NCERT Solutions for Class 11 Maths Chapter 5

Class: 11	Maths (English and Hindi Medium)
Chapter 5:	Complex Numbers and Quadratic Equations

11th Maths Chapter 5 Solutions

NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers Exercise 5.1 to 5.3 and miscellaneous exercise are given below in updated format for current academic session 2021-22. All Solutions are updated for CBSE Exam 2021 based on new CBSE Curriculum 2021-2021 for CBSE Board, MP Board, Gujrat UP Board, Uttarakhand Board and Bihar Board, who are following NCERT Books 2021-22 for their exams.

11th Maths Chapter 5 Solutions in English Medium
11th Maths Chapter 5 Solutions in Hindi Medium
11th Maths Chapter 5 Solutions in PDF

Class 11 Maths Chapter 5 Exercise 5.1 Solution in Videos

Class 11 Maths Chapter 5 Exercise 5.1 SolutionClass 11 Maths Exercise 5.1 Solution in Hindi

Class 11 Maths Chapter 5 Exercise 5.2 Solution in Videos

Class 11 Maths Chapter 5 Exercise 5.2 SolutionClass 11 Maths Exercise 5.2 Solution in Hindi

Class 11 Maths Chapter 5 Exercise 5.3 Solution in Videos

Class 11 Maths Chapter 5 Exercise 5.3 SolutionClass 11 Maths Exercise 5.3 Solution in Hindi

Class 11 Maths Chapter 5 Miscellaneous Exercise Solution in Videos

Class 11 Maths Chapter 5 Miscellaneous Exercise SolutionClass 11 Maths Chapter 5 Miscellaneous Solution in Hindi

What is meant by Argand plane?

The coordinate plane that represents the complex numbers is called the complex plane or the Argand plane.

Important Terms on Complex Numbers

1. 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).
2. a + ib = c + id if a = c, and b = d
3. If z1 = a + ib, z2 = c + id. In general, we cannot compare and say that z1> z2 or z1 c then z1> z2, i.e. we can compare two complex numbers only if they are purely real.
i). z = 0 + i 0 is additive identity of a complex number.
ii). –z = –a –ib is called the Additive Inverse or negative of z = a + ib.
iii). z = 1 + i 0 is multiplicative identity of complex number.

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Important Questions on 11th Maths Chapter 5

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

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\”, \”dateCreated\”: \”February 23, 2020\”, \”url\”: \”https://www.tiwariacademy.com/ncert-solutions/class-11/maths/chapter-5/\”, \”author\”: {\”@type\”:\”Person\”,\”name\”:\”Rakesh Tiwari\” }}, \”suggestedAnswer\”: {\”@type\”: \”Answer\”, \”upvoteCount\”: \”1\”, \”text\”: \”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\”, \”dateCreated\”: \”February 23, 2020\”, \”url\”: \”https://www.tiwariacademy.com/ncert-solutions/class-11/maths/chapter-5/\”, \”author\”: {\”@type\”:\”Person\”,\”name\”:\”Rakesh Tiwari\” }}, \”author\”: {\”@type\”:\”Person\”,\”name\”:\”Rakesh Tiwari\” }}]}]]>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 ]\”, \”dateCreated\”: \”February 23, 2020\”, \”url\”: \”https://www.tiwariacademy.com/ncert-solutions/class-11/maths/chapter-5/\”, \”author\”: {\”@type\”:\”Person\”,\”name\”:\”Rakesh Tiwari\” }}, \”suggestedAnswer\”: {\”@type\”: \”Answer\”, \”upvoteCount\”: \”1\”, \”text\”: \”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 ]\”, \”dateCreated\”: \”February 23, 2020\”, \”url\”: \”https://www.tiwariacademy.com/ncert-solutions/class-11/maths/chapter-5/\”, \”author\”: {\”@type\”:\”Person\”,\”name\”:\”Rakesh Tiwari\” }}, \”author\”: {\”@type\”:\”Person\”,\”name\”:\”Rakesh Tiwari\” }}]}]]>

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