Step-by-Step Solutions For Class 11 Maths Chapter 5 In Hindi - Free PDF Download
In NCERT Solutions Class 11 Maths Chapter 5 In Hindi, you'll dive into the world of complex numbers and quadratic equations. Here, you’ll learn what complex numbers are, how to use them in math problems, and how to solve different types of quadratic equations, even when the answers aren’t simple numbers. These NCERT solutions are explained step-by-step in simple Hindi, making it much easier to understand tricky concepts.
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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.
NCERT Solutions for Class 11 Maths Chapter 5 in Hindi medium PDF download are easily available on our official website (vedantu.com). Upon visiting the website, you have to register on the website with your phone number and email address. Then you will be able to download all the study materials of your preference in a click. You can also download the Class 11 Maths Complex Numbers and Quadratic Equations solution Hindi medium from Vedantu app as well by following the similar procedures, but you have to download the app from Google play store before doing that.
NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. These NCERT Solutions for Class 11 Maths Complex Numbers and Quadratic Equations in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language. The best feature of these solutions is a free download option. Students of Class 11 can download these solutions at any time as per their convenience for self-study purpose.
These solutions are nothing but a compilation of all the answers to the questions of the textbook exercises. The answers/ solutions are given in a stepwise format and very well researched by the subject matter experts who have relevant experience in this field. Relevant diagrams, graphs, illustrations are provided along with the answers wherever required. In nutshell, NCERT Solutions for Class 11 Maths in Hindi come really handy in exam preparation and quick revision as well prior to the final examinations.
FAQs on NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers And Quadratic Equations in Hindi - 2025-26
1. What is the standard method to express a complex number in the form a + ib, as shown in NCERT Solutions for Class 11 Maths Chapter 5?
To express a complex number in the standard form a + ib, you need to simplify the given expression by performing the required algebraic operations (addition, subtraction, multiplication, division). For division, the key step is to multiply both the numerator and the denominator by the conjugate of the denominator. This eliminates the imaginary part from the denominator. Finally, group the real parts together to get 'a' and the imaginary parts together to get 'b'.
2. How do you find the multiplicative inverse of a complex number in Exercise 5.1 of the NCERT textbook?
The step-by-step method to find the multiplicative inverse of a complex number z = a + ib is as follows:
- First, find the conjugate of z, which is z̄ = a - ib.
- Next, calculate the square of the modulus of z, which is |z|² = a² + b².
- The multiplicative inverse, denoted as z⁻¹, is given by the formula: z⁻¹ = z̄ / |z|².
- Finally, substitute the values and write the result in the standard form a + ib.
3. What is the significance of finding the modulus and argument of a complex number in NCERT Exercise 5.2?
Finding the modulus (r) and argument (θ) is crucial for converting a complex number from its algebraic form (a + ib) to its polar form (r(cosθ + isinθ)). The modulus represents the distance of the number from the origin on the Argand plane, while the argument represents the angle it makes with the positive real axis. This polar representation is fundamental for understanding De Moivre's theorem and solving higher-level problems involving complex number geometry and rotations, as per the CBSE 2025-26 syllabus.
4. How do you correctly determine the principal argument (θ) of a complex number while solving NCERT questions?
To find the correct principal argument (θ), which must lie in the range (-π, π], you should follow these steps precisely:
- First, find the acute angle α using the formula tan α = |b/a|.
- Identify the quadrant in which the complex number z = a + ib lies by checking the signs of 'a' (real part) and 'b' (imaginary part).
- If z is in Quadrant I (a>0, b>0), then θ = α.
- If z is in Quadrant II (a<0, b>0), then θ = π - α.
- If z is in Quadrant III (a<0, b<0), then θ = α - π.
- If z is in Quadrant IV (a>0, b<0), then θ = -α.
Correctly identifying the quadrant is the most critical step to avoid errors.
5. How are the quadratic equations in Exercise 5.3 of NCERT Class 11 Maths Chapter 5 solved?
The quadratic equations in Exercise 5.3 typically have a negative discriminant (D = b² - 4ac < 0), which results in complex roots. The solution is found using the standard quadratic formula: x = [-b ± √(b² - 4ac)] / 2a. Since the discriminant 'D' is negative, we rewrite √D as i√(-D) or i√|D|. The two roots of the equation will then be a pair of complex conjugates.
6. Why is it important to follow the step-by-step method provided in NCERT Solutions for solving these problems?
Following the step-by-step method as per the NCERT Solutions for the 2025-26 session is vital for two key reasons. Firstly, it aligns perfectly with the CBSE marking scheme, ensuring you write answers that secure full marks in exams. Secondly, it breaks down complex problems into logical, manageable parts. This systematic approach helps in building strong conceptual clarity and preventing common errors, especially in calculating the argument or solving quadratic equations with complex roots.
7. What types of problems are covered in the Miscellaneous Exercise of NCERT Class 11 Maths Chapter 5?
The Miscellaneous Exercise for Chapter 5 features higher-order thinking skill (HOTS) questions that integrate multiple concepts from the chapter. The NCERT solutions for these problems demonstrate how to:
- Evaluate expressions with high powers of 'i'.
- Prove identities involving the real and imaginary parts of complex number products.
- Convert complex fractions into the standard and polar forms.
- Solve equations involving the modulus of complex numbers.
- Apply properties of conjugates and moduli to solve complex proofs.
8. In the NCERT solutions, how does solving a quadratic equation with a negative discriminant differ from one with a positive discriminant?
The main difference lies in the nature of the roots obtained:
- A positive discriminant (D > 0) indicates that the quadratic equation has two distinct, real roots. The term √D is a real number.
- A negative discriminant (D < 0), as focused on in Chapter 5, indicates that the equation has no real roots. Instead, its roots are a pair of complex conjugates. The term √D is calculated as i√|D|, which introduces the imaginary unit 'i' into the solution.

















