NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations in Hindi

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NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations in Hindi PDF Download

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Access NCERT Solutions for Class 11 Mathematics Chapter 5 – समिश्र संख्या और द्विघातीय

प्रश्नवाली 5.1 

1. $\left( \text{5i} \right)\left( \text{ - }\dfrac{\text{3}}{\text{5}}\text{i} \right)$

उत्तर: 

हमे प्राप्त हैं, $\left( \text{5i} \right)$ $\left( \dfrac{\text{- 3}}{\text{ 5}}\text{i} \right)\text{ = 5}\left( \dfrac{\text{-3}}{\text{5}} \right)\text{  }\!\!\times\!\!\text{  i  }\!\!\times\!\!\text{  i = - 3}{{\text{i}}^{^{\text{2}}}}$

$\text{=  - 3  }\!\!\times\!\!\text{  }\left( \text{ - 1} \right)$

$\left[ \because \text{, }{{\text{i}}^{\text{2}}}\text{ = - 1} \right]$

$\text{= 3 = 3 + i0}$

2. ${{\text{i}}^{\text{9}}}\text{+}{{\text{i}}^{\text{19}}}$

उत्तर:  हमे प्राप्त हैं, ${{\text{i}}^{\text{9}}}\text{ + }{{\text{i}}^{\text{19}}}$ $\text{= }{{\left( {{\text{i}}^{\text{4}}} \right)}^{\text{2}}}\text{.i + }{{\left( {{\text{i}}^{\text{4}}} \right)}^{\text{4}}}\text{.}{{\text{i}}^{\text{2}}}\text{.i}$

${{\left( \text{1} \right)}^{\text{2}}}\text{.i + }{{\left( \text{1} \right)}^{\text{4}}}\text{.}\left( \text{ - 1} \right)\text{.i}$

$\left[ \because \text{,}{{\text{i}}^{\text{2}}}\text{ = - 1, }{{\text{i}}^{\text{4}}}\text{ = 1} \right]$

$\text{= i - i = 0 + i0}$

3. $\text{ i}{{\text{ }}^{\text{-39}}}$

उत्तर: हमें प्राप्त हैं, $\text{ i}{{\text{ }}^{\text{-39}}}$ $\text{= }\dfrac{\text{1}}{{{\text{i}}^{\text{39}}}}\text{ = }\dfrac{\text{i}}{{{\text{i}}^{\text{40}}}}\text{ = }\dfrac{\text{i}}{{{\left( {{\text{i}}^{\text{4}}} \right)}^{\text{10}}}}\text{ = i = 0 + i1}$          

$\left[ \because \text{,}{{\text{i}}^{\text{2}}}\text{ = -1} \right]$

4. $\text{3}\left( \text{7 + i7} \right)\text{ + i}\left( \text{7 + i7} \right)$ 

उत्तर:  हमें प्राप्त हैं,  $\text{3}\left( \text{7 + i7} \right)\text{ + i}\left( \text{7 + i7} \right)$

$\text{= 21 + i}\text{.21 + i}\text{.7 + }{{\text{i}}^{\text{2}}}\text{.7}$

$\text{= 21 + i}\text{.28 + }\left( \text{-1} \right)\text{.7}$

$\left[ \because \text{,}{{\text{i}}^{\text{2}}}\text{ = -1} \right]$

$\text{= 21 + 28i - 7}$

$\text{= 14 + 28i}$

5. $\left( 1-i \right)-\left( 1+i6 \right)$

उत्तर: हमें प्राप्त हैं, $\left( \text{1 - i} \right)\text{ - }\left( \text{ - 1 + i6} \right)\text{ = 1-i+1+ 6i = 2 - 7i}$

6. $\left( \dfrac{\text{1}}{\text{5}}\text{ + i}\dfrac{\text{2}}{\text{5}} \right)\text{ - }\left( \text{4 + i}\dfrac{\text{5}}{\text{2}} \right)$

उत्तर: हमें प्राप्त हैं,$\left( \dfrac{\text{1}}{\text{5}}\text{ + i}\dfrac{\text{2}}{\text{5}} \right)\text{ - }\left( \text{4 + i}\dfrac{\text{5}}{\text{2}} \right)\text{ = }\dfrac{\text{1}}{\text{5}}\text{ + i}\dfrac{\text{2}}{\text{5}}\text{ - 4 - i}\dfrac{\text{5}}{\text{2}}$ 

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

$\text{= }\left( \dfrac{\text{1 - 20}}{\text{5}} \right)\text{ + i}\left( \dfrac{\text{4 - 25}}{\text{10}} \right)$ 

$\text{= - }\dfrac{\text{19}}{\text{5}}\text{ - i}\dfrac{\text{21}}{\text{10}}$

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

उत्तर:  हमें प्राप्त हैं,

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

$\text{       = }\dfrac{\text{1}}{\text{3}}\text{ + i}\dfrac{\text{7}}{\text{3}}\text{ + 4 + i}\dfrac{\text{1}}{\text{3}}\text{ +}\dfrac{\text{4}}{\text{3}}\text{ - i}$

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

 $\text{ = }\left( \dfrac{\text{1 + 12 + 4}}{\text{3}} \right)\text{ + i}\left( \dfrac{\text{7 + 1 - 3}}{\text{3}} \right)$

$\text{          = }\dfrac{\text{17}}{\text{3}}\text{ + i}\dfrac{\text{5}}{\text{3}}$

8. ${{\left( \text{1 - i} \right)}^{\text{4}}}$ 

उत्तर: हमें प्राप्त हैं,

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

$\left[ \because \text{, }{{\text{i}}^{\text{2}}}\text{ = - 1} \right]$ 

$\text{ = }{{\left( \text{ - 2i} \right)}^{\text{2}}}$ 

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

$\text{= - 4 + i0}$ 

9. ${{\left( \dfrac{\text{1}}{\text{3}}\text{ + 3i} \right)}^{\text{3}}}$

उत्तर: हमें प्राप्त हैं,

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

$\text{ = }\dfrac{\text{1}}{\text{27}}\text{ + 27}{{\text{i}}^{\text{2}}}\text{.i + i + 9}{{\text{i}}^{\text{2}}}$ 

$\text{ = }\dfrac{\text{1}}{\text{27}}\text{ - 27i + i - 9}$

$\left[ \because \text{, }{{\text{i}}^{\text{2}}}\text{ = - 1} \right]$

$\text{= }\left( \dfrac{\text{1}}{\text{27}}\text{ - 9} \right)\text{ - 26i}$ 

$\text{= }\left( \dfrac{\text{1 - 243}}{\text{27}} \right)\text{ - 26i}$

$\text{ = - }\dfrac{\text{242}}{\text{27}}\text{ - 26i}$

10.  ${{\left( \text{- 2 - i}\dfrac{\text{1}}{\text{3}} \right)}^{\text{3}}}$

उत्तर:  हमें प्राप्त हैं,

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

$\text{= - 8 - }\dfrac{\text{1}}{\text{27}}{{\text{i}}^{\text{3}}}\text{ + 3  }\!\!\times\!\!\text{  4  }\!\!\times\!\!\text{  }\left( \text{ - }\dfrac{\text{1}}{\text{3}}\text{i} \right)\text{ + 3  }\!\!\times\!\!\text{  ( - 2)  }\!\!\times\!\!\text{  }\dfrac{\text{1}}{\text{9}}{{\text{i}}^{\text{2}}}\text{ }$ 

$\text{                            = - 8 - }\dfrac{\text{1}}{\text{27}}\text{ }{{\text{i}}^{\text{2}}}\text{  }\!\!\times\!\!\text{  i - 4 i - }\dfrac{\text{2}}{\text{3}}\text{ }{{\text{i}}^{\text{2}}}\text{ }$ 

$\text{                            = - 8 + }\dfrac{\text{1}}{\text{27}}\text{i - 4i + }\dfrac{\text{2}}{\text{3}}$ 

$\text{ }\!\![\!\!\text{ }\because \text{, }\left. {{\text{i}}^{\text{2}}}\text{ = - 1} \right]$ 

$\text{                            = }\left( \text{- 8 + }\dfrac{\text{2}}{\text{3}} \right)\text{ + i}\left( \dfrac{\text{1}}{\text{27}}\text{ - 47} \right)$ 

$\text{                            = }\left( \dfrac{\text{ - 24 + 2}}{\text{3}} \right)\text{ + i}\left( \dfrac{\text{1 - 108}}{\text{27}} \right)$ 

$\text{                            = - }\dfrac{\text{22}}{\text{3}}\text{ - }\dfrac{\text{107}}{\text{27}}\text{i}$

प्रश्न 11 से13 की सम्मिश्र संख्याओ में प्रत्येक का गुणात्मक प्रतिलोम ज्ञात कीजिए

11.   $\text{4 - 3i}$

उत्तर:  मान लिया, $\text{z = 4 - 3i}$

तब,${\bar{Z} = 4 + 3i}$ और $\text{ }\!\!|\!\!\text{ Z  }\!\!|\!\!\text{ = }\sqrt{{{\text{4}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}}}\text{ = }\sqrt{\text{16+9}}$ 

$\text{           = }\sqrt{\text{25}}\text{ = 5}$ 

इसलिए ,$\text{z = 4 - 3i}$ का गुणात्मक प्रतिलोम :

${{\text{Z}}^{\text{-1}}}\text{ = }\dfrac{{{\bar{z}}}}{\text{ }\!\!|\!\!\text{ z}{{\text{ }\!\!|\!\!\text{ }}^{\text{2}}}}\text{ = }\dfrac{\text{4 + 3i}}{{{\text{5}}^{\text{2}}}}\text{ = }\dfrac{\text{4 + 3i}}{\text{25}}\text{ = }\dfrac{\text{4}}{\text{25}}\text{ + i}\dfrac{\text{3}}{\text{25}}$

ऊपर दिया गया सारा हल निम्नलिखत ढंग से दिखाया जा सकता हैं 

${{\text{Z}}^{\text{-1}}}\text{ = }\dfrac{\text{1}}{\text{4 - 3i}}\text{ = }\dfrac{\text{(4 + 3i)}}{\text{(4 - 3i)(4 + 3i)}}\text{ = }\dfrac{\text{(4 + 3i)}}{{{\text{4}}^{\text{2}}}\text{ - (3i}{{\text{)}}^{\text{2}}}}\text{ = }\dfrac{\text{(4 + 3i)}}{\text{16  + 9}}\text{ =} $

$\dfrac{\text{(4 + 3i)}}{\text{25}}\text{ = }\dfrac{\text{4}}{\text{25}}\text{ + i}\dfrac{\text{3}}{\text{25}}$

12. $\sqrt{\text{5}}\text{ + 3i}$ 

उत्तर:  मान लिया ,$\text{z  = }\sqrt{\text{5}}\text{ + 3i}$

तब ,${\bar{Z} = }\sqrt{\text{5}}\text{ - 3i}$ और  $\text{ }\!\!|\!\!\text{ Z }\!\!|\!\!\text{  = }\sqrt{{{\sqrt{\text{5}}}^{\text{2}}}\text{ + }{{\text{3}}^{\text{2}}}}\text{ = }\sqrt{\text{5 + 9}}\text{ = }\sqrt{\text{14}}$

इसलिए ,$\text{z = }\sqrt{\text{5}}\text{ + 3i}$ का गुणात्मक प्रतिलोम :

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

ऊपर दिया गया सारा हल निम्नलिखत ढंग से दिखाया जा सकता हैं 

${{\text{Z}}^{\text{-1}}}\text{ = }\dfrac{\text{1}}{\sqrt{\text{5}}\text{ + 3i}}\text{ = }\dfrac{\sqrt{\text{5 }}\text{- 3i}}{\text{(}\sqrt{\text{5}}\text{ + 3i)(}\sqrt{\text{5 }}\text{- 3i)}}\text{ = }\dfrac{\sqrt{\text{5 }}\text{- 3i}}{{{\sqrt{\text{5}}}^{\text{2}}}\text{ - (3i}{{\text{)}}^{\text{2}}}}\text{ = }\dfrac{\text{(}\sqrt{\text{5 }}\text{- 3i)}}{\text{5 + 9}}$

$\text{= }\dfrac{\text{(}\sqrt{\text{5}}\text{ - 3i)}}{\text{14}}\text{ = }\dfrac{\sqrt{\text{5}}}{\text{14}}\text{ - i}\dfrac{\text{3}}{\text{14}}$

13. $\text{- i}$

उत्तर: मान लिया, $\text{z = 0 - i}$

तब, ${\bar{Z} = 0 + i}$ और $|Z|=\sqrt{{{0}^{2}}+{{1}^{2}}}=\sqrt{0+1}=1$

इसलिए , $\text{z = 0 - i}$ का गुणात्मक प्रतिलोम :

${{\text{Z}}^{\text{-1}}}\text{ = }\dfrac{{{\bar{z}}}}{\text{ }\!\!|\!\!\text{ z}{{\text{ }\!\!|\!\!\text{ }}^{\text{2}}}}\text{ = }\dfrac{\text{0 + i}}{{{\text{1}}^{\text{2}}}}\text{ = }\dfrac{\text{0 + i}}{\text{1}}\text{ = }\dfrac{\text{0}}{\text{1}}\text{ + }\dfrac{\text{i}}{\text{1}}\text{ = i}$  

ऊपर दिया गया सारा हल निम्नलिखत ढंग से दिखाया जा सकता हैं 

${{\text{Z}}^{\text{-1}}}\text{ = }\dfrac{\text{1}}{\text{0 - i}}\text{ = }\dfrac{\text{0 + i}}{\text{(0 - i)(0 + i)}}\text{ = }\dfrac{\text{0 + i}}{{{\text{0}}^{\text{2}}}\text{ - (i}{{\text{)}}^{\text{2}}}}\text{ = }\dfrac{\text{0 + i}}{\text{0 + 1}}\text{ = }\dfrac{\text{0 + l}}{\text{1}}\text{ = }\dfrac{\text{0}}{\text{1}}\text{ + }\dfrac{\text{l}}{\text{1}}\text{ = i}$

14. निम्नलिखत व्यंजक को $\text{a  + ib }$ के रूप मे व्यक्त कीजिए ।  

$\dfrac{\text{(3 + i}\sqrt{\text{5}}\text{) (3 - i}\sqrt{\text{5}}\text{)}}{\text{(}\sqrt{\text{3}}\text{ + i}\sqrt{\text{2}}\text{) - (}\sqrt{\text{3 }}\text{- i}\sqrt{\text{2}}\text{)}}$ 

उत्तर: हमें प्राप्त हैं,

$\dfrac{\text{(3 + i}\sqrt{\text{5}}\text{)(3 - i}\sqrt{\text{5}}\text{)}}{\text{(}\sqrt{\text{3}}\text{ + i }\sqrt{\text{2}}\text{) - (}\sqrt{\text{3}}\text{ - i}\sqrt{\text{2}}\text{)}}\quad $ 

$\text{           =}\dfrac{{{\text{3}}^{\text{2}}}\text{ - (i}\sqrt{\text{5}}{{\text{)}}^{\text{2}}}}{\sqrt{\text{3}}\text{ + i}\sqrt{\text{2 - }\sqrt{\text{3}}\text{ + i}\sqrt{\text{2}}}}\quad $ 

$\text{           = }\dfrac{\text{9 - 5}{{\text{i}}^{\text{2}}}}{\text{2}\sqrt{\text{2}}\text{i}}\quad \left[ \text{ }\because \text{,  }{{\text{i}}^{\text{2}}}\text{ = - 1} \right]\quad$ 

$\text{           = }\dfrac{\text{9 + 5}}{\text{2}\sqrt{\text{2}}\text{i}}$ 

$\text{           = }\dfrac{\text{14}}{\text{2}\sqrt{\text{2}}\text{i}}\text{ = }\dfrac{\text{7}}{\sqrt{\text{2}}\text{i}}\text{ = }\dfrac{\text{7}}{\sqrt{\text{2}}\text{i}}\text{  }\!\!\times\!\!\text{  }\dfrac{\sqrt{\text{2}}\text{i}}{\sqrt{\text{2}}\text{i}}\text{ = i}\dfrac{\text{7}\sqrt{\text{2}}}{\text{2}{{\text{i}}^{\text{2}}}}\text{ = i}\dfrac{\text{7}\sqrt{\text{2}}}{\text{2(-1)}}$ 

$\text{= - i}\dfrac{\text{7}\sqrt{\text{2}}}{\text{2}}\text{ = 0 - i}\dfrac{\text{7}\sqrt{\text{2}}}{\text{2}}$

प्रश्नावली 5.2

प्रश्न 1 से 2 तक सम्मिश्र संख्याओं में प्रत्येक का मापांक और कोणांकज्ञात कीजिये 

1. $\text{z = - 1 - i}\sqrt{\text{3}}$ 

उत्तर: मान लीजिए कि , 

$\text{z = - 1 - i}\sqrt{\text{3 }}\text{= r(cos }\!\!\theta\!\!\text{  + isin }\!\!\theta\!\!\text{ ) }$ 

$\text{r cos }\!\!\theta\!\!\text{  = - 1, r sin }\!\!\theta\!\!\text{  = - }\sqrt{\text{3}}$ 

वर्ग करके जोड़ने पर,

${{r}^{2}}=1+3=4$ 

$r=2$ 

$\cos \theta =-\dfrac{1}{\sqrt{2}}=-\cos \dfrac{\pi }{3}$ 

अतः 

$\text{ }\!\!\theta\!\!\text{  =  }\!\!\pi\!\!\text{  + }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{ = }\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}$ 

$\text{ }\!\!\theta\!\!\text{  = }\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\text{ - 2 }\!\!\pi\!\!\text{  = - }\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$ 

$\text{ }\!\!\theta\!\!\text{  = }\dfrac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\text{ - 2 }\!\!\pi\!\!\text{  = - }\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$

कोंणांक$\text{= - }\dfrac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$ और मापांक $\text{= 2}$

2.  $\text{z = - }\sqrt{\text{3}}\text{ + i}$

उत्तर:  मान लीजिए कि ,$\text{z = - }\sqrt{\text{3}}\text{ + i = r(cos }\!\!\theta\!\!\text{  + isin }\!\!\theta\!\!\text{ )}$

$\text{r cos }\!\!\theta\!\!\text{  = - }\sqrt{\text{3}}\text{, r sin }\!\!\theta\!\!\text{  = 1}$ 

वर्ग करके जोड़ने पर

${{\text{r}}^{\text{2}}}\text{ = 3 + 1 = 4}$ 

$\text{r = 2}$ 

$\dfrac{\text{r sin }\!\!\theta\!\!\text{ }}{\text{r cos }\!\!\theta\!\!\text{ }}\text{ = }\dfrac{\text{1}}{\text{- }\sqrt{\text{3}}}\text{ = - }\dfrac{\text{1}}{\sqrt{\text{3}}}$ 

$\text{- }\dfrac{\text{1}}{\sqrt{\text{3}}}\text{ = tan}\left( \text{ }\!\!\pi\!\!\text{  - }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$ 

$\text{= tan}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$ 

$\text{ }\!\!\theta\!\!\text{  = }\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

प्रश्न 3 से 8 तक सम्मिश्र संख्याओं में प्रत्येक को ध्रुवीय रूप में रूपांतररत किजीये:

3.  $\text{1 - i}$

उत्तर:  मान लीजिए कि,$\text{z = 1 - i = r}\left( \text{cos }\!\!\theta\!\!\text{  + i sin }\!\!\theta\!\!\text{ } \right)$ 

$\text{r cos }\!\!\theta\!\!\text{  = 1, r sin }\!\!\theta\!\!\text{  = - 1}$

वर्ग करके जोड़ने पर,

${{\text{r}}^{\text{2}}}\text{= 1+1 = 2}$ 

$\text{r = }\sqrt{\text{2}}$ 

$\text{tan }\!\!\theta\!\!\text{  = }\dfrac{\text{- 1}}{\text{1}}\text{ = - 1}$ 

$\text{= tan}\left( \text{2 }\!\!\pi\!\!\text{  - }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

$\text{= tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

$\text{ }\!\!\theta\!\!\text{  = - }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ 

$\text{z = }\sqrt{\text{2}}\left( \text{cos}\dfrac{\text{-  }\!\!\pi\!\!\text{ }}{\text{4}}\text{ + i sin}\dfrac{\text{-  }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

4.  $\text{z = - 1 + i}$

उत्तर:  मान लीजिए कि, $\text{z = - 1 + i = r}\left( \text{cos }\!\!\theta\!\!\text{  + i sin }\!\!\theta\!\!\text{ } \right)$

$\text{r cos }\!\!\theta\!\!\text{  = - 1, r sin }\!\!\theta\!\!\text{  = 1}$

वर्ग करके जोड़ने पर,

${{\text{r}}^{\text{2}}}\text{ = 1 + 1 = 2}$ 

$\text{r =}\sqrt{\text{2}}$ 

$\dfrac{\text{r sin }\!\!\theta\!\!\text{ }}{\text{r cos }\!\!\theta\!\!\text{ }}\text{ = tan }\!\!\theta\!\!\text{  = - 1}$ 

$\text{= tan}\left( \text{ }\!\!\pi\!\!\text{  - }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

$\text{= tan}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$ 

$\text{z = }\sqrt{\text{2}}\left( \text{cos}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{ + i sin}\dfrac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

5. $\text{- 1 - i}$

उत्तर:  मान लीजिए कि, $\text{z = - 1 - i = r}\left( \text{cos }\!\!\theta\!\!\text{  + i sin }\!\!\theta\!\!\text{ } \right)$

$\text{r cos }\!\!\theta\!\!\text{  = - 1,}\ \text{r sin }\!\!\theta\!\!\text{  = - 1}$

वर्ग करके जोड़ने पर

$\text{z = }\sqrt{\text{2}}\left( \text{cos }\dfrac{\text{- 3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{ + i sin}\dfrac{\text{ - 3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$         ${{\text{r}}^{\text{2}}}\text{ = (- 1}{{\text{)}}^{\text{2}}}\text{ + (- 1}{{\text{)}}^{\text{2}}}\text{ = 2}$ 

$\text{r = }\sqrt{\text{2 }}$ 

$\dfrac{\text{r sin  }\!\!\theta\!\!\text{ }}{\text{r cos  }\!\!\theta\!\!\text{ }}\text{ = tan  }\!\!\theta\!\!\text{  = }\dfrac{\text{- 1}}{\text{- 1}}\text{ = 1}$ 

$\text{= tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ 

$\text{= tan}\left( \text{ }\!\!\pi\!\!\text{  + }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

$\text{= tan}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{4}}\text{ = tan}\left( \dfrac{\text{- 3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$ 

$\text{ }\!\!\theta\!\!\text{  = }\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{4}}$ या $\dfrac{\text{- 3 }\!\!\pi\!\!\text{ }}{\text{4}}$

$\text{z = }\sqrt{\text{2}}\left( \text{cos}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{4}}\text{ + i sin}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

या

$\text{z = }\sqrt{\text{2}}\left( \text{cos }\dfrac{\text{- 3 }\!\!\pi\!\!\text{ }}{\text{4}}\text{ + i sin}\dfrac{\text{ - 3 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

6. $\text{z = - 3 = r(cos  }\!\!\theta\!\!\text{  + i sin  }\!\!\theta\!\!\text{ )}$

उत्तर:  मान लीजिए कि,$\text{z = - 3 = r(cos  }\!\!\theta\!\!\text{  + i sin  }\!\!\theta\!\!\text{ )}$

$\text{r cos  }\!\!\theta\!\!\text{  = - 3, r sin  }\!\!\theta\!\!\text{  = 0}$

वर्ग करके जोड़ने पर,

${{\text{r}}^{\text{2}}}\text{ = 9}$ 

$\text{r = 3}$ 

$\dfrac{\text{r sin  }\!\!\theta\!\!\text{ }}{\text{r cos  }\!\!\theta\!\!\text{ }}\text{ = }\dfrac{\text{0}}{\text{- 3}}\text{ = 0}$ 

$\text{ }\!\!\theta\!\!\text{  =  }\!\!\pi\!\!\text{ }$ 

$\text{z = 3}\left( \text{cos  }\!\!\pi\!\!\text{  + i sin  }\!\!\pi\!\!\text{ } \right)$ 

7. $\text{z = }\sqrt{\text{3}}\text{ + i = r(cos  }\!\!\theta\!\!\text{  + i sin  }\!\!\theta\!\!\text{ )}$

उत्तर:  मान लीजिए कि, $\text{z = }\sqrt{\text{3}}\text{ + i = r(cos  }\!\!\theta\!\!\text{  + i sin  }\!\!\theta\!\!\text{ )}$

$\text{r cos  }\!\!\theta\!\!\text{  = }\sqrt{\text{3}}\text{, r sin  }\!\!\theta\!\!\text{  = 1}$

वर्ग करके जोड़ने पर,

${{\text{r}}^{\text{2}}}\text{ = 4}$ 

$\text{r = 2}$ 

$\text{cos  }\!\!\theta\!\!\text{  = }\dfrac{\sqrt{\text{3}}}{\text{2}}\text{, sin  }\!\!\theta\!\!\text{  = }\dfrac{\text{1}}{\text{2}}$ 

$\dfrac{\text{r sin  }\!\!\theta\!\!\text{ }}{\text{r cos  }\!\!\theta\!\!\text{ }}\text{ = }\dfrac{\text{1}}{\sqrt{\text{3}}}$ 

$\text{tan  }\!\!\theta\!\!\text{  = }\dfrac{\text{1}}{\sqrt{\text{3}}}\text{ = tan}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{z = 2}\left( \text{cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\text{ + i sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

8. $\text{z = i}$

उत्तर:  मान लीजिए कि, $\text{z = i = r}\left( \text{cos  }\!\!\theta\!\!\text{  + i sin  }\!\!\theta\!\!\text{ } \right)$

$\text{r cos  }\!\!\theta\!\!\text{  = 0, r sin  }\!\!\theta\!\!\text{  = 1}$

वर्ग करके जोड़ने पर,

${{\text{r}}^{\text{2}}}\text{ = 0 + 1 = 1}$ 

$\text{r = 1}$ 

$\text{cos  }\!\!\theta\!\!\text{  = 0, sin  }\!\!\theta\!\!\text{  = 1}$ 

$\text{ }\!\!\theta\!\!\text{  = }\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ 

$\text{z = }\left( \text{cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{ + i sin}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$


प्रश्नावली 5.3

1. निम्नलिखत समीकरणो को हल करें ${{\text{x}}^{\text{2}}}\text{ + 3 = 0}$ 

उत्तर: इसकी तुलना $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$ से करने पर,$\text{a = 1, b = 0  }\!\!\And\!\!\text{  c = 3}$

इसीलिए  समीकरण का वीविक्तकार, 

${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{{\text{0}}^{\text{2}}}{{ -  4  \times  1  \times  3  =   -  12}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  0  \pm  }}\sqrt {{\text{  -  12}}} }}{{{\text{2x1}}}}{\text{  =  }}\dfrac{{{{ \pm  2}}\sqrt {\text{3}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{{\text{2}}}{{  =   \pm  }}\sqrt {\text{3}} {\text{i}}\quad {\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

2.  निम्नलिखत समीकरणो को हल करें, ${\text{2}}{{\text{x}}^{\text{2}}}{\text{  +  x  +  1  =  0}}$

उत्तर: इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  2, b  =  1} \&  \text{c  =  1}}$

इसीलिए  समीकरण का वीविक्तकार, 

${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{{\text{1}}^{\text{2}}}{{ -  4  \times  2  \times  1  =   -  7}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {{\text{  -  7}}} }}{{{\text{2 x 2}}}}{\text{  =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {\text{7}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{4}{\text{  =  }}\dfrac{{{\text{ -  1}}}}{{\text{4}}}{{  \pm  }}\dfrac{{\sqrt {\text{7}} }}{{\text{4}}}{\text{i}}\quad \;\;\;\;\;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

3. निम्नलिखत समीकरणो को हल करें, ${{\text{x}}^{\text{2}}}{\text{  +  3x  +  9  =  0}}$ 

उत्तर:  इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर, ${\text{a  =  1, b  =  3 \&  c  =  9}}$

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{{\text{3}}^{\text{2}}}{{ -  4  \times  1  \times  9  =   -  27}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  3  \pm  }}\sqrt {{\text{  -  27}}} }}{{{\text{2 x 1}}}}{\text{  =  }}\dfrac{{{{ -  3  \pm  }}\sqrt {{\text{27}}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{{\text{2}}}{\text{  =  }}\dfrac{{{{ -  3  \pm  3}}\sqrt {\text{3}} }}{{\text{2}}}\quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

4. निम्नलिखत समीकरणो को हल करें, ${\text{ - }}\;{{\text{x}}^{\text{2}}}{\text{  +  x  -  2   =  0}}$ 

उत्तर:  दिया गया द्विघात  है,    ${\text{ - }}\;{{\text{x}}^{\text{2}}}{\text{  +  x  -  2   =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =   - 1, b  =  1 &  c  =   - 2}}$

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{{\text{1}}^{\text{2}}}{{ -  4  \times   - 1  \times   - 2  =   -  7}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {{\text{  -  7}}} }}{{{\text{2 x  - 1}}}}{\text{  =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {\text{7}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{{{\text{ - 2}}}}{\text{  =  }}\dfrac{{{\text{ -  1}}}}{{\text{2}}}{{  \pm  }}\dfrac{{\sqrt {\text{7}} }}{{\text{2}}}{\text{i}}\quad \;\;\;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

5. निम्नलिखत समीकरणो को हल करें, ${{\text{x}}^{\text{2}}}{\text{  +  3x  +  5   =  0}}$ 

उत्तर:  दिया गया द्विघात  है,    ${{\text{x}}^{\text{2}}}{\text{  +  3x  +  5   =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  1, b  =  3 &  c  =  5}}$

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{{\text{3}}^{\text{2}}}{{ -  4  \times  1  \times  5  =   -  11}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  3  \pm  }}\sqrt {{\text{  -  11}}} }}{{{\text{2 x 1}}}}{\text{  =  }}\dfrac{{{{ -  3  \pm  }}\sqrt {{\text{11}}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{{\text{2}}}{\text{  =   }}\dfrac{{{\text{ -  3}}}}{{\text{2}}}{{  \pm  }}\dfrac{{\sqrt {\text{11}} }}{{\text{2}}}{\text{i}}\quad \;\;\;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

6. निम्नलिखत समीकरणो को हल करें, ${{\text{x}}^{\text{2}}}{\text{  -  x  +  2   =  0}}$ 

उत्तर: 

दिया गया द्विघात  है,    ${{\text{x}}^{\text{2}}}{\text{  -  x  +  2   =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  1, b  =   - 1 &  c  =  2}}$

इसीलिए  समीकरण का वीविक्तकार, 

${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{\left( {{\text{ - 1}}} \right)^{\text{2}}}{{  -  4  \times  1  \times  2  =   -  7}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{\text{ -  }}\left( {\;{\text{ - }}\;{\text{1}}} \right){{  \pm  }}\sqrt {{\text{  -  7}}} }}{{{\text{2 x 1}}}}{\text{  =  }}\dfrac{{{{ 1  \pm  }}\sqrt {\text{7}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{{\text{2}}}{\text{  =  }}\dfrac{{{\text{1}}\;{{ \pm }}\;\sqrt {\text{7}} }}{{\text{2}}}{\text{i}}\quad \;\;\;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

7. निम्नलिखत समीकरणो को हल करें, 

$\sqrt {\text{2}} {{\text{x}}^{\text{2}}}{\text{  +  x  +  }}\sqrt {\text{2}} {\text{  =  0}}$ 

उत्तर: दिया गया द्विघात  है,    $\sqrt {\text{2}} {{\text{x}}^{\text{2}}}{\text{  +  x  +  }}\sqrt {\text{2}} {\text{  =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  }}\sqrt {\text{2}} {\text{, b  =  1 &  c  =  }}\sqrt {\text{2}} $

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{{\text{1}}^{\text{2}}}{{ -  4  \times  }}\sqrt {\text{2}} {{  \times  }}\sqrt {\text{2}} {\text{  =   -  7}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {{\text{  -  7}}} }}{{{\text{2 x }}\sqrt {\text{2}} }}{\text{  =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {\text{7}} {{  \times  }}\sqrt {{\text{  -  1}}} }}{{{\text{2}}\sqrt {\text{2}} }}{\text{  =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {\text{7}} }}{{{\text{2}}\sqrt {\text{2}} }}{\text{i}}\quad \;\;\;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

8. निम्नलिखत समीकरणो को हल करें, $\sqrt 3 {{\text{x}}^{\text{2}}}{\text{  -  }}\sqrt 2 {\text{x  +  3}}\sqrt 3 {\text{  =  0}}$ 

उत्तर: दिया गया द्विघात  है,    $\sqrt {\text{3}} {{\text{x}}^{\text{2}}}{\text{  -  }}\sqrt {\text{2}} {\text{x  +  3}}\sqrt {\text{3}} {\text{  =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  }}\sqrt 3 {\text{, b  =  }}\sqrt 2 {\text{ &  c  =  3}}\sqrt 3 $

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{\left( {\sqrt {\text{2}} } \right)^{\text{2}}}{{  -  4  \times  }}\sqrt {\text{3}} {{  \times  3}}\sqrt {\text{3}} {\text{  =  2  -  36  =   -  34}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{\text{ -  (  -  }}\sqrt {\text{2}} {{)  \pm  }}\sqrt {{\text{  -  34}}} }}{{{\text{2 x }}\sqrt {\text{3}} }}{\text{  =  }}\dfrac{{\sqrt {\text{2}} {{  \pm  }}\sqrt {{\text{34}}} \sqrt {{\text{  -  1}}} }}{{{\text{2}}\sqrt {\text{3}} }}{\text{  =  }}\dfrac{{\sqrt {{\text{2 }}} {{ \pm  }}\sqrt {{\text{34}}} }}{{{\text{2}}\sqrt {\text{3}} }}{\text{i}}\quad \;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

9. निम्नलिखत समीकरणो को हल करें, ${{\text{x}}^{\text{2}}}{\text{  +  x  +  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{  =  0  =  >  }}\sqrt {\text{2}} {{\text{x}}^{\text{2}}}{\text{  +  }}\sqrt {\text{2}} {\text{x  +  1  =  0}}$ 

उत्तर: दिया गया द्विघात  है,    ${{\text{x}}^{\text{2}}}{\text{  +  x  +  }}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{  =  0  =  >  }}\sqrt {\text{2}} {{\text{x}}^{\text{2}}}{\text{  +  }}\sqrt {\text{2}} {\text{x  +  1  =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  }}\sqrt {\text{2}} {\text{, b  =  }}\sqrt {\text{2}} {\text{ \&  c  =  1}}$

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

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

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{\text{ -  (}}\sqrt {\text{2}} {{)  \pm  }}\sqrt {{\text{  -  2}}\left( {{\text{2}}\sqrt {\text{2}} {\text{  -  1}}} \right)} }}{{{\text{2 x }}\sqrt {\text{2}} }}{\text{  =  }}\dfrac{{{\text{ -  }}\sqrt {\text{2}} {{  \pm  }}\sqrt {{\text{2}}\left( {{\text{2}}\sqrt {\text{2}} {\text{  -  1}}} \right)} \sqrt {{\text{  -  1}}} }}{{{\text{2}}\sqrt {\text{2}} }}{\text{  =  }}\dfrac{{\sqrt {{\text{2 }}} {{ \pm  }}\sqrt {{\text{34}}} }}{{{\text{2}}\sqrt {\text{2}} }}{\text{i}}\quad \;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

10. निम्नलिखत समीकरणो को हल करें, ${{\text{x}}^{\text{2}}}{\text{  +  }}\dfrac{{\text{x}}}{{\sqrt {\text{2}} }}{\text{  +  1  =  0  =  >  }}\sqrt {\text{2}} {{\text{x}}^{\text{2}}}{\text{  +  x  +  }}\sqrt {\text{2}} {\text{  =  0}}$ 

उत्तर:  इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  }}\sqrt {\text{2}} {\text{, b  =  1 &  c  =  }}\sqrt 2 $

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{\left( {\text{1}} \right)^{\text{2}}}{{  -  4  \times  }}\sqrt {\text{2}} {{  \times  }}\sqrt {\text{2}} {\text{  =  2  -  4}}\sqrt {\text{2}} {\text{  =  1  -  8  =   -  7}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {{\text{  -  7}}} }}{{{\text{2 x }}\sqrt {\text{2}} }}{\text{  =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {\text{7}} \sqrt {{\text{  -  1}}} }}{{{\text{2}}\sqrt {\text{2}} }}{\text{  =  }}\dfrac{{{{ -  1  \pm  }}\sqrt {\text{7}} }}{{{\text{2}}\sqrt {\text{2}} }}{\text{i}}\quad \;\;\;\;{\text{\{ }}\sqrt {{\text{  -  1}}} {\text{  =  i\} }}$

प्रश्नावली A5

1.  ${\left[ {{{\text{i}}^{{\text{18}}}}{\text{  +  }}{{\left( {\dfrac{{\text{1}}}{{\text{i}}}} \right)}^{{\text{25}}}}} \right]^{\text{3}}}$ का मान ज्ञात कीजिए  

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

${\text{ =  }}{\left[ {{{\text{i}}^{\text{2}}}{\text{  +  }}\left\{ {\dfrac{{\text{1}}}{{\text{i}}}} \right\}} \right]^{\text{3}}}\quad \;\;\;\;\;\;\;\;\;\left[ {\because {\text{, }}{{\text{i}}^{\text{4}}}{\text{ = 1}}} \right]\quad$

${\text{ =  [ -  1  +  (  -  i)}}{{\text{]}}^{\text{3}}}\quad \;\;\;\;\;\;\;\;\;{\text{[}}\because {\text{, }}{{\text{i}}^{\text{2}}}{\text{  =   -  1   }}\;\&\;\dfrac{{\text{1}}}{{\text{i}}}{\text{  =   -  i ]}}$ 

2. किन्हीं दो सम्मिश्र संख्याओं ${{\text{z}}_{\text{1}}}$ और  ${{\text{z}}_{\text{2}}}$ के लिए, सिद्ध कीजिए: 

${\text{Re}}\left( {{{\text{z}}_{\text{1}}}{\text{ }}{{\text{z}}_{\text{2}}}} \right){\text{  =  Re}}{{\text{z}}_{\text{1}}}{\text{ Re}}{{\text{z}}_{\text{2}}}{\text{  -  }}\left| {{\text{m}}{{\text{z}}_{\text{1}}}} \right|{\text{ m}}{{\text{z}}_{\text{2}}}$

उत्तर:  ${{\text{z}}_{\text{1}}}{\text{  =  }}{{\text{a}}_{_{\text{1}}}}{\text{  +  i}}{{\text{b}}_{\text{1}}}$ 

और ${{\text{z}}_{\text{2}}}{\text{  =  }}{{\text{a}}_{{\text{2 }}}}{\text{  +  i}}{{\text{b}}_{\text{2}}}$

इसलिए,

${{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}{\text{  =  }}\left( {{{\text{a}}_{\text{1}}}{\text{  +  i}}{{\text{b}}_{\text{1}}}} \right){\text{ * }}\left( {{{\text{a}}_{\text{2}}}{\text{  +  i}}{{\text{b}}_{\text{2}}}} \right)$

${{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}{\text{  =  }}{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}{\text{  +  i}}{{\text{b}}_{\text{1}}}{{\text{a}}_{\text{2}}}{\text{  +  i}}{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{2}}}{\text{  -  }}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}$

${{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}{\text{  =  }}\left( {{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}{\text{  -  }}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}} \right){\text{  +  i}}\left( {{{\text{b}}_{\text{1}}}{{\text{a}}_{\text{2}}}{\text{  +  }}{{\text{a}}_{\text{1}}}{{\text{b}}_{\text{2}}}} \right)$

${\text{Re}}\left( {{{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}} \right){\text{  =  }}\left( {{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}{\text{  -  }}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}} \right)$ 

अतः, ${\text{Re}}\left( {{{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}} \right){\text{  =  Re}}{{\text{z}}_{\text{1}}}{\text{Re}}{{\text{z}}_{\text{2}}}{\text{  -  Im}}{{\text{z}}_{\text{1}}}\mid {\text{m}}{{\text{z}}_{\text{2}}}$

3. $\left( {\dfrac{{\text{1}}}{{{\text{1  -  4i}}}}{\text{  -  }}\dfrac{{\text{2}}}{{{\text{1  +  i}}}}} \right)\left( {\dfrac{{{\text{3  -  4i}}}}{{{\text{5  +  i}}}}} \right)$ को मानक के रूप में परिवर्तित कीजिये       

उत्तर:   $\left( {\dfrac{{\text{1}}}{{{\text{1  -  4i}}}}{\text{  -  }}\dfrac{{\text{2}}}{{{\text{1  +  i}}}}} \right)\left( {\dfrac{{{\text{3  -  4i}}}}{{{\text{5  +  i}}}}} \right)$

${\text{ =  }}\left( {\dfrac{{{\text{1  +  i  -  2(1  -  4i)}}}}{{{\text{(1  -  4i)(1  +  i)}}}}} \right)\left( {\dfrac{{{\text{3  -  4i}}}}{{{\text{5  +  i}}}}} \right)$

${\text{ =  }}\left( {\dfrac{{{\text{ -  1  +  9i}}}}{{{\text{(1  -  4i)(1  +  i)}}}}} \right)\left( {\dfrac{{{\text{3  -  4i}}}}{{{\text{5  +  i}}}}} \right)$ 

${\text{ =  }}\left( {\dfrac{{{\text{ -  1  +  9i}}}}{{{\text{1  -  3i  -  4}}{{\text{i}}^{\text{2}}}}}} \right)\left( {\dfrac{{{\text{3  -  4i}}}}{{{\text{5  +  i}}}}} \right)$

${\text{ =  }}\left( {\dfrac{{{\text{ -  3  +  4i  +  27i  -  36}}{{\text{i}}^{\text{2}}}}}{{{\text{25  +  5i  -  15i  -  3}}{{\text{i}}^{\text{2}}}}}} \right)$

${\text{ =  }}\left( {\dfrac{{{\text{33  +  31i}}}}{{{\text{28  -  10 i}}}}} \right)$ 

${\text{ =  }}\left( {\dfrac{{{\text{33  +  31i}}}}{{{\text{28  -  10 i}}}}} \right)\left( {\dfrac{{{\text{28 + 10i}}}}{{{\text{28 + 10i}}}}} \right){\text{307}}$

${\text{ =  }}\left( {\dfrac{{{\text{307  +  599i}}}}{{{\text{2}}{{\text{8}}^{\text{2}}}{\text{  +  1}}{{\text{0}}^{\text{2}}}}}} \right)$

${\text{ =  }}\left( {\dfrac{{{\text{307  +  599i}}}}{{{\text{442}}}}} \right)$

$= \;\left( {\dfrac{{{\text{307}}}}{{{\text{442}}}}{\text{  +  }}\dfrac{{{\text{599i}}}}{{{\text{442}}}}} \right)$ 

4. ${\text{x  -  iy  =  }}\sqrt {\dfrac{{{\text{a  -  ib}}}}{{{\text{c  -  id}}}}} $ सिद्ध कीजिए कि ${{\text{x}}^{\text{2}}}{\text{  +  }}{{\text{y}}^{\text{2}}}{\text{  =  }}\dfrac{{{{\text{a}}^{\text{2}}}{\text{  +  }}{{\text{b}}^{\text{2}}}}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}$

उत्तर:  ${\text{x  -  iy  =  }}\sqrt {\dfrac{{{\text{a  -  ib}}}}{{{\text{c  -  id}}}}} $

${\text{x  -  iy  =  }}\sqrt {\dfrac{{{\text{a  -  ib}}}}{{{\text{c  -  id}}}}{\text{  \times  }}\dfrac{{{\text{c  +  id}}}}{{{\text{c  +  id}}}}} $

${\text{x  -  iy  =  }}\sqrt {\dfrac{{{\text{ac  +  bd  +  i}}\left( {{\text{ad  -  bc}}} \right)}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}} $ 

दोनों ओर वर्ग करने पर

${\left( {{\text{x  -  iy}}} \right)^{\text{2}}}{\text{  =  }}\dfrac{{{\text{ac  +  bd  +  i}}\left( {{\text{ad  -  bc}}} \right)}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}$

${{\text{x}}^{\text{2}}}{\text{  -  }}{{\text{y}}^{\text{2}}}{\text{  -  i2xy  =  }}\dfrac{{{\text{ac  +  bd  +  i}}\left( {{\text{ad  -  bc}}} \right)}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}$

तुलना करने पर

${{\text{x}}^{\text{2}}}{\text{  -  }}{{\text{y}}^{\text{2}}}{\text{  =  }}\dfrac{{{\text{ac  +  bd}}}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}$

${\text{ -  2xy  =  }}\dfrac{{{\text{(ad  -  bc)}}}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}$

$\because {\text{, }}{\left( {{{\text{x}}^{\text{2}}}{\text{  +  }}{{\text{y}}^{\text{2}}}} \right)^{\text{2}}}{\text{  =  }}{\left( {{{\text{x}}^{\text{2}}}{\text{  -  }}{{\text{y}}^{\text{2}}}} \right)^{\text{2}}}{\text{  +  4}}{{\text{x}}^{\text{2}}}{{\text{y}}^{\text{2}}}$

${\left( {{{\text{x}}^{\text{2}}}{\text{  +  }}{{\text{y}}^{\text{2}}}} \right)^{\text{2}}}{\text{  =  }}{\left( {\dfrac{{{\text{ac  +  bd}}}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}} \right)^{\text{2}}}{\text{  +  }}{\left( {\dfrac{{{\text{(ad  -  bc)}}}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}} \right)^{\text{2}}}$

${\left( {{{\text{x}}^{\text{2}}}{\text{  +  }}{{\text{y}}^{\text{2}}}} \right)^{\text{2}}}{\text{  =  }}\dfrac{{{{\text{a}}^{\text{2}}}{\text{  +  }}{{\text{b}}^{\text{2}}}}}{{{{\text{c}}^{\text{2}}}{\text{  +  }}{{\text{d}}^{\text{2}}}}}s$ 

5.  निम्नलिखित को ध्रुवीय रूप में परिवर्तित:

$\left( {\text{i}} \right)$$\dfrac{{{\text{1  +  7i}}}}{{{{\left( {{\text{2  -  i}}} \right)}^{\text{2}}}}}$                             $\left( {{\text{ii}}} \right)$ $\dfrac{{{\text{1  +  3i}}}}{{{\text{1  -  2i}}}}$

उत्तर:  $\left( {\text{i}} \right)$$\dfrac{{{\text{1  +  7i}}}}{{{{\left( {{\text{2  -  i}}} \right)}^{\text{2}}}}}$ ${\text{ =  }}\dfrac{{{\text{1  +  7i}}}}{{{\text{4  +  }}{{\text{i}}^{\text{2}}}{\text{  -  4i}}}}{\text{  =  }}\dfrac{{{\text{1  +  7i}}}}{{{\text{4  -  1  -  4i}}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {{{\text{i}}^{\text{2}}}{\text{  =   -  1}}} \right]$

${\text{ =  }}\dfrac{{{\text{1  +  7i}}}}{{{\text{3  -  4i}}}}$

${\text{ =  }}\dfrac{{{\text{1  +  7i}}}}{{{\text{3  -  4i}}}}{{  \times  }}\dfrac{{{\text{3  +  4i}}}}{{{\text{3  +  4i}}}}$

${\text{ =  }}\dfrac{{{\text{ -  25  +  25i}}}}{{{\text{25}}}}$

${\text{ =   -  1  +  i}}$

${{\text{r}}^{\text{2}}}{\text{  =   -  1  +  i}}$ 

मापांक ${\text{r  =  }}\sqrt{\text{2}} $

${\theta \text{  =  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{{\text{ -  1}}}}} \right)$

कोणांक ${\theta \text{  =   -  4}}{{\text{5}}^{\text{0}}}$

से, यह स्पष्ट है कि दूसरे चतुरथनस में स्थित हैं

उत्तर: $\left( {{\text{ii}}} \right)$

$\dfrac{{{\text{1  +  3i}}}}{{{\text{1  -  2i}}}}$

${\text{ =  }}\dfrac{{{\text{1  +  3i}}}}{{{\text{1  -  2i}}}}{{  \times  }}\dfrac{{{\text{1  +  2i}}}}{{{\text{1  +  2i}}}}$

${\text{ =  }}\dfrac{{{\text{1  +  2i  +  3i  -  6}}}}{{{\text{1  +  4}}}}$

${\text{ =   -  1  +  i}}$

${{\text{r}}^{\text{2}}}{\text{  =  (  -  1}}{{\text{)}}^{\text{2}}}{\text{  +  (i}}{{\text{)}}^{\text{2}}}$                  

मापांक ${\text{r  =  }}\sqrt {\text{2}} $

$\theta{\text{=  ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{1}}}{{{\text{  -  1}}}}} \right)$

कोणांक $\theta {\text{=   -  4}}{{\text{5}}^{{^\circ }}}$

से, यह स्पष्ट है कि दूसरे चतुरथनस में स्थित हैं

6. निम्नलिखत समीकरण को हल करें, ${\text{3}}{{\text{x}}^{\text{2}}}{\text{  -  4x  +  }}\dfrac{{{\text{20}}}}{{\text{3}}}{\text{  =  0}}$

उत्तर:  दिया गया द्विघात  है,    ${\text{3}}{{\text{x}}^{\text{2}}}{\text{  -  4 x  +  }}\dfrac{{{\text{20}}}}{{\text{3}}}{\text{   =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  3 , b  =   -  4  &  c  =  }}\dfrac{{{\text{20}}}}{{\text{3}}}$

इसीलिए  समीकरण का वीविक्तकार, 

${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{\left( {{\text{  -  4}}} \right)^{\text{2}}}{{  \times   -  4  \times  3   \times  }}\dfrac{{{\text{20}}}}{{\text{3}}}{\text{  =  16  -  80  =   -  64 }}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{4  \pm  }}\sqrt {{\text{  -  64}}} }}{{\text{6}}}$

${\text{x  =  }}\dfrac{{{\text{4  +  8i}}}}{{\text{6}}}{\text{, x  =  }}\dfrac{{{\text{4  -  8i}}}}{{\text{6}}}$

${\text{x  =  }}\dfrac{{{\text{2  +  4i}}}}{{\text{3}}}{\text{, x  =  }}\dfrac{{{\text{2  -  4i}}}}{{\text{3}}}$ 

7.  निम्नलिखत समीकरण को हल करें,

${{\text{x}}^{\text{2}}}{\text{  -  2x  +  }}\dfrac{3}{2}{\text{  =  0}}$

उत्तर: दिया गया द्विघात  है,    ${{\text{x}}^{\text{2}}}{\text{  -  2x  +  }}\dfrac{3}{2}{\text{   =  0}}$

इसकी तुलना ${\text{a}}{{\text{x}}^{\text{2}}}{\text{  +  bx  +  c  =  0}}$ से करने पर,${\text{a  =  1, b  =   -  2 &  c  =  }}\dfrac{3}{2}$

इसीलिए  समीकरण का वीविक्तकार, ${\text{D  =  }}{{\text{b}}^{\text{2}}}{\text{  -  4ac}}$

${\text{ =  }}{\left( {{\text{  -  2}}} \right)^{\text{2}}}{{  \times   -  4  \times  1  \times  }}\dfrac{3}{2}{\text{  =  4  -  6  =   -  2}}$

इसलिए, समीकरण के लिए समाधान है,${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{x  =  }}\dfrac{{{{ -  b  \pm  }}\sqrt {\text{D}} }}{{{\text{2a}}}}$

${\text{ =  }}\dfrac{{{{2  \pm  }}\sqrt {{\text{  -  2}}} }}{2}$

${\text{x  =  }}\dfrac{{{\text{2  +  i}}\sqrt 2 }}{2}{\text{, x  =  }}\dfrac{{{\text{2  -  i}}\sqrt 2 }}{2}$

${\text{x  =  }}\dfrac{{{\text{2  +  i}}\sqrt 2 }}{2}{\text{, x  =  }}\dfrac{{{\text{2  -  i}}\sqrt 2 }}{2}$ 

8. निम्नलिखत समीकरण को हल करें,

${\text{27}}{{\text{x}}^{\text{2}}}{\text{  -  10x  +  1  =  0}}$

उत्तर:  दिया गया द्विघात  है,    $\text{27}{{\text{x}}^{\text{2}}}\text{ - 10x + 1 = 0}$

इसकी तुलना $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$ से करने पर,$\text{a = 27, b = - 10   }\!\!\And\!\!\text{  c = 1}$

इसीलिए  समीकरण का वीविक्तकार, 

$\text{D = }{{\text{b}}^{\text{2}}}\text{ - 4ac}$

$\text{= }{{\left( \text{ - 10} \right)}^{\text{2}}}\text{  - 4  }\!\!\times\!\!\text{  27  }\!\!\times\!\!\text{  1 = 100 - 108 = - 8}$

इसलिए, समीकरण के लिए समाधान है,$\text{x = }\dfrac{\text{- b  }\!\!\pm\!\!\text{  }\sqrt{\text{D}}}{\text{2a}}$

$\text{x = }\dfrac{\text{- b  }\!\!\pm\!\!\text{  }\sqrt{\text{D}}}{\text{2a}}$ 

$\text{= }\dfrac{\text{10  }\!\!\pm\!\!\text{  }\sqrt{\text{ - 8}}}{54}$ 

$\text{x = }\dfrac{\text{28 + i2}\sqrt{2}}{54}\text{, x = }\dfrac{\text{28 - i2}\sqrt{2}}{54}$ 

$\text{x = }\dfrac{\text{5 + i}\sqrt{2}}{27}\text{, x = }\dfrac{\text{5 - i}\sqrt{2}}{27}$

9. निम्नलिखत समीकरण को हल करें,

$\text{21}{{\text{x}}^{\text{2}}}\text{ - 28x + 10 = 0}$

उत्तर:  दिया गया द्विघात  है,    $\text{21}{{\text{x}}^{\text{2}}}\text{ - 28x + 10 = 0}$

इसकी तुलना $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$ से करने पर,$\text{a = 21, b = - 28  }\!\!\And\!\!\text{  c = 10}$

इसीलिए  समीकरण का वीविक्तकार, 

$\text{D = }{{\text{b}}^{\text{2}}}\text{ - 4ac}$

$\text{= }{{\left( \text{ - 28s} \right)}^{\text{2}}}\text{  - 4  }\!\!\times\!\!\text{  21  }\!\!\times\!\!\text{  10 = 784 - 840 = - 56}$

इसलिए, समीकरण के लिए समाधान है,$\text{x = }\dfrac{\text{- b  }\!\!\pm\!\!\text{  }\sqrt{\text{D}}}{\text{2a}}$

$\text{x = }\dfrac{\text{- b  }\!\!\pm\!\!\text{  }\sqrt{\text{D}}}{\text{2a}}$

$\text{= }\dfrac{\text{28  }\!\!\pm\!\!\text{  }\sqrt{\text{ - 56}}}{42}$

$\text{x = }\dfrac{\text{28 + i2}\sqrt{14}}{42}\text{, x = }\dfrac{\text{28 - i2}\sqrt{14}}{42}$

$\text{x = }\dfrac{\text{14 + i}\sqrt{14}}{21}\text{, x = }\dfrac{\text{14 - i}\sqrt{14}}{21}$ 

10. यदि${{\text{z}}_{\text{1}}}\text{ = 2 - i, }{{\text{z}}_{\text{2}}}\text{ = 1 + i; }\left| \dfrac{{{\text{z}}_{\text{1}}}\text{ + }{{\text{z}}_{\text{2}}}\text{ + 1}}{{{\text{z}}_{\text{1}}}\text{ -}{{\text{z}}_{\text{2}}}\text{ + 1}} \right|$ का मान ज्ञात कीजिए 

उत्तर: $\text{= }\left| \dfrac{\left( \text{2 - i} \right)\text{ + }\left( \text{1 + i} \right)\text{ + 1}}{\left( \text{2 - i} \right)\text{ - }\left( \text{1 + i} \right)\text{ + 1}} \right|$

$\text{= }\left| \dfrac{\text{4}}{\text{2 - 2i}} \right|$ 

$\text{=}\left| \dfrac{\text{2}}{\text{1 - i}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{1 + i}}{\text{1 + i}} \right|$ 

$\text{= }\left| \dfrac{\text{2 (1 + i)}}{\text{1 + 1}} \right|$ 

$\text{=  }\!\!|\!\!\text{ (1 + i) }\!\!|\!\!\text{ }$ 

$\text{= }\sqrt{\text{2}}$ 

11. $\text{a + ib = }\dfrac{{{\left( \text{x + i} \right)}^{\text{2}}}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}$, सिद्ध कीजिए कि, ${{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}\text{ = }\dfrac{{{\left( {{\text{x}}^{\text{2}}}\text{ + 1} \right)}^{\text{2}}}}{{{\left( \text{2}{{\text{x}}^{\text{2}}}\text{ + 1} \right)}^{\text{2}}}}$

उत्तर:  $\text{a + ib = }\dfrac{{{\left( \text{x + i} \right)}^{\text{2}}}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}$  

$\text{a + ib = }\dfrac{{{\text{x}}^{\text{2}}}\text{ + }{{\text{i}}^{\text{2}}}\text{ + i2x}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}$ 

$\text{a + ib = }\dfrac{{{\text{x}}^{\text{2}}}\text{ - 1}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}\text{ + i}\dfrac{\text{2x}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}$ 

तुलना करने पर 

$\text{a = }\dfrac{{{\text{x}}^{\text{2}}}\text{ - 1}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}$ 

$\text{b = }\dfrac{\text{2x}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}}$ 

अब, 

${{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}\text{ = }{{\left( \dfrac{{{\text{x}}^{\text{2}}}\text{ - 1}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}} \right)}^{\text{2}}}\text{ + }{{\left( \dfrac{\text{2x}}{\text{2}{{\text{x}}^{\text{2}}}\text{ + 1}} \right)}^{\text{2}}}$ 

${{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}\text{ = }\dfrac{{{\text{x}}^{\text{4}}}\text{ - 2}{{\text{x}}^{\text{2}}}\text{ + 1}}{{{\left( \text{2}{{\text{x}}^{\text{2}}}\text{ + 1} \right)}^{\text{2}}}}\text{ + }\dfrac{\text{4}{{\text{x}}^{\text{2}}}}{{{\left( \text{2}{{\text{x}}^{\text{2}}}\text{ + 1} \right)}^{\text{2}}}}$ 

${{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}\text{ = }\dfrac{{{\text{x}}^{\text{4}}}\text{ - 2}{{\text{x}}^{\text{2}}}\text{ +1 + 4}{{\text{x}}^{\text{2}}}}{{{\left( \text{2}{{\text{x}}^{\text{2}}}\text{ + 1} \right)}^{\text{2}}}}$ 

${{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}\text{ = }\dfrac{{{\text{x}}^{\text{4}}}\text{ + 2}{{\text{x}}^{\text{2}}}\text{ + 1}}{{{\left( \text{2}{{\text{x}}^{\text{2}}}\text{ + 1} \right)}^{\text{2}}}}$ 

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

12. माना ${{\text{z}}_{\text{1}}}\text{ = 2 - i, }{{\text{z}}_{\text{2}}}\text{ = - 2 + i}$ निम्न का मान निकालिए 

$\text{(i)  Re}\left( \dfrac{{{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}}{\overline{{{\text{z}}_{\text{1}}}}} \right)$                    $\text{(ii)   Im}\left( \dfrac{\text{1}}{{{\text{z}}_{\text{1}}}\overline{{{\text{z}}_{\text{1}}}}} \right)$

उत्तर:  $\text{(i)    Re}\left( \dfrac{{{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}}{\overline{{{\text{z}}_{\text{1}}}}} \right)$

$\left( \dfrac{{{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}}{\overline{{{\text{z}}_{\text{1}}}}} \right)$ 

$\text{= }\dfrac{\text{(2 - i)(- 2 + i)}}{\text{(2 + i)}}$ 

$\text{= }\dfrac{\text{- 3 + 4i}}{\text{(2 + i)}}$ 

$\text{= }\dfrac{\text{- 3 + 4i}}{\text{(2 + i)}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{2 - i}}{\text{2 - i}}\text{ = }\dfrac{\text{- 6 + 3i + 8i - 4}{{\text{i}}^{\text{2}}}}{{{\text{2}}^{\text{2}}}\text{ + }{{\text{i}}^{\text{2}}}}$ 

$\text{= }\dfrac{\text{- 2 + 11i}}{\text{5}}$ 

$\text{= }\dfrac{\text{- 2}}{\text{5}}\text{ + }\dfrac{\text{11i}}{\text{5}}$ 

अब, 

$\text{Re}\left( \dfrac{{{\text{z}}_{\text{1}}}{{\text{z}}_{\text{2}}}}{\overline{{{\text{z}}_{\text{1}}}}} \right)\text{ = - }\dfrac{\text{2}}{\text{5}}$

उत्तर: $\text{(ii)    Im}\left( \dfrac{\text{1}}{{{\text{z}}_{\text{1}}}\overline{{{\text{z}}_{\text{1}}}}} \right)$ 

$\left( \dfrac{\text{1}}{{{\text{z}}_{\text{1}}}\overline{{{\text{z}}_{\text{1}}}}} \right)$ 

$\text{= }\dfrac{\text{1}}{\text{(2 - i)(2 + i)}}$ 

$\text{= }\dfrac{\text{1}}{{{\text{2}}^{\text{2}}}\text{ - }{{\text{i}}^{\text{2}}}}$ 

$\text{= }\dfrac{\text{1}}{\text{5}}$ 

$\text{= }\dfrac{\text{1}}{\text{5}}\text{ + i0 }$ 

अब,

$\text{Im}\left( \dfrac{\text{1}}{{{\text{z}}_{\text{1}}}\overline{{{\text{z}}_{\text{1}}}}} \right)\text{ = 0}$

13.  सम्मिश्र संख्या $\dfrac{\text{1 + 2i}}{\text{1 - 3i}}$का मापांक और कोणांक ज्ञात कीजिए

उत्तर:  $\dfrac{\text{1 + 2i}}{\text{1 - 3i}}$

$\text{= }\dfrac{\text{1 + 2i}}{\text{1 - 3i}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{1 + 3i}}{\text{1 + 3i}}$ 

$\text{= }\dfrac{\text{1 + 3i + 2i + 6}{{\text{i}}^{\text{2}}}}{{{\text{1}}^{\text{2}}}\text{ + (3}{{\text{)}}^{\text{2}}}}$ 

$\text{= }\dfrac{\text{- 5 + 5i}}{\text{10}}$ 

$\text{= - }\dfrac{\text{1}}{\text{2}}\text{ + }\dfrac{\text{i1}}{\text{2}}$ 

${{\text{r}}^{\text{2}}}\text{ = }{{\left( \text{- }\dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}\text{ + }{{\left( \dfrac{\text{1}}{\text{2}} \right)}^{\text{2}}}$ 

मापांक $\text{r=}\dfrac{\text{1}}{\sqrt{\text{2}}}$

$\text{ }\!\!\theta\!\!\text{  = ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{- }\dfrac{\text{1}}{\text{2}}}{\dfrac{\text{1}}{\text{2}}} \right)$

कोणांक $\text{ }\!\!\theta\!\!\text{  = - 4}{{\text{5}}^{\text{ }\!\!{}^\circ\!\!\text{ }}}$

से, यह स्पष्ट है कक तीसरे चतुर्तांश में स्थित है

14. $\text{ale}\left( \text{x - iy} \right)\left( \text{3 + 5i} \right)\text{, - 6 - 24i}$ की सयुग्मी है तो वास्तविक संखिए $\text{x}$ और $\text{y}$ ज्ञात  कीजिए                        

उत्तर:  दिया है: $\left( \text{x - iy} \right)\left( \text{3 + 5i} \right)\text{, - 6 - 24i}$ की संयुग्मी है 

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

$\text{3x + i5x - i3y + 5y = - 6 + 24 i }$ 

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

तुलना करने पर

$\text{3x + 5y = - 6}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .....\left( \text{1} \right)$ 

$\text{5x - 3y = 24}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,\ .....\left( \text{2} \right)$ 

समीकरण (1) को 3 से, व  समीकरण (2) को 5 से गुणा करने पर और जोड़ने पर  

$\text{3}\left( \text{3x + 5y = - 6} \right)\text{+5}\left( \text{5x - 3y = 24} \right)$ 

$\ \ \ \ \text{34x = 102}$ 

$\ \ \ \ \ \ \text{x = 3}$ 

$\text{x}$ का मान समीकरण $\left( \text{1} \right)$ में रखने पर 

$\text{y = - 3}$

15. $\dfrac{\text{1 + i}}{\text{1 - i}}\text{ - }\dfrac{\text{1 - i}}{\text{1 + i}}$ का मापांक ज्ञात कीजिए।

उत्तर: $\dfrac{\text{1 + i}}{\text{1 - i}}\text{ - }\dfrac{\text{1 - i}}{\text{1 + i}}$

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

 $\text{= }\dfrac{\text{4i}}{\text{2}}$ 

 $\text{= 2i}$ 

इसलिए,

$\left| \text{ 2i } \right|\text{ = 2}$

16. $\text{afe (x + iy}{{\text{)}}^{\text{3}}}\text{ = u + iy}$ तो दर्शाये कि $\dfrac{\text{u}}{\text{x}}\text{ + }\dfrac{\text{v}}{\text{y}}\text{ = 4}\left( {{\text{x}}^{\text{2}}}\text{ - }{{\text{y}}^{\text{2}}} \right)$   

उत्तर: ${{\text{x}}^{\text{3}}}\text{ + (iy}{{\text{)}}^{\text{3}}}\text{ + 3}{{\text{x}}^{\text{2}}}\text{(iy) + 3x(iy}{{\text{)}}^{\text{2}}}\text{ = u + iv}$

${{\text{x}}^{\text{3}}}\text{ - i}{{\text{y}}^{\text{3}}}\text{ + 3}{{\text{x}}^{\text{2}}}\text{(iy) - 3x(y}{{\text{)}}^{\text{2}}}\text{ = u + iv}$ 

${{\text{x}}^{\text{3}}}\text{ - 3x}{{\text{y}}^{\text{2}}}\text{ - i}\left( {{\text{y}}^{\text{3}}}\text{ + 3}{{\text{x}}^{\text{2}}}\text{y} \right)\text{ = u + iv}$ 

तुलना करने पर 

$\text{u = }{{\text{x}}^{\text{3}}}\text{ - 3x}{{\text{y}}^{\text{2}}}$ 

$\text{v = 3}{{\text{x}}^{\text{2}}}\text{y - }{{\text{y}}^{\text{3}}}$ 

$\dfrac{\text{u}}{\text{x}}\text{ = }{{\text{x}}^{\text{2}}}\text{ - 3}{{\text{y}}^{\text{2}}}$ 

$\dfrac{\text{v}}{\text{y}}\text{ = 3}{{\text{x}}^{\text{2}}}\text{ - }{{\text{y}}^{\text{2}}}$ 

$\dfrac{\text{u}}{\text{x}}\text{ + }\dfrac{\text{v}}{\text{y}}\text{ = 4}\left( {{\text{x}}^{\text{2}}}\text{ - }{{\text{y}}^{\text{2}}} \right)$ 

17.  यदि $\text{ } \!\alpha \!\text{ }$ और $\text{ } \!\beta \!\text{ }$ भिन्न सम्मिश्र संख्याएँ हैं जहाँ $\left| \text{ } \!\beta \!\text{ } \right|\text{ + 1,}$  तब $\left| \dfrac{\text{ } \!\beta \!\text{  -  } \!\alpha \!\text{ }}{\text{1 - }\overline{\text{ } \!\alpha \!\text{ }}\text{ } \!\beta \!\text{ }} \right|$ का मान ज्ञात   कीजिए

उत्तर: ${{\left| \dfrac{\text{ } \!\beta \!\text{  -  } \!\alpha \!\text{ }}{{1 - \bar{ } \!\alpha \!\text{ } } \!\beta \!\text{ }} \right|}^{\text{2}}}\text{ = }\left( \dfrac{\text{ } \!\beta \!\text{  -  } \!\alpha \!\text{ }}{{1 - \bar{ } \!\alpha \!\text{ } } \!\beta \!\text{ }} \right)\overline{\left( \dfrac{\text{ } \!\beta \!\text{ - } \!\alpha \!\text{ }}{{1-\bar{ } \!\alpha \!\text{ } } \!\beta \!\text{ }} \right)}$   

${{\left| \dfrac{\text{ }\!\beta\!\text{  -  }\!\alpha\!\text{ }}{{1 - \bar{ }\!\alpha\!\text{ } }\!\beta\!\text{ }} \right|}^{\text{2}}}\text{ = }\left( \dfrac{\text{ }\!\beta\!\text{  -  }\!\alpha\!\text{ }}{{1 - \bar{ }\!\alpha\!\text{ } }\!\beta\!\text{ }} \right)\left( \dfrac{{\bar{ }\!\beta\!\text{ } - \bar{ }\!\alpha\!\text{ }}}{\text{1 -  }\!\alpha\!{ \bar{ }\!\beta\!\text{ }}} \right)$

$\text{= }\left( \dfrac{\text{ }\!\beta\!{ \bar{ }\!\beta\!\text{ } - \bar{ }\!\alpha\!\text{ } }\!\beta\!\text{  -  }\!\alpha\!{ \bar{ }\!\beta\!\text{ } +  }\!\alpha\!{ \bar{ }\!\alpha\!\text{ }}}{\text{1 -  }\!\alpha\!{ \bar{ }\!\beta\!\text{ } - \bar{ }\!\alpha\!\text{ } }\!\beta\!\text{  +  }\!\alpha\!{ \bar{ }\!\alpha\!\text{ } }\!\beta\!{ \bar{ }\!\beta\!\text{ }}} \right)\quad$

$\text{= }\left( \dfrac{{1 - \bar{ }\!\alpha\!\text{ } }\!\beta\!\text{  -  }\!\alpha\!{ \bar{ }\!\beta\!\text{ } + }\!|\!\text{   }\!\alpha\!\text{ }{{\text{ }\!|\!\text{ }}^{\text{2}}}}{\text{1 -  }\!\alpha\!{ \bar{ }\!\beta\!\text{ } - \bar{ }\!\alpha\!\text{ } }\!\beta\!\text{  +  }\!|\!\text{ a}{{\text{ }\!|\!\text{ }}^{\text{2}}}} \right)$

$\text{= 1}$

18. समीकरण ${{\left| \text{1 - i} \right|}^{\text{x}}}\text{ = }{{\text{2}}^{\text{x}}}$ के शान्यतेर पूर्णांक मूलों कि संख्या ज्ञात कीजिए 

उत्तर: $\left( \sqrt{{{\text{1}}^{\text{2}}}\text{ + }{{\text{1}}^{\text{2}}}} \right)\text{ = }{{\text{2}}^{\text{x}}}$

${{\left( \sqrt{\text{2}} \right)}^{\text{x}}}\text{ - }{{\text{2}}^{\text{x}}}$

${\text{2}}^{\dfrac{\text{x}}{\text{2}}}\text{ = }{{\text{2}}^{\text{x}}}$

$\dfrac{\text{x}}{\text{2}}\text{ = x}$

$\text{x = 0}$

इस प्रकार, समीकरण का केवल एक ही मूल है जो कि$\text{ 0}$ है|

19. $\text{ale}\left( \text{a + ib} \right)\left( \text{c + id} \right)\left( \text{e + if} \right)\left( \text{g + ih} \right)\text{ = A + iB}$ तो दर्शाइए कि

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

उत्तर: $\text{ale}\left( \text{a + ib} \right)\left( \text{c + id} \right)\left( \text{e + if} \right)\left( \text{g + ih} \right)\text{ = A + iB}$

$\left| \left( \text{a + ib} \right)\left( \text{c + id} \right)\left( \text{e + if} \right)\left( \text{g + ih} \right) \right|\text{ = A + iB}$ 

$\left| \text{a + ib} \right|\left| \text{c + id} \right|\left| \text{e + if} \right|\left| \text{g + ih} \right|\text{ = }\left| \text{A + iB} \right|$ 

$\sqrt{{{\text{a}}^{\text{2}}}\text{ + }{{\text{b}}^{\text{2}}}}\text{.}\sqrt{{{\text{c}}^{\text{2}}}\text{ + }{{\text{d}}^{\text{2}}}}\text{.}\sqrt{{{\text{e}}^{\text{2}}}\text{ + }{{\text{f}}^{\text{2}}}}\text{.}\sqrt{{{\text{g}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}\text{ = }\sqrt{{{\text{A}}^{\text{2}}}\text{ + }{{\text{B}}^{\text{2}}}}$ 

दोनों तरफ वर्ग करने पर 

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

20. यदि${{\left( \dfrac{\text{1 + i}}{\text{1 - i}} \right)}^{\text{m}}}\text{ = 1, m}$ कीजिए 

उत्तर: ${{\left( \dfrac{\text{1 + i}}{\text{1 - i}} \right)}^{\text{m}}}\text{ = 1}$

${{\left( \dfrac{\text{1 + i}}{\text{1 - i}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{1 + i}}{\text{1 + i}} \right)}^{\text{m}}}\text{ = 1}$ 

${{\left( \dfrac{\text{2i}}{\text{2}} \right)}^{\text{m}}}\text{ = 1}$ 

${{\text{i}}^{\text{m}}}\text{ = 1}$ 

$\text{m = 4,8,12}......\text{  }$ 

अतः, का न्यूनतम पूर्णांक मान $\text{4}$ है|

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations in Hindi

Chapter-wise NCERT Solutions are provided everywhere on the internet with an aim to help the students to gain a comprehensive understanding. Class 11 Maths Chapter 5 solution Hindi mediums are created by our in-house experts keeping the understanding ability of all types of candidates in mind. NCERT textbooks and solutions are built to give a strong foundation to every concept. These NCERT Solutions for Class 11 Maths Chapter 5 in Hindi ensure a smooth understanding of all the concepts including the advanced concepts covered in the textbook.

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