NCERT Solutions for Class 11 Maths Chapter 4 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.

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Chapter-Specific NCERT Solutions for Class 11 Maths

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