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NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 4 - Complex Number and Quadratic Equations

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Last updated date: 14th Jul 2024
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NCERT Solutions for Class 11 Maths Chapter 4 Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 4 Complex Number And Quadratic Equations includes Solutions to all Miscellaneous Exercise problems. The Miscellaneous Exercise NCERT Solutions for Maths Class 11 are based on the ideas presented in Maths Chapter 4. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the CBSE Class 11 Maths Syllabus and practice them offline.

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Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 4 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Class 11 Maths Complex Number and Quadratic Equations Miscellaneous Exercise
3. Class 11 Maths Chapter 4: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 4 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs
Competitive Exams after 12th Science
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Access NCERT Class 11 Maths Complex Number and Quadratic Equations 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) $


Conclusion

NCERT Class 11 Maths Complex Numbers Miscellaneous Solutions is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing Solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.


Class 11 Maths Chapter 4: Exercises Breakdown

Exercise

Number of Questions

Exercise 4.1

14 Questions & Solutions


CBSE Class 11 Maths Chapter 4 Other Study Materials


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 for Class 11 Maths Miscellaneous Exercise Chapter 4 - Complex Number and Quadratic Equations

1. What are the polar forms of complex numbers in NCERT Solutions of Class 11 Maths Chapter 4 Miscellaneous Solutions?

A complex number can be expressed in polar form as z=r(cosθ+isinθ), where 𝑟 is the modulus and 𝜃 is the argument.. This form is particularly useful for operations like multiplication, division, and finding powers and roots of complex numbers, as it simplifies the calculations by using trigonometric functions.

2. What is De Moivre's Theorem in NCERT Solutions of Class 11 Maths Chapter 4 Miscellaneous Solutions?

In NCERT Solutions of Class 11 Maths Chapter 4 Miscellaneous Solutions, De Moivre's Theorem states that in finding the powers and roots of complex numbers. It simplifies calculations by converting the multiplication of complex numbers into trigonometric form, making it easier to compute higher powers and extract roots.

3. How should I approach the NCERT Solutions of Class 11 Maths Chapter 4 Miscellaneous Solutions?

The NCERT Class 11 Maths Complex Numbers Miscellaneous Solutions at the end of this chapter consolidates all the concepts learned. It includes a variety of problems, from basic to advanced, covering complex number operations, quadratic equations, and their applications. Practice diligently to build a strong foundation and enhance problem-solving skills.

4. What is the significance of the modulus of a complex number in NCERT Solutions of Class 11 Maths Ch 4 Miscellaneous Exercise?

The modulus of a complex number in NCERT Solutions of Class 11 Maths Ch 4 Miscellaneous Exercise represents the distance of the complex number from the origin in the complex plane. This distance is a measure of the magnitude of the complex number and is always a non-negative real number. The modulus is crucial in comparing the sizes of complex numbers and is used in various operations, such as finding the argument and converting it to polar form.

5. What is the purpose of the Class 11 Maths Ch 4 Miscellaneous Exercise?

The Class 11 Maths Ch 4 Miscellaneous Exercise is designed to provide a comprehensive review of all the concepts covered in the chapter. It helps students apply their knowledge in various situations and enhances their problem-solving skills by presenting different questions.

6. How can I approach solving problems in the Complex Numbers Class 11 Miscellaneous Exercise?

To solve problems in the Complex Numbers Class 11 Miscellaneous Exercise:

  • Start by carefully reading the problem to understand what is being asked.

  • Identify the concepts and methods from the chapter that can be applied.

  • Break down the problem into smaller, manageable parts if needed.

  • Solve each part step-by-step and combine the results to get the final answer.

  • Check your solution to ensure it is logical and correct.

7. What should I do if I get stuck on a problem in the Complex Numbers Class 11 Miscellaneous Exercise?

If you get stuck on a problem Complex Numbers Class 11 Miscellaneous Exercise:

  • Re-read the problem to make sure you understand it correctly.

  • Review the relevant concepts and examples from the chapter.

  • Try to simplify the problem or approach it from a different angle.

  • Discuss the problem with classmates or ask your teacher for help.

  • Practice similar issues to build your confidence and skills.

9. How does practising the Class 11 Complex Numbers Miscellaneous Solutions help in exams?

Practising the Class 11 Complex Numbers Miscellaneous Solutionshelps in exams by:

  • Reinforcing your understanding of the chapter's concepts.

  • Improving your ability to tackle different types of questions.

  • Enhancing your problem-solving speed and accuracy.

  • Building your confidence to handle complex problems under exam conditions.

  • Providing a thorough revision of the chapter.

10. Are the solutions provided in the NCERT Class 11 Complex Numbers Miscellaneous Solutions easy to understand?

Yes, the solutions provided in the Class 11 Maths Complex Numbers Miscellaneous Solutions are designed to be easy to understand. They offer clear, step-by-step explanations and are written in simple language to help students grasp the concepts easily. These solutions often include tips and strategies to make problem-solving more efficient.