NCERT Solutions Class 11 Maths Chapter 5

NCERT Solutions Class 11 Mathematics Chapter 5- Complex Numbers and Quadratic Equations

The students must not underestimate the importance of revision. They must plan their study schedule in advance so that they can cover all the concepts essential for the examination. For students studying Mathematics, it is advised to revise all the important questions, theorems, derivation, and formulae and practice several problems related to each concept regularly. The NCERT Solutions Class 11 Mathematics Chapter 5 is an ultimate set of study material which covers all the necessary information and formulas.

The NCERT Solutions Class 11 Mathematics Chapter 5 talks about various concepts related to Complex numbers and Quadratic equations. To gain expertise in this chapter, students must refer to the NCERT Solutions Class 8, NCERT Solutions Class 9 and NCERT Solutions Class 10 to recall the concepts related to linear equations and quadratic equations. 

The NCERT Solutions Class 11 Mathematics Chapter 5 help students gain conceptual understanding and attain high scores. With the help of these CBSE revision notes, students can comprehend all important concepts and theorems from this chapter. Students must consider using several academic notes provided by Extramarks to enhance their preparation. 

 

Key Topics Covered In NCERT Solutions Class 11 Mathematics Chapter 5

In Extramarks NCERT Solutions for Class 11 Mathematics Chapter 5, students can be assured that all the concepts are defined and explained in detail. 

The subject matter experts at Extramarks have covered the following key topics in the ch 5 Mathematics Class 11 notes:

Exercise Topic
5.1 Introduction
5.2 Complex numbers
5.3 Algebra of Complex Numbers
5.4 Modulus and the Conjugate
5.5 Argand Plane and Polar Representation
5.6 Quadratic Equations

 

5.1 Introduction

This chapter helps students to understand the types of numbers: Imaginary and Complex. The chapter includes the basic introduction to the algebra of complex numbers, the power of i, modulus, conjugate and polar representation of complex numbers. Using the NCERT Solutions Class 11 Mathematics Chapter 5, students can easily ace their exams. Students will learn about the argand plane and solution of quadratic equations so that they can prepare for ch 5 Mathematics Class 11 thoroughly.  

 

5.2 Complex numbers

In this section, the definition of complex numbers and their representation in the general form are given. Also, the concepts of the Real part (Re z) and the Imaginary part (Img z) are explained here. Solved examples have also been given for better understanding of the concept. 

Refer to the NCERT Solutions Class 11 Mathematics Chapter 5 to understand each concept thoroughly. 

 

5.3 Algebra of Complex Numbers

This section is the most important in the entire chapter. Students will learn to develop algebraic operations for complex numbers. Firstly, the addition of two numbers is explained. The properties under addition are given below: Closure, the commutative law, the associative law, the existence of additive identity and the existence of an additive inverse. 

This section in the NCERT Solutions Class 11 Mathematics Chapter 5 further explains the concept of the difference between two complex numbers and the properties it holds. Furthermore, students will gain information about the multiplication of given complex numbers. The properties of multiplication are explained in detail using proofs. The properties of multiplication are the closure law, the commutative law, the associative law, the existence of a multiplicative identity, the existence of multiplicative inverse and the distributive law. This section in the NCERT Solutions Class 11 Mathematics Chapter 5 further explains the division of two complex numbers. 

Students will also learn about the Power and Values of i. In this section, the square roots of negative real numbers are only mentioned. 

Several solved examples are used to help students understand the special identities of complex numbers along. These identities are proved in a stepwise manner. 

 

5.4 Modulus and the Conjugate

The modulus and conjugate of the complex number z = a + ib are explained in this topic. The modulus is denoted by |z|, whereas the conjugate is z. Also, the multiplicative inverse of the complex number z is mentioned in this section. 

The properties of modulus and conjugate are given below: 

  1. |z1.z2|= |z1| .|z2|
  2. z1z2=z1z2, where |z2| 0
  3. z1.z2 = z1 . z2
  4. z1z2 = z1 z2
  5. (z1z2) = z1z2, where z2 0

Unlimited problems are included in the NCERT Solutions Class 11 Mathematics Chapter 5 for students to practice. 

 

5.5  Argand Plane and Polar Representation

In this section, the Argand plane is defined. This section geometrically represents the complex

number x + iy corresponding to the ordered pair (x, y). The x and y-axes in the Argand plane are known as the real axis and the imaginary axis, respectively. The mirror image of a point is also explained in this section. In the Class 11 Mathematics chapter 5 notes, students will learn how to represent a complex number in the polar form. Concepts such as the principal arguments and amplitude of a complex number are explained in the NCERT Solutions. Students have to answer several questions based on this topic. 

 

5.6 Quadratic Equations

In this section, students will learn about quadratic or polynomial equations with degree n having n roots. Several solved examples are included in the NCERT Solutions Class 11 Mathematics Chapter 5. The Fundamental Theorem of Algebra is also mentioned in this topic. 

 

List of NCERT Solutions Class 11 Mathematics Chapter 5 Exercise & Solutions

Students can access the NCERT Solutions Class 11 Mathematics Chapter 5 Complex numbers and Quadratic equations on the Extramarks mobile application and web portal. The notes include a detailed explanation of all the concepts, several practice problems and solved examples for students to understand each topic clearly. Chapter 5 Class 11 Mathematics notes enable the students to quickly revise and glance through all important formulas and derivations of the chapter. Using these NCERT Solutions Class 11 Mathematics Chapter 5, students will be able to tackle all questions and solve them on their own. 

Click on the below links to gain full details about the NCERT Solutions Class 11 Mathematics Chapter 5: 

 

Students can also explore other study materials such as CBSE sample papers, mock tests, etc. Students preparing for the board exams can use the NCERT Solutions Class 12 and also the latest edition of NCERT reference books on our Extramarks website: 

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7

 

NCERT Exemplar Class 11 Mathematics

The NCERT Exemplar Mathematics notes are one of the most important reference materials. Extramarks have provided the NCERT Exemplar for students to gain an in-depth understanding of any particular topic. It includes several tricky and challenging questions which enhance the thinking and problem-solving ability of students. Along with the NCERT Exemplar, all students can also refer to the NCERT Solutions Class 11 Mathematics Chapter 5 to prepare effectively for both board exams and entrance examinations. 

Extramarks aims to provide the best learning experience to every student. Therefore, we provide guidance and support through our academic notes and all-inclusive NCERT Solutions for Class 11 Mathematics chapter 5. Students are also advised to solve several question papers and sample papers for extra practice. 

Key Features of NCERT Solutions Class 11 Mathematics Chapter 5

  • In the NCERT Solutions Class 11 Mathematics Chapter 5, every problem has been solved using reasoning and logic. 
  • The information in the notes strictly adheres to the CBSE syllabus, marking system, and weightage allotted for each concept. 
  • Studying with the help of the NCERT Solutions Class 11 Mathematics Chapter 5 aids students to boost confidence so that they can establish a strong command of the subject. 
  • Students will also learn tips, tricks and shortcut techniques essential for solving Mathematics problems without making any silly mistakes. 
  • Students can effectively prepare for their annual examinations using the NCERT Solutions Class 11 Mathematics Chapter 5. 
  • These notes prepare them to tackle any question in the exam easily. 

Q.1 Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

1.5i35i2.„i9+i193.„i394.„3(7+i7)+i(7+i7)5.„(1i)(1+i6)6.„15+i254+i527.„13+i73+4+i1343+i8.„(1i)49.13+3i310.213i3

Ans

1.

(5i)(35i)=5×35i2=3(1)[∵  i2=1]=3+0i,   which is in the form of a+ib.

2.

     i9+ i19=i4×2+1+i4×4+3=(i4)2i+(i4)4i3=(1)2×i+(1)4(i)[∵i4=1 and i3= i]=ii=0+0i, which is in the form of a+ib.

3.

  i 39=1i39 =1i4×9+3=1(i4)9.i3=1(1)9.(i)=1i×ii=ii2=i1=0+i, which is in the form of a+ib.

4.
3(7 + i 7)+ i(7 + i 7) = 21 + 21i + 7i + 7i2
= 21 + 28i + 7(– 1)
= 21 + 28i – 7
= 14 + 28i, which is in the form of a + ib.

5.
(1 – i) – (–1 + i 6) = 1 – i + 1 – i 6
= 2 – 7i, which is in the form of a + ib.

6.

(15+i25)(4+i52)=(154)+i(2552)=195+i(42510)=195i2110,which is in the form of a+ib.

7.

[(13+i73)+(4+i13)](43+i)=(13+4+43)+i(73+131)    =(1+12+43)+i(7+133)    =(173)+i(53),Which is in the form of a+ib.

8.

       (1i)4={(1i)2}2=(12i+i2)2 =(12i1)2=(2i)2=4i2=4(1)=4+i0,which is in the form of a+ib.

9.

(13+3i)3=(13)3+3(13)2(3i)+3(13)(3i)2+(3i)3[∵(a+b)3=a3+3a2b+3ab2+b3]=127+i+i29+i327=127+i9i27=(1279)+i(127)=(124327)+i(26)=(24227)i26which is in the form of a+ib.

10.

(213i)3=(2)3+3(2)2(13i)+3(2)(13i)2+(13i)3[∵(a+b)3=a3+3a2b+3ab2+b3]=84i23i2127i3=84i23(1)127(i)[∵i2=1 and i3=i]=8+234i+127i=223(108127)i=223(10727)i,which is in the form of a+ib.

Q.2 Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. 4 – 3i
12.

5+3i

13. – i

Ans

11.

Multiplicative inverse of (43i)=1(43i)×(4+3i)(4+3i)=(4+3i)(429i2) [∵ (ab)(a+b)=a2b2]=(4+3i)(16+9)=425+325iThus, the multiplicative inverse of (43i) is (425+325i).

12.

Multiplicative inverse of (5+3i)=1(5+3i)×(53i)(53i)=(53i){(5)2(3i)2}=(53i){59i2}=(53i)(5+9)=514i314Thus, the multiplicative inverse of (5+3i) is (514i314).

13.

Multiplicative inverse of (i)=1(i)×ii=ii2=i(1)=i1=i,Thus, the multiplicative inverse of i is i.

Q.3

Express the following expression in the form of a+ib:3+i53-i53+2i32i

Ans

(3+i5)(3i5)(3+2i)(32i)=32(i5)23+2i3+2i[∵(ab)(a+b)             =a2b2]      =(9i25)22i      ={9(1)5}22i       =1422i      =72i×2i2i      =72ii2(2)      =72i(1)2[∵i2=1]      =072i2,  which in the form of a + ib.

Q.4 Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

1.z=1i32.„z=3+i

Ans

1.

z=1i3Comparing with z=r(cosθ+isinθ), we getrcosθ=1 and rsinθ=3Squaring and adding, we getr2(cos2θ+sin2θ)=(1)2+(3)2r2=1+3  r=2So,‹        modulus=2

   cosθ=12 and sinθ=32  cosθ=cosπ3  and sinθ=sinπ3Since, sinθ and cosθ are negative in III quadrant.So, θ=(ππ3)=2π3Thus, modulus and argument of (1i3) is 2 and 2π3 respectively.

2.

z=3+iComparing with z=r(cosθ+isinθ), we getrcosθ=3 and rsinθ=1Squarring and adding, we getr2(cos2θ+sin2θ)=(3)2+(1)2r2=3+1=4  r=2So,‹         modulus=2   cosθ=32 and sinθ=12  cosθ=cosπ6  and sinθ=sinπ6Since, sinθ is positive and cosθ is negative in II quadrant.

So, θ=(ππ6)=5π6Thus, modulus and argument of (3+i) is 2 and 5π6respectively.

Q.5 Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

3. 1 – i
4. – 1 + i
5. – 1 – i
6. – 3
7.

3+i

8. i

Ans

3.

Letz=1iComparing with z=r(cosθ+isinθ), we getrcosθ=1 and rsinθ=1Squaring and adding, we getr2(cos2θ+sin2θ)=(1)2+(1)2r2=1+1  r=2   cosθ=12 and sinθ=12  cosθ=cosπ4  and sinθ=sinπ4Since, cosθ is positive and sinθ is negative in IV quadrant.So, θ=π4[∵π<θπ]Thus, polar form of z=2{cos(π4)+isin(π4)}.

4.

Letz=1+iComparing with z=r(cosθ+isinθ), we getrcosθ=1 and rsinθ=1Squaring and adding, we getr2(cos2θ+sin2θ)=(1)2+(1)2 r2=1+1  r=2   cosθ=12 and sinθ=12  cosθ=cosπ4  and sinθ=sinπ4Since, sin θ is positive and cos θ is negative in II quadrant.So, θ=ππ4=3π4Thus, polar form of z=2{cos(3π4)+isin(3π4)}.

5.

Letz=1iComparing with z=r(cosθ+isinθ), we getrcosθ=1 and rsinθ=1Squaring and adding, we getr2(cos2θ+sin2θ)=(1)2+(1)2r2=1+1  r=2   cosθ=12 and sinθ=12  cosθ=cosπ4  and sinθ=sinπ4Since, sin θ and cos θ is negative in III quadrant. So, θ=(ππ4)=3π4[∵π<θπ]Thus, polar form of z=2{cos(3π4)+isin(3π4)}.

6.

Letz=3+0iComparing with z=r(cosθ+isinθ), we getrcosθ=3 and rsinθ=0Squaring and adding, we getr2(cos2θ+sin2θ)=(3)2+(0)2r2=9  r=3   cosθ=33=1 and sinθ=0  cosθ=cos0  and sinθ=sin0Since, sin θ is positive and cos θ is negative in II quadrant.So, θ=(π0)=πThus, polar form of z=3{cosπ+isinπ}.

7.

Letz=3+iComparing with z=r(cosθ+isinθ), we getrcosθ=3 and rsinθ=1Squaring and adding, we get r2(cos2θ+sin2θ)=(3)2+(1)2r2=4  r=2   cosθ=32 and sinθ=12  cosθ=cosπ6  and sinθ=sinπ6So, θ=  π6Thus, polar form of z=2{cosπ6+isinπ6}.

8.

Letz=0+iComparing with z=r(cosθ+isinθ), we getrcosθ=0 and rsinθ=1Squaring and adding, we getr2(cos2θ+sin2θ)=(0)2+(1)2r2=1  r=1   cosθ=0 and sinθ=1  cosθ=cosπ2  and sinθ=sinπ2So, θ=  π2 Thus, polar form of z=1{cosπ2+isinπ2}.

Q.6 Solve the following equation:
x2 + 3 = 0

Ans

x 2 +3 =0 comparing with ax 2 +bx+c=0, we get a=1,b=0 and c=3 By using quadratic formula, x= b± b 2 4ac 2a = 0± 0 2 4( 1 )( 3 ) 2( 1 ) =± 12 2 =± 2i 3 2 [ ∵ 1 =i ] x=±i 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@CACA@

Q.7 Solve the following equation:
2x2 + x + 1 = 0

Ans

2x 2 +x+1=0 comparing with ax 2 +bx+c=0, we get a=2,b=1 and c=1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6DD7@ By using quadratic formula, x= b± b 2 4ac 2a = 1± 1 2 4( 2 )( 1 ) 2( 2 ) = 1± 18 4 = 1± 7 4 x= 1±i 7 4 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9F8C@

Q.8 Solve the following equation:
x2 + 3x + 9 = 0

Ans

x 2 +3x+9=0 comparing with ax 2 +bx+c=0, we get a=1,b=3 and c=9 By using quadratic formula, x= b± b 2 4ac 2a = 3± 3 2 4( 1 )( 9 ) 2( 1 ) = 3± 936 2 = 3± 27 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BDCB@ x= 3±3 3 i 2 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpdaWcaaqaaiabgkHiTiaaiodacqGHXcqScaaIZaWaaOaaaeaacaaIZaaaleqaaOGaaGPaVlaadMgaaeaacaaIYaaaaiaaxMaacaWLjaWaamWaaeaacqWI1isudaGcaaqaaiabgkHiTiaaigdaaSqabaGccqGH9aqpcaWGPbaacaGLBbGaayzxaaaaaa@5372@

Q.9 Solve the following equation:
– x2 + x – 2 = 0

Ans

x 2 +x2=0 comparing with ax 2 +bx+c=0, we get a=1,b=1 and c=2 By using quadratic formula, x= b± b 2 4ac 2a = 1± 1 2 4( 1 )( 2 ) 2( 1 ) = 1± 18 2 = 1± 7 2 x= 1±i 7 2 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD93@

Q.10 Solve the following equation:
x2 + 3x + 5 = 0

Ans

x 2 +3x+5=0 comparing with ax 2 +bx+c=0, we get a=1,b=3 and c=5 By using quadratic formula, MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@85D8@ x= b± b 2 4ac 2a = 3± 3 2 4( 1 )( 5 ) 2( 1 ) = 3± 920 2 = 1± 11 2 x= 3±i 11 2 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8A60@

Q.11 Solve the following equation:
x2 – x + 2 = 0

Ans

x 2 x+2=0 Comparing with ax 2 +bx+c=0, we get a=1,b=1 and c=2 By using quadratic formula, x= b± b 2 4ac 2a = ( 1 )± ( 1 ) 2 4( 1 )( 2 ) 2( 1 ) = 1± 18 2 = 1± 7 2 x= 1± 7 i 2 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabIhadaahaaWcbeqaaiaabkdaaaGccaGGtaIaaeiEaiabgUcaRiaabkdacqGH9aqpcaaIWaaabaGaam4qaiaad+gacaWGTbGaamiCaiaadggacaWGYbGaamyAaiaad6gacaWGNbGaaeiiaiaabEhacaqGPbGaaeiDaiaabIgacaqGGaGaaeyyaiaabIhadaahaaWcbeqaaiaabkdaaaGccqGHRaWkcaWGIbGaamiEaiabgUcaRiaadogacqGH9aqpcaaIWaGaaiilaiaabccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzaiaabshaaeaacaqGHbGaeyypa0JaaGymaiaacYcacaaMc8UaaGPaVlaadkgacqGH9aqpcqGHsislcaaIXaGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGJbGaeyypa0JaaGOmaaqaaiaadkeacaWG5bGaaeiiaiaabwhacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabghacaqG1bGaaeyyaiaabsgacaqGYbGaaeyyaiaabshacaqGPbGaae4yaiaabccacaqGMbGaae4BaiaabkhacaqGTbGaaeyDaiaabYgacaqGHbGaaeilaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabIhacqGH9aqpdaWcaaqaaiabgkHiTiaadkgacqGHXcqSdaGcaaqaaiaadkgadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamyyaiaadogaaSqabaaakeaacaaIYaGaamyyaaaaaeaacaWLjaGaeyypa0ZaaSaaaeaacqGHsisldaqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGHXcqSdaGcaaqaamaabmaabaGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaisdadaqadaqaaiaaigdaaiaawIcacaGLPaaadaqadaqaaiaaikdaaiaawIcacaGLPaaaaSqabaaakeaacaaIYaWaaeWaaeaacaaIXaaacaGLOaGaayzkaaaaaaqaaiaaxMaacqGH9aqpdaWcaaqaaiaaigdacqGHXcqSdaGcaaqaaiaaigdacqGHsislcaaI4aaaleqaaaGcbaGaaGOmaaaacqGH9aqpdaWcaaqaaiaaigdacqGHXcqSdaGcaaqaaiabgkHiTiaaiEdaaSqabaaakeaacaaIYaaaaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamiEaiabg2da9maalaaabaGaaGymaiabgglaXoaakaaabaGaaG4naaWcbeaakiaaykW7caWGPbaabaGaaGOmaaaacaWLjaGaaCzcaiaaxMaacaWLjaWaamWaaeaacqWI1isudaGcaaqaaiabgkHiTiaaigdaaSqabaGccqGH9aqpcaWGPbaacaGLBbGaayzxaaaaaaa@D92C@

Q.12 Solve the following equation:

2 x 2 +x+ 2 =0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaGcaaqaaGqabiaa=jdaaSqabaGccaWF4bWaaWbaaSqabeaacGaMa+Nmaaaakiaa=TcacaWF4bGaa83kamaakaaabaGaa8NmaaWcbeaakiaa=1dacaWFWaaaaa@4189@

Ans

2 x 2 +x+ 2 =0 Comparing with ax 2 +bx+c=0, we get a= 2 ,b=1 and c= 2 By using quadratic formula, x= b± b 2 4ac 2a = 1± 1 2 4( 2 )( 2 ) 2( 2 ) = 1± 18 2 2 = 1± 7 2 2 x= 1±i 7 2 2 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaamaakaaabaGaaGOmaaWcbeaakiaadIhadaahaaWcbeqaaiacyciIYaaaaOGaey4kaSIaamiEaiabgUcaRmaakaaabaGaaGOmaaWcbeaakiabg2da9iaaicdaaeaacaWGdbGaam4Baiaad2gacaWGWbGaamyyaiaadkhacaWGPbGaamOBaiaadEgacaqGGaGaae4DaiaabMgacaqG0bGaaeiAaiaabccacaqGHbGaaeiEamaaCaaaleqabaGaaeOmaaaakiabgUcaRiaadkgacaWG4bGaey4kaSIaam4yaiabg2da9iaaicdacaGGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaabggacqGH9aqpdaGcaaqaaiaaikdaaSqabaGccaGGSaGaaGPaVlaaykW7caWGIbGaeyypa0JaaGymaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaae4yaiabg2da9maakaaabaGaaGOmaaWcbeaaaOqaaiaadkeacaWG5bGaaeiiaiaabwhacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabghacaqG1bGaaeyyaiaabsgacaqGYbGaaeyyaiaabshacaqGPbGaae4yaiaabccacaqGMbGaae4BaiaabkhacaqGTbGaaeyDaiaabYgacaqGHbGaaeilaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabIhacqGH9aqpdaWcaaqaaiabgkHiTiaadkgacqGHXcqSdaGcaaqaaiaadkgadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamyyaiaadogaaSqabaaakeaacaaIYaGaamyyaaaaaeaacaWLjaGaeyypa0ZaaSaaaeaacqGHsislcaaIXaGaeyySae7aaOaaaeaacaaIXaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGinamaabmaabaWaaOaaaeaacaaIYaaaleqaaaGccaGLOaGaayzkaaWaaeWaaeaadaGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaSqabaaakeaacaaIYaWaaeWaaeaadaGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaaaabaGaaCzcaiabg2da9maalaaabaGaeyOeI0IaaGymaiabgglaXoaakaaabaGaaGymaiabgkHiTiaaiIdaaSqabaaakeaacaaIYaWaaOaaaeaacaaIYaaaleqaaaaakiabg2da9maalaaabaGaeyOeI0IaaGymaiabgglaXoaakaaabaGaeyOeI0IaaG4naaWcbeaaaOqaaiaaikdadaGcaaqaaiaaikdaaSqabaaaaaGcbaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG4bGaeyypa0ZaaSaaaeaacqGHsislcaaIXaGaeyySaeRaamyAamaakaaabaGaaG4naaWcbeaaaOqaaiaaikdadaGcaaqaaiaaikdaaSqabaaaaOGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaWaamWaaeaacqWI1isudaGcaaqaaiabgkHiTiaaigdaaSqabaGccqGH9aqpcaWGPbaacaGLBbGaayzxaaaaaaa@DB95@

Q.13 Solve the following equation:

3 x 2 2 x+3 3 =0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaGcaaqaaGqabiaa=ndaaSqabaGccaWF4bWaaWbaaSqabeaacGaMa+NmaaaakiabgkHiTmaakaaabaGaa8NmaaWcbeaakiaa=HhacaWFRaGaa83mamaakaaabaGaa83maaWcbeaakiaa=1dacaWFWaaaaa@4358@

Ans

3 x 2 2 x+3 3 =0 Comparing with ax 2 +bx+c=0, we get a= 3 ,b= 2 and c=3 3 By using quadratic formula, x= b± b 2 4ac 2a MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A12A@ = ( 2 )± ( 2 ) 2 4( 3 )( 3 3 ) 2( 3 ) = 2 ± 236 2 3 = 2 ± 34 2 3 x= 2 ± 34 i 2 3 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7C74@

Q.14 Solve the following equation:

x 2 +x+ 1 2 =0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWF4bWaaWbaaSqabeaacGaMa+Nmaaaakiaa=TcacaWF4bGaa83kamaalaaabaGaaGymaaqaamaakaaabaGaa8NmaaWcbeaaaaGccaWF9aGaa8hmaaaa@417C@

Ans

x 2 +x+ 1 2 =0 Comparing with ax 2 +bx+c=0, we get a=1,b=1 and c= 1 2 By using quadratic formula, x= b± b 2 4ac 2a = ( 1 )± ( 1 ) 2 4( 1 )( 1 2 ) 2( 1 ) = 1± 12 2 2 = 1± 12 2 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C4B1@ x= 1±i 2 2 1 2 [ ∵ 1 =i ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpdaWcaaqaaiabgkHiTiaaigdacqGHXcqScaWGPbWaaOaaaeaacaaIYaWaaOaaaeaacaaIYaaaleqaaOGaeyOeI0IaaGymaaWcbeaaaOqaaiaaikdaaaGaaCzcaiaaxMaadaWadaqaaiablwJirnaakaaabaGaeyOeI0IaaGymaaWcbeaakiabg2da9iaadMgaaiaawUfacaGLDbaaaaa@53B0@

Q.15 Solve the following equation:

x2+x2+1=0

Ans

x2+x2+1=0Comparing with ax2+bx+c=0, we geta=1,  b=12 and c=1By using quadratic formula,x=b±b24ac2a =(12)±(12)24(1)(1)2(1) =12±1242 =12±722     x=1±i722 [∵ 1=i]

Q.16 Find the square root of the following: – 15 – 8i

Ans

Let‹   x+iy=158iSquarring both sides, we get      (x+iy)2=158ix2+2ixy+(iy)2=158i    x2+2ixyy2=158i    x2y2+2ixy=158iComparing real and imaginary parts from both sides, we get    x2y2=15 …(i)  and     xy=4(x2+y2)2=(x2y2)2+4(xy)2=(15)2+4(4)2=225+64=289  x2+y2=17      …(ii)Adding equation (i) and equation(ii)  2x2=2x=±1and  putting x=1 in xy=4, we get (1)y=4    y=4putting x=1 in xy=4, we get(1)y=4    y=4 x+iy=14i or 1+4i158i=14i or 1+4i

Q.17 Find the square root of the following: – 8 – 6i.

Ans

Let‹   x+iy= 86iSquaring both sides, we get      (x+iy)2= 86ix2+2ixy+(iy)2= 86i    x2+2ixyy2= 86i    x2y2+2ixy= 86iComparing real and imaginary parts from both sides, we get    x2y2=8 ...(i)   and     xy=3(x2+y2)2=(x2y2)2+4(xy)2=(8)2+4(3)2=64+36=100  x2+y2=10     ...(ii)Adding equation (i) and equation(ii)   2x2=2x=±1and  putting x=1 in xy=6, we get(1)y=3    y=3putting x=1 in xy=3, we get(1)y=3    y=3 x+iy=13i or 1+3i86i=13i or 1+3i

Q.18 Find the square root of the following: 1 – i

Ans

Let‹   x+iy=1iSquarring both sides, we get      (x+iy)2= 1ix2+2ixy+(iy)2= 1i    x2+2ixyy2=1i    x2y2+2ixy=1iComparing real and imaginary parts from both sides, we get    x2y2=1 …(i)and     xy=12(x2+y2)2=(x2y2)2+4(xy)2=(1)2+4(12)2 =1+1=2  x2+y2=2     …(ii)Adding equation (i) and equation(ii)  2x2=1+2x=±1+22and  putting x=1+22 in xy=12, we get(1+22)y=12   y=12×22+1   y=2+12(2+1)×2121   y=2+12Putting x=1+22 in xy=12, we get(1+22)y=12   y=12×22+1     y=2+12(2+1)×2121    y=2+12 x+iy=1+222+12i or 1+22+2+12i86i=1+222+12i or 1+22+2+1286i=±1+22∓2+12i

Q.19 Find the square root of the following: –i

Ans

Letx+iy=iSquaring both sides, we get(x+iy)2=0ix2+2ixy+(iy)2=0ix2+2ixyy2=0ix2y2+2ixy=0iComparing real and imaginary parts from both sides, we getx2y2=0 ...(i)and xy=12(x2+y2)2=(x2y2)2+4(xy)2 ‹‹‹‹‹‹‹‹‹‹„„„„„„„‹y=2+122+1×2121‹‹‹‹‹‹‹‹‹‹‹„„„„„„„y=2+12„„„x+iy=1+222+12i„or1+22+2+12i1i=1+222+12i„or1+22+2+12i1i=±1+22∓2+12i

Q.20 Find the square root of the following: i

Ans

Let‹   x+iy=iSquaring both sides, we get      (x+iy)2= 0+ix2+2ixy+(iy)2= 0+i    x2+2ixyy2=0+i    x2y2+2ixy=0+iComparing real and imaginary parts from both sides, we get      x2y2=0 …(i)and      xy=12(x2+y2)2=(0)2+4(12)2=1  x2+y2=1      …(ii)Adding equation (i) and equation(ii)  2x2=1x=±12and  putting x=12 in xy=12, we get12y=12      y=12 Putting x=12 in xy=12, we get  (12)y=12         y=12 x+iy=1212i or 1212ii=12+12i or 1212ii=±12±12i

Q.21 Find the square root of the following: 1 + i

Ans

Let‹   x+iy=1+iSquaring both sides, we get      (x+iy)2= 1+ix2+2ixy+(iy)2= 1+i    x2+2ixyy2=1+i    x2y2+2ixy=1+iComparing real and imaginary parts from both sides, we get    x2y2=1 …(i)and     xy=12(x2+y2)2=(x2y2)2+4(xy)2 (x2+y2)2=(1)2+4(12)2=1+1    x2+y2=2 ...(ii)Adding equation(i) and equation(ii), we get2x2=1+2        x2=1+22        x=±2+12Putting value of x in xy=12, we gety=12x   =12×±22+1            =±122+1×2121            =±212(21)        y=±212 x+iy=±2+12±212i1+i=±2+12±212i

Q.22

Evaluate:   [i18+(1i)25]3

Ans

   [i18+(1i)25]3=[i4×4+2+(1i4×6+1)]3=[(i4)4i2+(1i4)61i]3=[(1)4(1)+(11)6×1i]3 [∵ i4=1]=[1+1i×ii]3=[1+ii2]3=[1i]3 [∵ i2=1]=(1)3+3(1)2(i)+3(1)(i)2+(i)3[∵ (a+b)3=a3+3a2b+3ab2+b3] =13i+3+i [∵ (i)2=1,(i)3=i]=22i

Q.23 For any two complex numbers, z1 and z2, prove that
Re (z1 z2) = Re z1 Re z2 – Imz1 Imz2.

Ans

Let‹ z1=a+ib and z2=c+idThen, z1.z2=(a+ib)(c+id)   =ac+iad+ibc+i2bd   =ac+i(ad+bc)bd   =(acbd)+i(ad+bc)    Re(z1z2)  =acbd   =(a)(c)(b)(d)   =Re(z1)Re(z2)Im(z1)Im(z2)[∵Im(z1)=b,    Im(z2)=d]Thus,Re(z1z2)  =Re(z1)Re(z2)Im(z1)Im(z2)

Q.24

Reduce  11 – 4i21 + i3 – 4i5 + i to the standard form.

Ans

(114i21+i)(34i5+i)=(114i×1+4i1+4i21+i×1i1i)(34i5+i×5i5i)       =(1+4i1(4i)22(1i)12i2)((34i)(5i)52i2)       =(1+4i116i22(1i)12(1))((1520i3i+4i2)52(1))       ={1+4i116(1)2(1i)1+1}[{1520i3i+4(1)}25+1]       ={1+4i1+162(1i)2}[(1123i)26]       ={1+4i17(1i)}[(1123i)26]       =117×26{1+4i17+17i}(1123i)       =117×26(16+21i)(1123i)       =1442(176+368i+231i483i2)       =1442(176+599i+483)(114i21+i)(34i5+i)=1442(307+599i)

Q.25

If xiy=aibcidprove that (x2+y2)2=a2+b2c2+d2.

Ans

Given:‹‹xiy=aibcidSquaring both sides, we getxiy2=aibcid2x22i‹„xy+i2y2=aibcid×c+idc+id„„x22i‹„xyy2=aibc+idc2i2d2„„x2y22i‹„xy=acibc+iadi2bdc21d2‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹„„„„„„„„„„„„„„„„„„„„„„=acibc+iad+bdc2+d2„‹‹‹‹Z‹i2=1„„„„„„„„„„„„„„„„„„„„„„‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=ac+bdc2+d2ibcadc2+d2Comparingrealpartofbothsides,wegetx2y2=ac+bdc2+d2We know that x2+y22=x2y22+2xy2So x2+y22=ac+bdc2+d22+bcadc2+d22‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=a2c2+b2d2+2abcdc2+d22+b2c2+a2d22abcdc2+d22‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=a2c2+b2d2+2abcd+b2c2+a2d22abcdc2+d22‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=a2c2+b2d2+b2c2+a2d2c2+d22‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=a2c2+d2+b2c2+d2c2+d22‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=a2+b2c2+d2c2+d22‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹‹=a2+b2c2+d2x2+y22=a2+b2c2+d2Hence proved.

Q.26

Convert the following in the polar form:i1+7i2i2ii1+3i12i

Ans

(i)  1+7i(2i)2=1+7i(44i+i2)         =1+7i(44i1) =1+7i(34i)×(3+4i)(3+4i) =3+4i+21i+28i29(4i)2 =3+25i+28(1)9(4)2i2 =25+25i916(1) =25+25i9+16 =2525+2525i =1+iLet         z=1+iComparing with z=r(cosθ+isinθ), we getrcosθ=1 and rsinθ=1Squarring and adding, we getr2(cos2θ+sin2θ)=(1)2+(1)2r2=1+1   r=2   cosθ=12 and sinθ=12  cosθ=cosπ4  and sinθ=sinπ4 Since, sin θ is positive and cos θ is negative in II quadrant.So, θ=ππ4=3π4Thus, polar form of z=2{cos(3π4)+isin(3π4)}.(ii) 1+3i12i==1+3i12i×1+2i1+2i =1+2i+3i+6i214i2 =1+2i+3i+6(1)14(1) =1+5i61+4 =55+5i5 =1+iLet         z=1+iComparing with z=r(cosθ+isinθ), we getrcosθ=1 and rsinθ=1Squarring and adding, we getr2(cos2θ+sin2θ)=(1)2+(1)2r2=1+1   r=2   cosθ=12 and sinθ=12   cosθ=cosπ4  and sinθ=sinπ4Since, sin θ is positive and cos θ is negative in II quadrant.So, θ=ππ4=3π4Thus, polar form of z=2{cos(3π4)+isin(3π4)}.

Q.27 Solve each of the equation in Exercises 6 to 9.

6.3x24x+203=07.„x22x+32=08.„27x210x+1=09.„21x228x+10=0

Ans

6.

3x24x+203=0comparing with ax2+bx+c=0, we geta=3,  b=4 and c=203By using quadratic formula,x=b±b24ac2a =(4)±(4)24(3)(203)2(3) =4±16806=4±646 x=4±8i6 [∵ 1=i] =23±43i

7.

We  have,x22x+32=0comparing with ax2+bx+c=0, we geta=1,  b=2 and c=32By using quadratic formula,x=b±b24ac2a =(2)±(2)24(1)(32)2(1) =2±462=2±22x=2±2i2 [∵ 1=i] =1±22i

8.

We  have,27x210x+1=0 Comparing with ax2+bx+c=0, we geta=27,  b=10 and c=1By using quadratic formula,x=b±b24ac2a =(10)±(10)24(27)12(27) =10±10010854=10±854x=10±22i54 [∵ 1=i] =5±2i27x=527±227i

9.

We  have,21x228x+10=0Comparing with ax2+bx+c=0, we geta=21,  b=28 and c=10By using quadratic formula,x=b±b24ac2a =(28)±(28)24(21)(10)2(21) =28±78484042=28±5642x=28±214i42 [∵ 1=i] =14±14i21 =1421±1421ix=23±1421i

Q.28

Ifz1=2i,z2=1+i, find |z1+z2+1z1z2+1|.

Ans

Given:z1=2i,z2=1+iz1+z2+1z1z2+1=2i+1+i+12i1i+1=422i=21i×1+i1+i=2+2i1i2=2+2i1(1)=2+2i2=1+i|z1+z2+1z1z2+1|=|1+i|=(1)2+(1)2=1+1=2

Q.29

If a + ib = x + iy22x2+1, prove that a2+b2 = x2+122x2+12.

Ans

Given:  a+ib=(x+iy)22x2+1           =x2+2xyi+i2y22x2+1           =x2+2xyiy22x2+1           =x2y22x2+1+2xy2x2+1iTaking modulus of both sides, we get    |a+ib|=|x2y22x2+1+2xy2x2+1i|     a2+b2=(x2y22x2+1)2+(2xy2x2+1)2Squaring both sides, we get      a2+b2=(x2y22x2+1)2+(2xy2x2+1)2           =(x2y2)2+4x2y2(2x2+1)2           =x42x2y2+y4+4x2y2(2x2+1)2           =x4+2x2y2+y4(2x2+1)2           =(x2+y2)2(2x2+1)2Thus,      a2+b2=(x2+y2)2(2x2+1)2.

Q.30

Let z1= 2 – i,  z2 = -2 + i. Findi Rez1z2z¯1, ii Im 1z1z¯1

Ans

We„have,„„„„„„z1=2i,z1¯=2+i and z2=2+i„z1.z2=2i2+i„„„„„„„„„„=4+2i+2ii2„„„„„„„„„„=4+4i1„„„„„„„„„„=3+4iNow,z1z21=3+4i2+i„„„„„„„„„„„„„„„=3+4i2+i×2i2i„„„„„„„„„„„„„„„=3+4i2i22i2„„„„„„„„„„„„„„„=6+3i+8i4i24+1„„„„„„„„„„„„„„„=6+3i+8i415„„„„„„„„„„„„„„„=2+11i5„„„„„„„„„„„„„„„=25+115iRe z1z21=25„„„„„„„„„z11=2i2+i„„„„„„„„„„„„„„„=4i2„„„„„„„„„„„„„„„=41„„„„„„„„„„„„„„„=4+1„„„„„„„„„„„„„„„=5„„„„„„„„„1z11=15+0iIm1z11=0

Q.31

Find the modulus and argument of the complex number1+2i1-3i.

Ans

We„have,1+2i13i=1+2i13i×1+3i1+3i„„„„„„„„„=1+3i+2i+6i213i2„„„„„„„„„=1+5i+61191„„„„„„„„„=1+5i61+9„„„„„„„„„=5+5i10„„„„„„„„„=510+510i„„„„„„„„„=12+12iModulus12+12i„„„„„„„„„„=122+122„„„„„„„„„„=14+14„„„„„„„„„„=24„„„„„„„„„„=22argumentθ=tan11212If z=a+ibθ=tan1ba„„„„„„„„„„„„„„„„„„=tan11212„„„„„„„„„„„„„„„„„„=tan11=3π4

Q.32 Find the real numbers x and y if
(x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Ans

(xi y) (3+5i)=3x+5xi3yi5yi2  =3x+5xi3yi+5y  =(3x+5y)+(5x3y)iConjugate of (3x+5y)+(5x3y)i=(3x+5y)(5x3y)i   ...(i)Comparing equation(i) with 624i, we get3x+5y=6 and (5x3y)=243x+5y=6   ...(ii)5x3y=24 ...(iii)Multiplying equation(ii) by 3 and equation(iii) by 5, we get   9x+15y=18 ...(iv)25x15y=120 ...(v)Adding equation (iv) and equation (v), we get34 x=102x=10234=3Putting value of x in equation (ii),we get3(3)+5y=6   5y=69     y=155=3Thus, the values of x and y are 3 and 3 respectively.

Q.33

Find the modulus of  1 + i1 – i 1 i1 + i.

Ans

1+i1i1i1+i=1+i1i×1+i1+i1i1+i×1i1i =1+2i+i212i212i+i212i2 =1+2i11(1)12i11(1) =2i22i2 =i+i =2i |1+i1i1i1+i|=|0+2i|=0+4=2

Q.34

If (x + iy)3 = u + iv, then show that   ux+vy = 4x2 – y2.

Ans

Given:(x + iy)3 = u + ivx3+3x2(iy)+3x(iy)2+(iy)3=u + iv            x3+3ix2y3xy2iy3=u + iv      (x33xy2)+(3x2yy3)i=u + iv Comparing real and imaginary parts of both sides,we get      u=x33xy2 and v=3x2yy3ux=x23y2 and vy=3x2y2L.H.S.=ux+vy=x23y2+3x2y2=4x24y2=4(x2y2)=R.H.S.Hence proved.

Q.35

If α and β are different complex numbers with β =1, then find β-α1-α¯ β.

Ans

Since,|z|2=z.z¯ |βα1α¯β|2=(βα1α¯β)(βα1α¯β)¯=(βα1α¯β)(β¯α¯1α¯¯β¯)=(βα1α¯β)(β¯α¯1αβ¯) [∵ z¯¯=z]=ββ¯βα¯αβ¯+αα¯1αβ¯α¯β+α¯βαβ¯ =|β|2βα¯αβ¯+|α|21αβ¯α¯β+(αα¯)(ββ¯)[∵ zz¯=|z|2]=|β|2βα¯αβ¯+|α|21αβ¯α¯β+|α|2|β|2=(12)βα¯αβ¯+|α|21αβ¯α¯β+|α|2(12)=1βα¯αβ¯+|α|21αβ¯α¯β+|α|2.1=1βα¯αβ¯+|α|21αβ¯α¯β+|α|2=1|βα1α¯β|=1=1

Q.36

Find the number of non-zero integral solutions of theequation 1 – ix= 2x.

Ans

„„„„„„„„„„„„„„„1ix=2x12+12x=2x„„„„„„„„„„„„„2x=2x Squaring both sides, we get„„„„22x=22x„„„„„„„2x=22x„„„„„„„„x=2x„„„„„„x=0„„„„„„„„„x=0Thus, x=0.

Q.37 If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2

Ans

Given:          (a+ib) (c+id) (e+if) (g+ih)=A+iBTaking modulus of both sides, we get                 |(a+ib) (c+id) (e+if) (g+ih)|=|A+iB||(a+ib)|×|(c+id)|×|(e+if)|×|(g+ih)|=|A+iB|(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2Squaring both sides, we get      (a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2Hence proved.

Q.38

If 1 + i1 – im=1, then find the least positive integral„value of m.

Ans

We have,    (1+i1i)m=1   (1+i1i×1+i1+i)m=1     (1+2i+i21i2)m=1     (1+2i11(1))m=1            (2i2)m=1                      im=1                      im=i4k                     m=4kFor least positive integral value of m, k=1.So, m=4.

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FAQs (Frequently Asked Questions)

1. Which study material should I refer to for preparing for NCERT Solutions Class 11 Mathematics Chapter 5?

Students are advised to use the best academic notes provided by the Extramarks platform. Along with NCERT Solutions Class 11 Mathematics Chapter 5, students are advised to use the NCERT Exemplar notes and several reference books. They can also appear for several test series and solve CBSE sample papers and various CBSE past years’ question papers to assess themselves based on their preparation level. Below we have listed some important reference books. 

  1. Mathematics: for Class 11 by RS Agarwal
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  5. Higher Algebra by Hall & Knight

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Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 4 Principle of Mathematical Induction

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight Lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 15 Statistics

Chapter 16 Probability