# NCERT Solutions for Class 11 Maths Chapter 5

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.1 or Exercise 5.2 or Exercise 5.3 or Miscellaneous Exercises in English Medium and सम्मिश्र संख्याएँ और द्विघात समीकरण की प्रश्नावली 5.1 or प्रश्नावली 5.2 or प्रश्नावली 5.3 or विविध प्रश्नावली 5 in हिंदी मीडियम is available to or view online or download in PDF. Join the Discussion Forum to share your views.

 Class: 11 Subject: Maths – गणित Chapter 5: Complex Numbers and Quadratic Equations

## NCERT Solutions for Class 11 Maths Chapter 5

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### Class 11 Maths Chapter 5 Complex Numbers Solutions

All Solutions are updated for CBSE Exam 2019 based on new CBSE Curriculum 2019-2020 for CBSE Board, MP Board, Gujrat UP Board, Uttarakhand Board and Bihar Board, who are following NCERT Books for their exams.

##### 11 Maths Chapter 5 Exercise 5.1 Sols

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers Exercise 5.1 is given below. For other questions, please visit to Exercise 5.2 or Exercise 5.3 or Miscellaneous Exercises or go for हिंदी मीडियम Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 5 Exercise 5.2 Sols

NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.2 is given below. For other questions, please visit to Exercise 5.1 or Exercise 5.3 or Miscellaneous Exercises or go for हिंदी मीडियम Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 5 Exercise 5.3 Sols

Class 11 Maths Chapter 5 Exercise 5.1 Solutions is given below. For other questions, please visit to Exercise 5.1 or Exercise 5.2 or Miscellaneous Exercises or go for हिंदी मीडियम Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 5 Miscellaneous Exercise 5 Sols

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise 5 is given below. For other questions, please visit to Exercise 5.1 or Exercise 5.2 or Exercise 5.3 or go for हिंदी मीडियम Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित पाठ 5 प्रश्नावली 5.1 के हल

NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 in Hindi is given below. For other questions, please visit to प्रश्नावली 5.2 or प्रश्नावली 5.3 or विविध प्रश्नावली 5 or go for English Medium Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित पाठ 5 प्रश्नावली 5.2 के हल

11 Maths Chapter 5 Exercise 5.2 solutions in Hindi is given below. For other questions, please visit to  प्रश्नावली 5.1 or प्रश्नावली 5.3 or विविध प्रश्नावली 5 or go for English Medium Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित पाठ 5 प्रश्नावली 5.3 के हल

Class 11 Maths Chapter 5 Exercise 5.3  sols in Hindi is given below. For other questions, please visit to  प्रश्नावली 5.1 or प्रश्नावली 5.2 or प्रश्नावली 5.3 or विविध प्रश्नावली 5 or go for English Medium Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 गणित पाठ 5 विविध प्रश्नावली 5 के हल

NCERT Solutions for Class 11 Maths Chapter 5 Miscellaneous Exercise 5 in Hindi is given below. For other questions, please visit to  प्रश्नावली 5.1 or प्रश्नावली 5.2 or प्रश्नावली 5.3 or go for English Medium Solutions. Visit to Class 11 Maths main page or Top of the page.

Visit to Class 11 Maths main page or Top of the page

###### Important Terms on Complex Numbers
• A number of the form z = a + ib, where a, b are real numbers, is called a complex number. a is called the real part of z, denoted by Re(z) and b is called the imaginary part of z, denoted by Im(z).
• a + ib = c + id if a = c, and b = d
• If z1 = a + ib, z2 = c + id. In general, we cannot compare and say that z1> z2 or z1< z2 but if b, d = 0 and a > c then z1> z2, i.e. we can compare two complex numbers only if they are purely real.
• z = 0 + i 0 is additive identity of a complex number.
• –z = –a –ib is called the Additive Inverse or negative of z = a + ib.
• z = 1 + i 0 is multiplicative identity of complex number.
• The coordinate plane that represents the complex numbers is called the complex plane or the Argand plane.

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#### Express the following complex number in the form a + ib: i^9+i^19

i^9+i^19
= (i^4 )^2.i + (i^4 )^4.i^2.i
=(1)^2.i + (1)^4.(-1).i
[∵i^4 = 1 and i^2 = – 1] = i – i
= 0
= 0 + i0

#### निम्नलिखित सम्मिश्र संख्या का मापांक और कोणांक ज्ञात कीजिए: Z= -1 -i√3

माना Z=-1-i√3=r(cos⁡θ+i sin⁡θ )
तुलना करने पर
r cos⁡θ=-1 …(1)
r sin⁡θ=-√3 …(2)
समीकरण (1) और (2) को वर्ग करके जोड़ने पर
r^2 cos^2⁡θ+r^2 sin^2⁡θ = 1 + 3
⇒ r^2 (cos^2⁡θ + sin^2⁡θ ) = 4
⇒ r^2 = 4
⇒ r = 2
[∵r=|Z|>0] इसलिए, मापांक = 2
अब, समीकरण (2) को (1) से भाग देने पर
(r sin⁡θ)/(r cos⁡θ )=(-√3)/(-1)
⇒tan⁡θ=√3 …(3)
समीकरण (1), (2) और (3) से, यह स्पष्ट है कि sin⁡θ और cos⁡θ ऋणात्मक हैं जबकि tan⁡θ धनात्मक है इसलिए, θ तीसरे चतुर्थांश में स्थित है।
अतः,
कोणांक θ=-(π-π/3)=-2π/3

#### Solve the following equation: 2x^2 + x + 1 = 0.

The given quadratic equation is 2x^2 + x + 1 = 0.
On comparing the given equation with 〖ax〗^2+bx+c=0,
we obtain a = 2, b = 1, and c = 1
Therefore, the discriminant of the given equation is given by
D = b^2-4ac
= 1^2-4×2×1
=-7
Therefore, the required solutions are
x = (-b±√D)/2a=(-1±√(-7))/(2×2)
= (-1±√7.√(-1))/4
= (-1±√7 i)/4 [∵ √(-1)=i]

#### निम्नलिखित सम्मिश्र संख्या को ध्रुवीय रूप में रूपांतरित कीजिए: -3

माना Z=-3+0i=r(cos⁡θ+i sin⁡θ )
तुलना करने पर
r cos⁡θ =-3 …(1)
r sin⁡θ =0 …(2)
समीकरण (1) और (2) को वर्ग करके जोड़ने पर
r^2 cos^2⁡θ+r^2 sin^2⁡θ=9+0
⇒r^2 (cos^2⁡θ+sin^2⁡θ )=9
⇒r^2=9
⇒r=3
[∵r=|Z|>0] इसलिए, मापांक = 3
अब, समीकरण (2) को (1) से भाग देने पर
(r sin⁡θ)/(r cos⁡θ )=0/(-3)
⇒tan⁡θ=0 …(3)
समीकरण (1), (2) और (3) से, यह स्पष्ट है कि sin⁡θ और tan⁡θ दोनों ही 0 हैं जबकि cos⁡θ ऋणात्मक है।
अतः,
कोणांक θ=π
इसलिए, सम्मिश्र संख्या Z=-3 का ध्रुवीय रूप में रूपांतरण निम्नलिखित है:
3[cos⁡π+i sin⁡π ]

#### किन्हीं दो सम्मिश्र संख्याओं z_1 और z_2 के लिए, सिद्ध कीजिए: Re(z_1 z_2 )=Re z_1 Re z_2-Im z_1 Im z_2.

माना z_1=x_1+iy_1 और z_2=x_2+iy_2
इसलिए,z_1 z_2=(x_1+iy_1 )(x_2+iy_2 )=x_1 x_2+ix_1 y_2+ix_2 y_1+i^2 y_1 y_2
= x_1 x_2+ix_1 y_2+ix_2 y_1-y_1 y_2
[∵ i^2=-1] =(x_1 x_2-y_1 y_2 )+i(x_1 y_2+x_2 y_1 )
अतः,Re(z_1 z_2 ) = x_1 x_2-y_1 y_2=Re z_1 Re z_2-Im z_1 Im z_2