\!\!\text{ } \\ & \mathrm{=2-2i} \\ \end{align}

The expression is evaluated

2. For any two complex numbers  ${{\mathrm{z}}_{\mathrm{1}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }{{\mathrm{z}}_{\mathrm{2}}}$ , prove that $\mathrm{Re}\left( {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right)\mathrm{=Re}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Re}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{-Im}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Im}{{\mathrm{z}}_{\mathrm{2}}}$

Ans:

Let ${{\mathrm{z}}_{\mathrm{1}}}\mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}}$

$\begin{matrix} \mathrm{ }\!\!\!\!\text{ }{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=}\left( {{\mathrm{x}}_{\mathrm{1}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}} \right)\left( {{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} \right) \\ \mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}\left( {{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} \right)\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}\left( {{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} \right) \\ \mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{2}}}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \\ \end{matrix}$

\begin{align} & \mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \\ & \mathrm{=}\left( {{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \right)\mathrm{+i}\left( {{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}} \right) \\ & \mathrm{Re}\left( {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right)\mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \\ & \mathrm{Re}\left( {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right)\mathrm{=Re}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Re}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{-Im}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Im}{{\mathrm{z}}_{\mathrm{2}}} \\ \end{align}

Hence, proved

3. Reduce  $\left( \dfrac{\mathrm{1}}{\mathrm{1-4i}}\mathrm{-}\dfrac{\mathrm{2}}{\mathrm{1+i}} \right)\left( \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right)$  to the standard form

Ans:

Expression

\begin{align} & \left( \dfrac{\mathrm{1}}{\mathrm{1-4i}}\mathrm{-}\dfrac{\mathrm{2}}{\mathrm{1+i}} \right)\left( \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right)\mathrm{=}\left[ \dfrac{\mathrm{(1+i)-2(1-4i)}}{\mathrm{(1-4i)(1+i)}} \right]\left[ \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right] \\ & \mathrm{=}\left[ \dfrac{\mathrm{1+i-2+8i}}{\mathrm{1+i-4i-4}{{\mathrm{i}}^{\mathrm{2}}}} \right]\left[ \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right]\mathrm{=}\left[ \dfrac{\mathrm{-1+9i}}{\mathrm{5-3i}} \right]\left[ \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right] \\ & \mathrm{=}\left[ \dfrac{\mathrm{-3+4i+27i-36}{{\mathrm{i}}^{\mathrm{2}}}}{\mathrm{25+5i-15i-3}{{\mathrm{i}}^{\mathrm{2}}}} \right]\mathrm{=}\dfrac{\mathrm{33+31i}}{\mathrm{28-10i}}\mathrm{=}\dfrac{\mathrm{33+31i}}{\mathrm{2(14-5i)}} \\ \end{align}

\begin{align} & \mathrm{=}\dfrac{\mathrm{(33+31i)}}{\mathrm{2(14-5i)}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{(14+5i)}}{\mathrm{(14+5i)}}\mathrm{ }\!\!~\!\!\text{ }\!\![\!\!\text{ On multiplying numerator and denominator by(14+5i) }\!\!]\!\!\text{ } \\ &\mathrm{=}\dfrac{\mathrm{462+165i+434i+155}{{\mathrm{i}}^{\mathrm{2}}}}{\mathrm{2}\left[ {{\mathrm{(14)}}^{\mathrm{2}}}\mathrm{-(5i}{{\mathrm{)}}^{\mathrm{2}}} \right]}\mathrm{=}\dfrac{\mathrm{307+599i}}{\mathrm{2}\left( \mathrm{196-25}{{\mathrm{i}}^{\mathrm{2}}} \right)} \\ &\mathrm{=}\dfrac{\mathrm{307+599i}}{\mathrm{2(221)}}\mathrm{=}\dfrac{\mathrm{307+599i}}{\mathrm{442}}\mathrm{=}\dfrac{\mathrm{307}}{\mathrm{442}}\mathrm{+}\dfrac{\mathrm{599i}}{\mathrm{442}} \\ \end{align}

This is the required standard form

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

Ans:

Expression

\begin{align} & \mathrm{x-iy=}\sqrt{\dfrac{\mathrm{a-ib}}{\mathrm{c-id}}} \\ & \left. \mathrm{=}\sqrt{\dfrac{\mathrm{a-ib}}{\mathrm{c-id}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{c+id}}{\mathrm{c+id}}}\quad \mathrm{ }\!\!~\!\!\text{ }\!\![\!\!\text{ On multiplying numerator and denominator by }\!\!~\!\!\text{ (c+id)} \right] \\ &\mathrm{=}\sqrt{\dfrac{\mathrm{(ac+bd)+i(ad-bc)}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}} \\ \end{align}

\begin{align} & \mathrm{ }\!\!\!\!\text{ (x-iy}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{(ac+bd)+i(ad-bc)}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \\ &{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{-2ixy=}\dfrac{\mathrm{(ac+bd)+i(ad-bc)}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \\ \end{align}

On comparing

\begin{align} &{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{ac+bd}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}\mathrm{,-2xy=}\dfrac{\mathrm{ad-bc}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}\mathrm{}...\mathrm{(1)} \\ & {{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}\mathrm{=}{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}\mathrm{+4}{{\mathrm{x}}^{\mathrm{2}}}{{\mathrm{y}}^{\mathrm{2}}} \\ & \mathrm{=}{{\left( \dfrac{\mathrm{ac+bd}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \right)}^{\mathrm{2}}}\mathrm{+}\left( \dfrac{\mathrm{ad-bc}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \right) \\ &\mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+2acbd+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}\mathrm{-2adbc}}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\ \end{align}

\begin{align} &\mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\ & \mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\ & \mathrm{=}\dfrac{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\left( {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}} \right)}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\ & \mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \\ \end{align}

Hence, proved

5. If ${{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=1+i}$ Find $\left| \dfrac{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{+}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}}{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{-}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}} \right|$

Evaluate

Ans:

Complex numbers

${{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=1+i}$

$\begin{matrix} \mathrm{ }\!\!\!\!\text{ }\left| \dfrac{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{+}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}}{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{-}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}} \right|\mathrm{=}\left| \dfrac{\mathrm{(2-i)+(1+i)+1}}{\mathrm{(2-i)-(1+i)+1}} \right| \\ \mathrm{=}\left| \dfrac{\mathrm{4}}{\mathrm{2-2i}} \right|\mathrm{=}\left| \dfrac{\mathrm{4}}{\mathrm{2(1-i)}} \right| \\ \mathrm{=}\left| \dfrac{\mathrm{2}}{\mathrm{1-i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{1+i}}{\mathrm{1+i}} \right|\mathrm{=}\left| \dfrac{\mathrm{2(1+i)}}{\left( {{\mathrm{1}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{i}}^{\mathrm{2}}} \right)} \right| \\ \mathrm{=}\left| \dfrac{\mathrm{2(1+i)}}{\mathrm{1+1}} \right|\quad \left[ {{\mathrm{i}}^{\mathrm{2}}}\mathrm{=-1} \right] \\ \mathrm{=}\left| \dfrac{\mathrm{2(1+i)}}{\mathrm{2}} \right| \\ \end{matrix}$

$\mathrm{= }\!\!|\!\!\text{ 1+i }\!\!|\!\!\text{ =}\sqrt{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}}\mathrm{=}\sqrt{\mathrm{2}}$

Thus, the value of $\left| \dfrac{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{+}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}}{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{-}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}} \right|$  is  $\sqrt{\mathrm{2}}$

6. If  $\mathrm{a+ib=}\dfrac{{{\mathrm{(x+i)}}^{\mathrm{2}}}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}}$

Prove that   ${{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{=}\dfrac{{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}$

Ans:

Expression

$\mathrm{a+ib=}\dfrac{{{\mathrm{(x+i)}}^{\mathrm{2}}}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}}$

\begin{align} &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2xi}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-1+i2x}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-1}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}}\mathrm{+i}\left( \dfrac{\mathrm{2x}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \right) \\ \end{align}

On comparing

\begin{align} & \mathrm{ }\!\!\!\!\text{ }{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{=}{{\left( \dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-1}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \right)}^{\mathrm{2}}}\mathrm{+}{{\left( \dfrac{\mathrm{2x}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \right)}^{\mathrm{2}}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{4}}}\mathrm{+1-2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+4}{{\mathrm{x}}^{\mathrm{2}}}}{{{\mathrm{(2x+1)}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{4}}}\mathrm{+1+2}{{\mathrm{x}}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} \\ & \mathrm{=}\dfrac{{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} \\ & \mathrm{ }\!\!\!\!\text{ }{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{=}\dfrac{{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} \\ \end{align}

Hence, proved

7. Let ${{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=-2+i}$

Find    \begin{align} & \mathrm{Re}\left( \dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right) \\ & \mathrm{Im}\left( \dfrac{\mathrm{1}}{{{\mathrm{z}}_{\mathrm{1}}}{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right) \\ \end{align}

Ans:

Complex numbers \begin{align} &{{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=-2+i} \\ &{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=(2-i)(-2+i)=-4+2i+2i-}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{=-4+4i-(-1)=-3+4i} \\ & \overline{{{\mathrm{z}}_{\mathrm{1}}}}\mathrm{=2+i} \\ & \mathrm{ }\!\!\!\!\text{ }\dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{\overline{{{\mathrm{z}}_{\mathrm{1}}}}}\mathrm{=}\dfrac{\mathrm{-3+4i}}{\mathrm{2+i}} \\ \end{align}

On multiplying numerator and denominator by  $\left( 2-i \right)$ , we obtain

\begin{align} &\dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{\overline{{{\mathrm{z}}_{\mathrm{1}}}}}\mathrm{=}\dfrac{\mathrm{(-3+4i)(2-i)}}{\mathrm{(2+i)(2-i)}}\mathrm{=}\dfrac{\mathrm{-6+3i+8i-4}{{\mathrm{i}}^{\mathrm{2}}}}{{{\mathrm{2}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{-6+11i-4(-1)}}{{{\mathrm{2}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{\mathrm{-2+11i}}{\mathrm{5}}\mathrm{=}\dfrac{\mathrm{-2}}{\mathrm{5}}\mathrm{+}\dfrac{\mathrm{11}}{\mathrm{5}}\mathrm{i} \\ \end{align}

On comparing real parts, we obtain

\begin{align} & \mathrm{Re}\left( \dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right)\mathrm{=}\dfrac{\mathrm{-2}}{\mathrm{5}} \\ & \mathrm{ }\!\!~\!\!\text{ }\dfrac{\mathrm{1}}{{{\mathrm{z}}_{\mathrm{1}}}{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}}\mathrm{=}\dfrac{\mathrm{1}}{\mathrm{(2-i)(2+i)}}\mathrm{=}\dfrac{\mathrm{1}}{{{\mathrm{(2)}}^{\mathrm{2}}}\mathrm{+(1}{{\mathrm{)}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{1}}{\mathrm{5}} \\ \end{align}

On comparing imaginary parts, we obtain

$\mathrm{Im}\left( \dfrac{\mathrm{1}}{{{\mathrm{z}}_{\mathrm{1}}}{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right)\mathrm{=0}$

Hence, solved

8. Find the real numbers  $\mathrm{x }\!\!\And\!\!\text{ y}$  if  $\left( \mathrm{x-iy} \right)\left( \mathrm{3+5i} \right)$  is the conjugate of  $\mathrm{-6-24i}$

Ans:

Let  $\mathrm{z=}\left( \mathrm{x-iy} \right)\left( \mathrm{3+5i} \right)$

\begin{align} &\mathrm{z=3x+5xi-3yi-5y}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{=3x+5xi-3yi+5y=(3x+5y)+i(5x-3y)} \\ & \mathrm{ }\!\!\!\!\text{ \bar{z}=(3x+5y)-i(5x-3y)} \\ \end{align}

It is given that,  $\overline{\mathrm{z}}\mathrm{=-6-24i}$

$\mathrm{ }\!\!\!\!\text{ (3x+5y)-i(5x-3y)=-6-24i}$

Equating real and imaginary parts, we obtain

$\begin{matrix} \mathrm{3x+5y=-6}\quad \mathrm{}..\mathrm{(i)} \\ \mathrm{5x-3y=24}...\mathrm{(ii)} \\ \end{matrix}$

On solving we will get

\begin{align} & \mathrm{3(3)+5y=-6} \\ & \mathrm{5y=-6-9=-15} \\ & \mathrm{y=-3} \\ \end{align}

Thus, the values of  $\mathrm{x and y are 3 and -3}$ respectively

9. Find the modulus of  $\dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{-}\dfrac{\mathrm{1-i}}{\mathrm{1+i}}$

Evaluate

Ans:

Expression

\begin{align} &\dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{-}\dfrac{\mathrm{1-i}}{\mathrm{1+i}}\mathrm{=}\dfrac{{{\mathrm{(1+i)}}^{\mathrm{2}}}\mathrm{-(1-i}{{\mathrm{)}}^{\mathrm{2}}}}{\mathrm{(1-i)(1+i)}} \\ &\mathrm{=}\dfrac{\mathrm{1+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2i-1-}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2i}}{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{\mathrm{4i}}{\mathrm{2}}\mathrm{=2i} \\ & \left| \dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{-}\dfrac{\mathrm{1-i}}{\mathrm{1+i}} \right|\mathrm{= }\!\!|\!\!\text{ 2i }\!\!|\!\!\text{ =}\sqrt{{{\mathrm{2}}^{\mathrm{2}}}}\mathrm{=2} \\ \end{align}

10. Find the modulus of  ${{\mathrm{(x+iy)}}^{\mathrm{3}}}\mathrm{=u+iv}$

Than show that    $\dfrac{\mathrm{u}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{v}}{\mathrm{y}}\mathrm{=4}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right)$

Ans:

\begin{align} & {{\mathrm{(x+iy)}}^{\mathrm{3}}}\mathrm{=u+iv} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{+(iy}{{\mathrm{)}}^{\mathrm{3}}}\mathrm{+3 }\!\!\times\!\!\text{ x }\!\!\times\!\!\text{ iy(x+iy)=u+iv} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{3}}}{{\mathrm{y}}^{\mathrm{3}}}\mathrm{+3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{yi+3x}{{\mathrm{y}}^{\mathrm{2}}}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{=u+iv} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{-i}{{\mathrm{y}}^{\mathrm{3}}}\mathrm{+3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{yi-3x}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{=u+iv} \\ & \mathrm{}\left( {{\mathrm{x}}^{\mathrm{3}}}\mathrm{-3x}{{\mathrm{y}}^{\mathrm{2}}} \right)\mathrm{+i}\left( \mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{y-}{{\mathrm{y}}^{\mathrm{3}}} \right)\mathrm{=u+iv} \\ \end{align}

On equating real and imaginary

\begin{align} &\mathrm{u=}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{-3x}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{,v=3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{y-}{{\mathrm{y}}^{\mathrm{3}}} \\ &\dfrac{\mathrm{u}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{v}}{\mathrm{y}}\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{3}}}\mathrm{-3x}{{\mathrm{y}}^{\mathrm{2}}}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{y-}{{\mathrm{y}}^{\mathrm{3}}}}{\mathrm{y}} \\ & \mathrm{=}\dfrac{\mathrm{x}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-3}{{\mathrm{y}}^{\mathrm{2}}} \right)}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{y}\left( \mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right)}{\mathrm{y}} \\ &\mathrm{=}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-3}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{+3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \\ &\mathrm{=4}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-4}{{\mathrm{y}}^{\mathrm{2}}} \\ & \mathrm{=4}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right) \\ &\dfrac{\mathrm{u}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{v}}{\mathrm{y}}\mathrm{=4}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right) \\ \end{align}

Hence, proved

11. If  $\mathrm{ }\!\!\alpha\!\!\text{ and }\!\!\beta\!\!\text{ }$  are different complex numbers with  $\left| \mathrm{ }\!\!\beta\!\!\text{ } \right|\mathrm{=1}$  , then find  $\left| \dfrac{\mathrm{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ }}{\mathrm{1-}\overline{\mathrm{ }\!\!\alpha\!\!\text{ }}\mathrm{ }\!\!\beta\!\!\text{ }} \right|\mathrm{=1}$

Ans:

Let  $\mathrm{ }\!\!\alpha\!\!\text{ =a+ib }\!\!\And\!\!\text{ }\!\!\beta\!\!\text{ =x+iy}$

It is given that,  $\left| \mathrm{ }\!\!\beta\!\!\text{ } \right|\mathrm{=1}$

\begin{align} &\sqrt{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}}}\mathrm{=1} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{=1}..\left| \dfrac{\mathrm{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ }}{\mathrm{1-\bar{ }\!\!\alpha\!\!\text{ }}} \right|\mathrm{=}\left| \dfrac{\mathrm{(x+iy)-(a+ib)}}{\mathrm{1-(a-ib)(x+iy)}} \right| \\ & \mathrm{=}\left| \dfrac{\mathrm{(x-a)+i(y-b)}}{\mathrm{1-(ax+aiy-ibx+by)}} \right| \\ & \mathrm{=}\left| \dfrac{\mathrm{(x-a)+i(y-b)}}{\mathrm{(1-ax-by)+i(bx-ay)}} \right| \\ & \mathrm{=}\left| \dfrac{\mathrm{(x-a)+i(y-b)}}{\mathrm{(1-ax-by)+i(bx-ay)}} \right| \\ \end{align}

\begin{align} & \mathrm{=}\dfrac{\sqrt{{{\mathrm{(x-a)}}^{\mathrm{2}}}\mathrm{+(y-b}{{\mathrm{)}}^{\mathrm{2}}}}}{\sqrt{{{\mathrm{(1-ax-by)}}^{\mathrm{2}}}\mathrm{+(bx-ay}{{\mathrm{)}}^{\mathrm{2}}}}} \\ &\mathrm{=}\dfrac{\sqrt{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}\mathrm{-2ax+}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-2by}}}{\sqrt{\mathrm{1+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{-2ax+2abxy-2by+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{-2abxy}}} \\ & \mathrm{=}\dfrac{\sqrt{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-2ax-2by}}}{\sqrt{\mathrm{1+}{{\mathrm{a}}^{\mathrm{2}}}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\left( {{\mathrm{y}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{x}}^{\mathrm{2}}} \right)\mathrm{-2ax-2by}}} \\ \end{align}

$\left| \dfrac{\mathrm{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ }}{\mathrm{1-}\overline{\mathrm{ }\!\!\alpha\!\!\text{ }}\mathrm{ }\!\!\beta\!\!\text{ }} \right|\mathrm{=1}$

12. Find the number of non-zero integral solutions of the equation  ${{\left| \mathrm{1-i} \right|}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}}$

Ans:

Equation

\begin{align} & \mathrm{ }\!\!|\!\!\text{ 1-i}{{\mathrm{ }\!\!|\!\!\text{ }}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\ & {{\left( \sqrt{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+(-1}{{\mathrm{)}}^{\mathrm{2}}}} \right)}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\ & {{\mathrm{(}\sqrt{\mathrm{2}}\mathrm{)}}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\ & {{\mathrm{2}}^{\mathrm{x/2}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\ & \dfrac{\mathrm{x}}{\mathrm{2}}\mathrm{=x} \\ & \mathrm{x=2x} \\ & \mathrm{x=0} \\ \end{align}

Thus,  $\mathrm{0}$ is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is  $\mathrm{0}$.

13. If  $\mathrm{(a+ib)(c+id)(e+if)(g+ih)=A+iB}$ Then show that $\left( {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}} \right)\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\left( {{\mathrm{e}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{f}}^{\mathrm{2}}} \right)\left( {{\mathrm{g}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{h}}^{\mathrm{2}}} \right)\mathrm{=}{{\mathrm{A}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{B}}^{\mathrm{2}}}$

Ans:

Expression

\begin{align} & \mathrm{(a+ib)(c+id)(e+if)(g+ih)=A+iB} \\ & \mathrm{ }\!\!\!\!\text{ }\!\!|\!\!\text{ (a+ib)(c+id)(e+if)(g+ih) }\!\!|\!\!\text{ = }\!\!|\!\!\text{ A+iB }\!\!|\!\!\text{ } \\ & \mathrm{ }\!\!|\!\!\text{ (a+ib) }\!\!|\!\!\text{ }\!\!\times\!\!\text{ }\!\!|\!\!\text{ (c+id) }\!\!|\!\!\text{ }\!\!\times\!\!\text{ }\!\!|\!\!\text{ (e+if) }\!\!|\!\!\text{ }\!\!\times\!\!\text{ }\!\!|\!\!\text{ (g+ih) }\!\!|\!\!\text{ = }\!\!|\!\!\text{ A+iB }\!\!|\!\!\text{ }\quad \mathrm{Q}\left[ \left| {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right|\mathrm{=}\left| {{\mathrm{z}}_{\mathrm{1}}} \right|\left| {{\mathrm{z}}_{\mathrm{2}}} \right| \right] \\ &\sqrt{{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{{{\mathrm{e}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{f}}^{\mathrm{2}}}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{{{\mathrm{g}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{h}}^{\mathrm{2}}}}\mathrm{=}\sqrt{{{\mathrm{A}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{B}}^{\mathrm{2}}}} \\ \end{align}

By squaring

$\left( {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}} \right)\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\left( {{\mathrm{e}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{f}}^{\mathrm{2}}} \right)\left( {{\mathrm{g}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{h}}^{\mathrm{2}}} \right)\mathrm{=}{{\mathrm{A}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{B}}^{\mathrm{2}}}$

Hence, proved

14. If  ${{\left( \dfrac{\mathrm{1+i}}{\mathrm{1-i}} \right)}^{\mathrm{m}}}\mathrm{=1}$

Then find the least positive integral value of  $m$

Ans

\begin{align} & {{\left( \dfrac{\mathrm{1+i}}{\mathrm{1-i}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ & {{\left( \dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{1+i}}{\mathrm{1+i}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ & {{\left( \dfrac{{{\mathrm{(1+i)}}^{\mathrm{2}}}}{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ & {{\left( \dfrac{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2i}}{\mathrm{2}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ \end{align}

\begin{align} & {{\left( \dfrac{\mathrm{1-1+2i}}{\mathrm{2}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ & {{\left( \dfrac{\mathrm{2i}}{\mathrm{2}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ & {{\mathrm{i}}^{\mathrm{m}}}\mathrm{=1} \\ & {{\mathrm{i}}^{\mathrm{m}}}\mathrm{=}{{\mathrm{i}}^{\mathrm{4k}}} \\ \end{align}

$\mathrm{m=4k}$  , where  $\mathrm{k}$  is some integer

Therefore, the least positive is one

Thus, the least positive integral value of  $\mathrm{m}$  is  $\mathrm{4=}\left( \mathrm{4 }\!\!\times\!\!\text{ 1} \right)$

## Overview of Deleted Syllabus for CBSE Class 11 Maths Complex Numbers and Quadratic Equations

 Chapter Dropped Topics Complex Numbers and Quadratic Equations 4.4.1 Representation of a complex number 4.6 Quadratic Equation Example 11, 13, 15, 16 Exercise 4.3 Miscellaneous Exercise - 5,8,9,13 Summary - last three points 4.7 Square - root of a Complex Number

## Class 11 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions EXERCISE 4.1 14 Questions and Solutions Miscellaneous Exercise 14 Questions and Solutions

## Conclusion

NCERT Class 11 Maths Chapter 4 Solutions for Class 11 on Complex Numbers and Quadratic Equations provided by Vedantu are comprehensive and designed to help students understand these key mathematical concepts. The solutions cover every exercise in the NCERT textbook, offering clear explanations and step-by-step methods for solving problems. Important points to focus on include understanding the fundamental properties of complex numbers, operations like addition, subtraction, multiplication, and division of complex numbers, and solving quadratic equations. These concepts are essential for mastering the topic and are often tested in exams. From previous year question papers, typically around 3-4 questions are asked from this chapter. These questions test students' grasp of theoretical concepts as well as their problem-solving skills.

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

1. What are the exercises and topics covered under NCERT Complex Number Class 11 Maths Chapter 5?

The Ch 5 Maths Class 11 NCERT Solutions consist of solved exercises that cover critical equations related to complex numbers and quadratic equations. These NCERT Solutions provide clarity on the theorems and concepts of Complex Numbers. There are three sets of exercises in this chapter for which the solutions are given in this PDF. Exercise 5.1 has 14 questions with answers and covers topics like finding the multiplicative inverse or expressing a set of numbers into complex numbers.

Next is Exercise 5.2 that comprises 8 questions and covers topics like finding the modulus and argument of a given set of numbers. Lastly, Exercise 5.3 comprises 10 questions and provides step by step solutions for solving various quadratic equations.

2. What is the marks distribution for Class 11 Maths?

The marking distribution for Class 11 Maths is discussed here. There are six units in which the mark weightage is distributed. First is Sets and Functions which have a marks weightage of 60 marks, the second unit is Algebra and has a marks weightage of 30.

Unit three covers the topic of Coordinate geometry and has a marks allocation of 10. Unit 4 covers Calculus and has a marks weightage of 30.

This is followed by unit 5 and unit 6 comprising mathematical reasoning, statistics, and probability with a mark’s distribution of 10 and 30 respectively. The total period under which all these chapters are divided is 240.

3. On which website can we get the most reliable Complex Number Class 11 NCERT Solutions?

The most reliable and accurate NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are available on Vedantu. Students do not need to pay any additional charges for downloading the solutions if they already have a registration. These solutions are prepared and compiled by subject-matter experts who have considerable years of experience in handling the CBSE syllabus and therefore offer the most advanced and detailed solutions to the exercises.

4. How can I ace Chapter 5 of Class 11 Maths?

You can easily ace Chapter - Complex Numbers and Quadratic Equations of Class 11 Maths. What you need is a strategy that you must follow consistently to get good grades. The best strategy is using NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations available on Vedantu. The NCERT Solutions PDF is very comprehensive and covers all the questions in detail. Practice all the exercises from the PDF to get full marks.

5. What are the important sub-topics that can come for exams from Class 11 Maths Chapter 5?

The important topics that need to be covered in NCERT Class 11 Maths Chapter 5 are -  Complex Numbers,

• Addition of Two Complex Numbers,

• Difference Between Two Complex Numbers,

• Multiplication and Division of Two Complex Numbers,

• Power of i (iota),

• Identities,

• Modulus and Conjugate of Complex Numbers,

• Argand Plane

Solve all the examples and exercise questions thoroughly to complete your preparation for your test.

6. What are some of the properties of the multiplication of complex numbers?

Some multiplication properties of complex numbers are closure law (the product of two complex numbers is also a complex number), commutative law (product of x1 and x2= product of x2 and x1), associative law (for complex numbers x1, x2,x3, (x1 x2) x3 = x1 (x2x3)), multiplicative identity (for every complex number x, x multiplied by 1 is 1), multiplicative identity (x.1/x is 1), and distributive law (x1(x2,x3)=  x1x2, x1x3).

7. How should I prepare for NCERT Class 11 Maths Chapter 5?

You can download the NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations from Vedantu and practice each and every example and solved exercises including the miscellaneous questions. The solutions to all questions are given step by step so that you understand everything very clearly. All the answers have been provided by experts who have curated precise content for you to practice. These solutions are available at free of cost on Vedantu’s website(vedantu.com) and mobile app.

8. What is an argand plane?

An argand plane or a complex plane is a graphical representation of complex numbers that are plotted along x-axis and y-axis. The x-axis in an argand plane is understood as the real axis and the y-axis in an argand plane is called an imaginary axis. To get answers to more such questions from this chapter, you can get access to the NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations available only on Vedantu.

9. What are complex numbers, and why are they important?

Complex numbers are numbers that comprise a real part and an imaginary part (in the form a + bi). They are essential in mathematics, especially in solving equations that have no real solutions, like quadratic equations with negative discriminants.

10. What is the conjugate of a complex number?

The conjugate of a complex number (a + bi) is (a - bi), where the sign of the imaginary part is changed. Conjugates are important in operations like division and in finding the modulus of complex numbers.

11. What is the modulus of a complex number?

The modulus (or absolute value) of a complex number (a + bi) is the distance of the complex number from the origin on the complex plane. It is denoted by |a + bi| and calculated as the square root of the sum of the squares of the real and imaginary parts: √(a² + b²).

12. What is the discriminant of a quadratic equation, and how does it determine the nature of roots?

The discriminant of a quadratic equation (Δ = b² - 4ac) determines the nature of its roots. If Δ > 0, the equation has two distinct real roots. If Δ = 0, it has one real root (a repeated root). If Δ < 0, it has two complex roots.

13. How do you multiply and divide complex numbers?

To multiply complex numbers, use the distributive property and remember that i2=−1i^2 = -1i2=−1. To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator to rationalize it.