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

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NCERT Solutions for Class 11 Maths Chapter 4 - FREE PDF Download

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers 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.

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Access Exercise wise NCERT Solutions for Chapter 4 Maths Class 11

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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 $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

We get the final answer

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} $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

We get the final answer

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 $ 

Here we got final answer

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} $  

Here we got final answer

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} $  

Here we got final answer

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} $  

Here we got final answer

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} $  

Here we get the answer

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.


Other Study Material for CBSE Class  11 Maths Chapter 4



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.



Important Related Links for CBSE Class 11 Maths

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

1. What key topics do the NCERT Solutions for Class 11 Maths Chapter 4 cover for the 2025-26 session?

The NCERT Solutions for Class 11 Maths Chapter 4, updated for the CBSE 2025-26 syllabus, provide step-by-step methods for the following topics:

  • The concept of the imaginary unit (iota, i) and its powers.
  • Algebraic operations on complex numbers: addition, subtraction, multiplication, and division.
  • Finding the modulus and conjugate of a complex number.
  • Calculating the multiplicative inverse.
  • Solving quadratic equations that have complex roots (where the discriminant is negative).
  • Expressing all solutions in the standard form a + ib.

2. How do the NCERT Solutions explain the method for solving quadratic equations with complex roots?

The NCERT Solutions demonstrate a clear, step-by-step method. When a quadratic equation's discriminant (D = b² - 4ac) is negative, the solutions show how to use the quadratic formula by expressing the square root of the negative discriminant in terms of 'i'. For instance, √(-D) is written as i√(D). This allows you to find the two complex roots and present them in the standard a + ib form, as required by the CBSE pattern.

3. What is the correct method used in the NCERT Solutions to find the multiplicative inverse of a complex number?

The NCERT Solutions use a standard formula to find the multiplicative inverse of a complex number z = a + ib. The inverse, denoted as z⁻¹, is calculated using its conjugate (z̄ = a - ib) and its modulus (|z| = √(a² + b²)). The step-by-step method shown is: z⁻¹ = z̄ / |z|², which simplifies to (a - ib) / (a² + b²). This ensures the denominator becomes a real number, and the final answer is presented in the a + ib form.

4. Why do the NCERT Solutions frequently use the conjugate when dividing complex numbers?

The conjugate is a critical tool used in the solutions to rationalise the denominator. When dividing one complex number by another, the denominator is a complex number. To convert it into a real number, the solutions show how to multiply both the numerator and the denominator by the conjugate of the denominator. This process, based on the property (a + ib)(a - ib) = a² + b², simplifies the expression into the standard a + ib form, making the final answer clear and correct.

5. How should a student use these NCERT Solutions to approach the Miscellaneous Exercise for Chapter 4?

The Miscellaneous Exercise contains more challenging, application-based problems. The NCERT Solutions for this section are designed to help you break down complex questions. It's best to first attempt the problems yourself, then use the solutions to:

  • Understand the sequence of steps required for multi-step problems.
  • Identify which properties of complex numbers (like modulus, conjugate, or distributive law) are needed.
  • Verify if your final answer, expressed in the a + ib form, is correct.

6. What is the principle behind simplifying high powers of 'i' (iota) as shown in the exercise solutions?

The solutions demonstrate that any power of 'i' can be simplified by using its cyclic pattern of four values: i, -1, -i, and 1. The core principle shown is to express any high power of 'i', say iⁿ, in the form of (i⁴)ᑫ · iʳ, where 'n' is divided by 4 to get a quotient 'q' and a remainder 'r'. Since i⁴ = 1, the expression simplifies to 1ᑫ · iʳ = iʳ. The answer is then determined by the remainder (0, 1, 2, or 3), which is a method consistently applied across the solutions.

7. Which topics and questions from Chapter 4 have been deleted for the CBSE 2025-26 exams?

Yes, as per the latest CBSE syllabus for 2025-26, students should skip the following topics and corresponding questions while using the NCERT book and its solutions:

  • Polar representation of complex numbers and the concept of argument.
  • The square root of a complex number.
  • All questions in Exercise 4.3.
  • From the Miscellaneous Exercise, questions 5, 8, 9, and 13 are excluded.

The provided NCERT solutions focus only on the topics included in the current syllabus.

8. How do the step-by-step answers in the NCERT solutions help prevent common mistakes in this chapter?

The detailed solutions help prevent common errors by reinforcing correct procedures. For example, they show how to:

  • Consistently substitute i² = -1 during multiplication to avoid sign errors.
  • Always combine real parts with real parts and imaginary parts with imaginary parts separately.
  • Properly apply the formula for the multiplicative inverse instead of just taking the reciprocal.
  • Ensure the final answer is always simplified to the standard a + ib form, a common requirement in exams.

9. What is the importance of expressing the final answer in the standard 'a + ib' form, as shown in all NCERT solutions?

Expressing the answer in the standard form a + ib is crucial because it clearly distinguishes the real part (a) from the imaginary part (b). This format is a mandatory requirement in CBSE examinations for full marks. The NCERT solutions consistently follow this convention to train students to present their answers in a clear, standardized, and accurate manner, which is essential for both understanding and scoring well.

10. What are the key properties of multiplication of complex numbers demonstrated in the NCERT Solutions?

The solutions implicitly or explicitly demonstrate several key properties of complex number multiplication through solved examples. These include:

  • Closure Law: The product of two complex numbers is always another complex number.
  • Commutative Law: For any two complex numbers z₁ and z₂, z₁z₂ = z₂z₁.
  • Associative Law: For any three complex numbers, (z₁z₂)z₃ = z₁(z₂z₃).
  • Distributive Law: Multiplication distributes over addition, i.e., z₁(z₂ + z₃) = z₁z₂ + z₁z₃.

Following the solution steps helps in understanding how these properties are applied.