How Can Complex Numbers And Quadratic Equations Class 11 Questions And Answers Help You Prepare For Exams ?
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.


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 |
2 | NCERT Solutions of Class 11 Maths Complex Numbers and Quadratic Equations Miscellaneous Exercise |
How Can Complex Numbers And Quadratic Equations Class 11 Questions And Answers Help You Prepare For Exams With CBSE Aligned Solutions
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.
S. No | NCERT Solutions Class 11 Maths All Chapters |
1 | |
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 | |
9 | |
10 | Chapter 11 - Introduction to Three Dimensional Geometry Solutions |
11 | |
12 | |
13 |
Important Related Links for CBSE Class 11 Maths
S.No. | Important Study Material for Maths Class 11 |
1 | |
2 | |
3 | |
4 | |
5 | |
6 | |
7 | |
8 | |
9 | |
10 |
FAQs on NCERT Solutions for Class 11 Maths Chapter Chapter 4 Complex Numbers And Quadratic Equations
1. What are NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations?
NCERT Solutions for Class 11 Maths Chapter 4 are comprehensive step-by-step answers to all questions covering complex numbers, their operations, and quadratic equations with complex roots.
These solutions help students understand the algebraic structure of complex numbers and their applications in solving quadratic equations that have no real solutions.
2. How can students access the free PDF of Class 11 Maths Chapter 4 solutions?
Students can download the free PDF of Class 11 Maths Chapter 4 solutions from Vedantu's website, which provides offline access to all solved exercises and examples.
Having offline access ensures uninterrupted study sessions and allows students to practice problems even without internet connectivity.
3. What is the standard form of a complex number in Class 11?
The standard form of a complex number is z = a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit with i² = -1.
Standard form allows easy identification of real and imaginary components, essential for performing arithmetic operations and solving equations.
4. How do you add and subtract complex numbers?
To add or subtract complex numbers, combine the real parts separately and the imaginary parts separately using the formula (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
5. What is the conjugate of a complex number and its properties?
The conjugate of a complex number z = a + bi is z̄ = a - bi, obtained by changing the sign of the imaginary part.
Complex conjugates are crucial for division operations and finding modulus, as z × z̄ always gives a real number equal to |z|².
6. How do you multiply complex numbers step by step?
To multiply complex numbers, use the distributive property (a + bi)(c + di) = ac + adi + bci + bdi², then substitute i² = -1 to get (ac - bd) + (ad + bc)i.
Steps: Multiply each term in the first complex number by each term in the second Combine like terms. Replace i² with -1. Write the result in standard form a + bi.
7. What is the process for dividing complex numbers?
To divide complex numbers, multiply both numerator and denominator by the conjugate of the denominator, then simplify to get a real denominator.
Division by complex numbers requires rationalization to eliminate imaginary terms from the denominator, similar to rationalizing radical expressions.
8. How are quadratic equations with complex roots solved?
Quadratic equations with complex roots are solved using the quadratic formula x = [-b ± √(b² - 4ac)]/2a, where a negative discriminant produces complex solutions.
When the discriminant (b² - 4ac) is negative, the equation has no real solutions but has two complex conjugate roots.
9. What are the key formulas covered in NCERT Solutions for complex numbers?
The key formulas include i² = -1, |z| = √(a² + b²) for modulus, z̄ = a - bi for conjugate, and z × z̄ = |z|² for the product of complex conjugates.
Formula list: Powers of i: i¹ = i, i² = -1, i³ = -i, i⁴ = 1
Argument: arg(z) = tan⁻¹(b/a)
Polar form: z = r(cos θ + i sin θ)

















