NEETClass 11thClass 12thClass 12th PlusJEEClass 11thClass 12thClass 12th PlusClass 6-10Class 6thClass 7thClass 8thClass 9thClass 10thOnline CoursesDistance LearningInternational OlympiadNEETClass 11thClass 12thClass 12th PlusJEE (Main+Advanced)Class 11thClass 12thClass 12th PlusJEE MainClass 11thClass 12thClass 12th PlusClass 6-10Class 6thClass 7thClass 8thClass 9thClass 10thKCET/MHT-CETKCETMHT-CETNEET20262025202420232022JEE20262025202420232022Class 6-1020262025JEE MainPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DatePercentile PredictorAnswer KeyCounsellingEligibilityExam PatternJEE MathsJEE ChemistryJEE PhysicsJEE AdvancedPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DateAnswer KeyEligibilityExam PatternRank PredictorNEETPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DateCollege PredictorAnswer KeyRank PredictorCounsellingEligibilityExam PatternBiologyNCERT SolutionsClass 6Class 7Class 8Class 9Class 10Class 11Class 12TextbooksCBSEClass 12Class 11Class 10Class 9Class 8Class 7Class 6SubjectsSyllabusNotesSample PapersQuestion PapersICSEClass 10Class 9Class 8Class 7Class 6State BoardBiharKarnatakaMadhya PradeshMaharashtraTamilnaduWest BengalUttar PradeshOlympiadMathsScienceEnglishSocial ScienceNSOIMONMTCTALLENTEXASATInstant Online ScholarshipAIOT(NEET)ALLEN for SchoolsAbout ALLENBlogsNewsCareersRequest a call backBook a demo
  • Classroom Courses
  • NEW
  • ALLEN E-Store
NCERT Solutions
Class 11
Maths
Chapter 4 Complex Numbers and Quadratic Equations

Frequently Asked Questions

They provide the necessary foundation for understanding imaginary numbers, which are essential for solving higher-order equations and understanding electrical circuits in Physics.

These solutions cover the fundamental properties of i, modulus, and conjugate, which are core components of the Algebra section in JEE and are strictly followed in Board marking schemes.

The Argand plane is a geometric representation of complex numbers where the horizontal axis represents real parts and the vertical axis represents imaginary parts.

Yes they emphasise on conceptual clarity and speed building techniques for algebraic manipulation of complex numbers which are very important for JEE and BITSAT.

Complex numbers help learners comprehend numbers beyond the real number system and quadratic formulas help students to find roots when real solutions are not possible . These topics are important for higher level of mathematics and preparing for competitive exams.

Join ALLEN!

(Session 2026 - 27)

Choose class
Choose your goal
Preferred Mode
Choose State
  • About
    • About us
    • Blog
    • Allen News
    • Privacy policy
    • Public notice
    • Careers
    • Dhoni Inspires NEET Aspirants
    • Dhoni Inspires JEE Aspirants
  • Help & Support
    • Refund policy
    • Transfer policy
    • Terms & Conditions
    • Contact us
  • Popular goals
    • NEET Coaching
    • JEE Coaching
    • 6th to 10th
  • Courses
    • Classroom Courses
    • Online Courses
    • Distance Learning
    • Online Test Series
    • International Olympiads Online Course
    • NEET Test Series
    • JEE Test Series
    • JEE Main Test Series
  • Centers
    • Kota
    • Bangalore
    • Indore
    • Delhi
    • More centres
  • Exam information
    • JEE Main
    • JEE Advanced
    • NEET Exam
    • CBSE Notes
    • NIOS
    • NCERT Solutions
    • Olympiad
    • JEE Counselling
    • NEET Counselling
    • JEE Main Syllabus

ALLEN Career Institute Pvt. Ltd. © All Rights Reserved.

ISO

NCERT Solutions Class 11 Maths Chapter 4 – Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 4 (Complex Numbers and Quadratic Equation) makes students familiar with the numbers beyond the real number system. Students learn that the square root of a negative number is not a real number, but this chapter introduces the imaginary unit iota (i) to be able to solve such equations. Chapter 4 NCERT Solutions provides a clear understanding of the Complex Numbers and Quadratic Equations and offers detailed solutions to all NCERT textbook questions, helping students strengthen their concepts and practise effectively at their own pace.



The ALLEN NCERT Solutions for Class 11 Maths Chapter 4 PDF helps learners comprehend complex numbers with the help of clear explanations and step by step method. The solutions include Algebraic operations, Argand plane, Geometric representation of complex numbers and quadratic formulas with complex roots.

Competitive exams like JEE Main, JEE Advanced, BITSAT, etc. require a good understanding of complex numbers and Argand plane. These solutions cover important topics like modulus and conjugate of a complex number and powers of i which are frequently asked in exams

1.0Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations: Key Concepts

This chapter is concerned with the extension of the number system to include imaginary parts and with the solving of equations that do not have real solutions. Key take-aways are:

  • Complex Numbers: The form z = a + ib, with a representing the real part and b representing the imaginary part.
  • The Imaginary Unit (i): Understanding i = −1​ and its cyclic powers (i2=−1,i3=−i,i4=1).
  • Algebra of Complex Numbers: Rules for addition, subtraction, multiplication, and division (rationalization) of complex numbers.
  • Modulus and Conjugate:
  • Conjugate (zˉ): If z = a + ib, then zˉ = a - ib.
  • Modulus (|z|): The distance of the point from the origin, ∣z∣=a2+b2​.
  • Argand Plane and Polar Representation: Representing complex numbers as points (a, b) in a coordinate plane where the y-axis is imaginary.
  • Quadratic Equations: Solving ax2+bx+c=0 where the discriminant D < 0 using the formula:

x=2a−b±i∣D∣​​.

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 4

EXERCISE - 4.1

Ques 1. Express the given complex number in the Q. 1 to 10 in the form a+ib.

  1. (5i)(−53​i)

Sol.

(5i)(−53​i)​=−5×53​×i×i=−3i2=−3(−1)[i2=1]=3​

  1. i9+i19 Sol.

i9+i19​=i4×2+1+i4×4+3=(i4)2⋅i+(i4)4⋅i3=1×i+1×(−i)[i4=1,i3=−i]=i+(−i)=0​

  1. i−39 Sol.

i−39​=i−4×9−3=(i4)−9⋅i−3=(i)−9⋅i−3=i31​=−i1​=i−1​×ii​=i2−i​=−1−i​=i​​[i4=1][i3=−i][i3=−1]​

  1. 3(7+i7)+i(7+i7)

Sol.

​3(7+i7)+i(7+i7)=21+21i+7i+7i2=21+28i+7×(−1)[∵i2=−1]=14+28i​

  1. (1−i)−(−1+i6) Sol. (1−i)−(−1+i6)=1−i+1−6i=2−7i


  1. (51​+i52​)−(4+i25​)

Sol.

​(51​+i52​)−(4+i25​)=51​+52​i−4−25​i=(51​−4)+i(52​−25​)=5−19​+i(10−21​)=5−19​−1021​i​

  1. [(31​+i37​)+(4+i31​)]−(−34​+i)

Sol.

​[(31​+i37​)+(4+i31​)]−(−34​+i)=31​+37​i+4+31​i+34​−i=(31​+4+34​)+i(37​+31​−1)=317​+i35​​

  1. (1−i)4

Sol.

​(1−i)4=[(1−i)2]2=[12+i2−2i]2=[1−1−2i]2=(−2i)2=(−2i)×(−2i)=4i2=−4[i2=−1]​

  1. (31​+3i)3

Sol.

(31​+3i)3

​=(31​)3+(3i)3+3(31​)(3i)(31​+3i)=271​+27i3+3i(31​+3i)=271​+27(−i)+i+9i2[i3=−i]=271​−27i+i−9[i2=−1]=(271​−9)+i(−27+1)=27−242​−26i​

  1. (−2−31​i)3 Sol.

​(−2−31​i)3=(−1)3(2+31​i)3=−[23+(3i​)3+3(2)(3i​)(2+3i​)]=−[8+27i3​+2i(2+3i​)]​

​=−[8−27i3​+4i+32i3​][i3=−i]=−[8−27i3​+4i−32​][i2=−i]=−[322​+27107i​]=−322​−27107​i​

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

  1. 4-3i Sol. Let z=4−3i Then, z=4+3i and

∣z∣2=42+(−3)2=16+9=25

Therefore, the multiplicative inverse of 4−3i is given by

z−1=∣z∣2zˉ​=254+3i​=254​+253​i

  1. 5​+3i Sol. Let z=5​+3i Then, z=5​−3i and ∣z∣2=(5​)2+32=5+9=14 Therefore, the multiplicative inverse of 5​+3i

z−1​=∣z∣2zˉ​=145​−3i​=145​​−143i​​

  1. -i Sol. Let z=−i Then, z=i and ∣z∣2=12=1 Therefore, the multiplicative inverse of -i is given by z−1=∣z∣2z​=1i​=i
  2. Express the following expression in the form

 of a+ib⋅(3​+2​i)−(3​−i2​)(3+i5​)(3−i5​)​

Sol. (3​+2​i)−(3​−i2​)(3+i5​)(3−i5​)​

=3​+2​i−3​+2​i(3)2−(i5​)2​

​[(a+b)(a−b)=a2−b2]=22​i9−5i2​=22​i9−5(−1)​[i2=1]=22​i9+5​×ii​=22​i214i​=22​(−1)14i​=2​−7i​×2​2​​=2−72​i​​

MISCELLANEOUS EXERCISE

  1. Evaluate : [i18+(i1​)25]3 Sol. [i18+(i1​)25]3=[i4×4+2+i4×6+11​]3

=[(i4)4⋅i2+(i4)6⋅i1​]3=[i2+i1​]3=[−i+i1​×ii​]3=[−1+i2i​]3=[−1−i]3=(−1)3[1+i]3=−[13+i3+3⋅1⋅i(1+i)]=−[1+i3+3i+3i2]=−[1−i+3i−3]=−[−2+2i]=2−2i​[i2=1][i2=−1]​

  1. For any two complex numbers z1​ and z2​, prove that

Re(z1​z2​)=Rez1​Rez2​−Imz1​Imz2​

Sol. Let z1​=x1​+iy1​ and z2​=x2​+iy2​

∴z1​z2​=(x1​+iy1​)(x2​+iy2​)

​=x1​(x2​+iy2​)+iy1​(x2​+iy2​)=x1​x2​+ix1​y2​+iy1​x2​+i2y1​y2​=x1​x2​+ix1​y2​+iy1​x2​−y1​y2​[i2=−1]=(x1​x2​−y1​y2​)+i(x1​y2​+y1​x2​)​

⇒Re(z1​z2​)=x1​x2​−y1​y2​ ⇒Re(z1​z2​)=Rez1​Rez2​−Imz1​Imz2​

Hence Proved

  1. Reduce (1−4i1​−1+i2​)(5+i3−4i​) to the standard form. Sol. (1−4i1​−1+i2​)(5+i3−4i​)

​=[(1−4i)(1+i)(1+i)−2(1−4i)​][5+i3−4i​]=[1+i−4i−4i21+i−2+8i​][5+i3−4i​]=[5−3i−1+9i​][5+i3−4i​]=[25+5i−15i−3i2−3+4i+27i−36i2​]=[28−10i33+31i​]=2(14−5i)33+31i​=2(14−5i)(33+31i)​×(14+5i)(14+5i)​​

[On multiplying numerator and denominator by (14+5i) ]

​=5[(14)2−(5i)2]462+165i+434i+155i2​=2(196−25i2)307+599i​=2(221)307+599i​=442307+599i​=442307​+442599i​​

This is the required standard form.

4. If x−iy=c−ida−ib​​ prove that (x2+y2)2=c2+d2a2+b2​

Sol. Given

x−iy​=c−ida−ib​​=c−ida−ib​×c+idc+id​​​

[On multiplying numerator and denominator by (c+id)]

=c2+d2(ac+bd)+i(ad−bc)​​∴(x−iy)2=c2+d2(ac+bd)+i(ad−bc)​⇒x2−y2−2ixy=c2+d2(ac+bd)+i(ad−bc)​​

On comparing real and imaginary parts, we obtain

​x2−y2=c2+d2ac+bd​,−2xy=c2+d2ad−bc​…..(1)(x2+y2)2=(x2−y2)2+4x2y2=(c2+d2ac+bd​)2+(c2+d2ad−bc​)2[ Using (1) ]=(c2+d2)2a2c2+b2d2+2acbd+a2d2+b2c2−2adbc​=(c2+d2)2a2c2+b2d2+a2d2+b2c2​=(c2+d2)2a2(c2+d2)+b2(c2+d2)​=(c2+d2)2(c2+d2)(a2+b2)​=c2+d2a2+b2​​

Hence, proved

  1. If z1​=2−i,z2​=1+i, find ​z1​−z2​+iz1​+z2​+1​​ Sol. Given, z1​=2−i,z2​=1+i

∴​​z1​−z2​+iz1​+z2​+1​​=​(2−i)−(1+i)+1(2−i)+(1+i)+1​​=​2−2i4​​=​2(1−i)4​​=​1−i2​×1+i1+i​​=​12−i22(1+i)​​=​1+12(1+i)​​[i2=−1]=​22(1+i)​​=∣1+i∣=12+12​=2​​

Thus, the value of ​z1​−z2​+1z1​+z2​+1​​ is 2​.

6. If a+ib=2x2+1(x+i)2​, prove that a2+b2=(2x+1)2(x2+1)2​ Sol. a+ib=2x2+1(x+i)2​=2x2+1x2+i2+2xi​

=2x2+1x2−1+i2x​=2x2+1x2−1​+i(2x2+12x​)

On comparing real and imaginary parts, we obtain

a=2x2+1x2−1​ and b=2x2+12x​

∴a2+b2=(2x2+1x2−1​)2+(2x2+12x​)2

=(2x2+1)2x4+1−2x2+4x2​

=(2x2+1)2x4+1+2x2​=(2x2+1)2(x2+1)2​

∴a2+b2=(2x2+1)2(x2+1)2​

Hence, proved

  1. Let z1​=2−i,z2​=−2+i. Find (i) Re(z1​z1​z2​​) (ii) Im(Z1​Z1​1​)

Sol. z1​=2−i,z2​=−2+i

(i)

z1​z2​z1​=​=(2−i)(−2+i)=−4+2i+2i−i2=−4+4i−(−1)=−3+4i2+i​

∴z1​z1​z2​​=2+i−3+4i​

On multiplying numerator and denominator by (2−i), we obtain

zˉ1​z1​z2​​​=(2+i)(2−i)(−3+4i)(2−i)​=22+12−6+3i+8i−4i2​=22+12−6+11i−4(−1)​=5−2+11i​=5−2​+511​i​

On comparing real parts, we obtain

Re(zˉ1​z1​z2​​)=5−2​

(ii) z1​z1​1​=(2−i)(2+i)1​=(2)2+(1)21​=51​ On comparing imaginary parts we obtain

Im(z1​zˉ1​1​)=0

  1. Find the real numbers x and y if (x−iy)(3+5i) is the conjugate of −6−24i. Sol. Let z=(x−iy)(3+5i)

z​=3x+5xi−3yi−5yi2=3x+5xi−3yi+5y=(3x+5y)+i(5x−3y)​

∴zˉ=(3x+5y)−i(5x−3y)

It is given that,

z=−6−24i

∴(3x+5y)−i(5x−3y)=−6−24i

Equating real and imaginary parts, we obtain

​3x+5y=−65x−3y=24​

Multiplying equation (1) by 3 and equation (2) by 5 and then adding them, we obtain

9x+15y=−1825x−15y=12034x=102​​

∴x=34102​=3

Putting the value of x in equation (1), we obtain

3(3)+5y=−6

⇒5y=−6−9=−15 ⇒y=−3 Thus, the values of x and y are 3 and -3 respectively.

9. Find the modulus of 1−i1+i​−1+i1−i​.

Sol. 1−i1+i​−1+i1−i​

​=(1−i)(1+i)(1+i)2−(1−i)2​=12+121+i2+2i−1−i2+2i​=24i​=2i​

∴​1−i1+i​−1+i1−i​​=∣2i∣=22​=2

  1. If (x+iy)3=u+iv, then show that :

xu​+yv​=4(x2−y2)

Sol. (x+iy)3=u+iv

⇒⇒⇒⇒​x3+(iy)3+3⋅x⋅iy(x+iy)=u+ivx3+i3y3+3x2yi+3xy2i2=u+ivx3−iy3+3x2yi−3xy2=u+iv(x3−3xy2)+i(3x2y−y3)=u+iv​

On equating real and imaginary parts, we obtain

u=x3−3xy2,v=3x2y−y3

∴xu​+yv​∴xu​+yv​​=xx3−3xy2​+y3x2y−y3​=xx(x2−3y2)​+yy(3x2y−y2)​=x2−3y2+3x2−y2=4x2−4y2=4(x2−y2)=4(x2−y2)​

Hence, proved.

11. If α and β are different complex numbers with ∣β∣=1, then find ​1−αˉββ−α​​.

Sol. Let α=a+ib and β=x+iy It is given that, ∣β∣=1

∴⇒​x2+y2​=1x2+y2=1​

===​​1−αˉββ−α​​=​1−(a−ib)(x+iy)(x+iy)−(a+ib)​​​1−(ax+aiy−ibx+by)(x−a)+i(y−b)​​​(1−ax−by)+i(bx−ay)(x−a)+i(y−b)​​∣(1−ax−by)+i(bx−ay)∣∣(x−a)+i(y−b)∣​[​z2​z1​​​=​z2​z1​​​]​

=(1−ax−by)2​+(bx−ay)2(x−a)2+(y−b)2​​ =1+a2x2+b2y2−2ax+2abxy−2by+b2x2+a2y2−2abxy​x2+a2−2ax+y2+b2−2by​​

=1+a2(x2+y2)+b2(y2+x2)−2ax−2by​(x2+y2)+a2+b2−2ax−2by​​

=1+a2+b2−2ax−2by​1+a2+b2−2ax−2by​​[ Using (1) ] ∴​1−αˉββ−α​​=1

  1. Find the number of non-zero integral solutions of the equation ∣1−i∣x=2x. Sol. ∣1−i∣x=2x⇒(12+(−1)2​)x=2x

⇒(2​)x=2x⇒2x​=x⇒2x−x=0​⇒22x​=2x⇒x=2x⇒x=0​

Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0 .

13. If (a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that:

(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2.

Sol. (a+ib)(c+id)(e+if)(g+ih)=A+iB

∴⇒​∣(a+ib)(c+id)(e+if)(g+ih)∣=∣A+iB∣∣(a+ib)∣×∣(c+id)∣×∣(e+if)∣×∣(g+ih)∣=∣A+iB∣[∣z1​z2​∣=∣z1​∣∣z2​∣]​

⇒a2+b2​×c2+d2​×e2+f2​×g2+h2​=A2+B2​ On squaring both sides, we obtain

(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2.

Hence proved.

14. If (1−i1+i​)m=1 then find the least positive integral value of m. Sol. (1−i1+i​)m=1⇒(1−i1+i​×1+i1×i​)m=1

⇒(12+i2(1+i)2​)m=1⇒(212+i2+2i​)m=1

⇒(21−1+2i​)m=1⇒(22i​)m=1 ⇒im=1 ∴m=4k, where k is some integer. Therefore, the least positive integer is 1. Thus, the least positive integral value of m is 4 (= 4 × 1).

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Prepare each Class 11 Maths chapter with NCERT Solutions, including exercise answers, important formulas, and clear explanations to help you understand and solve textbook questions.

Chapter Number

Chapterwise NCERT Solutions

Chapter 1

Sets

Chapter 2

Relations and Functions

Chapter 3

Trigonometric Functions

Chapter 5

Linear Inequalities

Chapter 6

Permutations and Combinations

Chapter 7

Binomial Theorem

Chapter 8

Sequence and Series

Chapter 9

Straight Lines

Chapter 10

Conic Sections

Chapter 11

Introduction to Three-dimensional Geometry

Chapter 12

Limits and Derivatives

Chapter 13

Statistics

Chapter 14

Probability

4.0Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations: Exercise-wise Questions and Topics

Exercise

Number of Questions

Important Topics Covered

Exercise 4.1

14 Questions and Solutions

Complex numbers in the form (a+ib) and multiplicative inverse of complex numbers

Miscellaneous Exercise

14 Questions and Solutions

Operations on complex numbers, conjugates, modulus, argument, and quadratic equations

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 4 (Complex Numbers and Quadratic Equations)

  • Easy Understanding of the Argand Plane: The solutions explain how complex numbers are represented as points in the Argand plane. Students can understand the real part, imaginary part, modulus, and addition of complex numbers more easily.
  • Systematic Power of i Calculations: Detailed explanations on how to simplify large powers of iota (i^n) by using the periodicity of 4, reducing complex expressions to their simplest form.
  • Step-by-Step Quadratic Solving: Every quadratic equation with a negative discriminant is solved using a clear, methodical approach that replaces −1​ with i, building confidence in handling complex roots.
  • Property-Based Conjugate and Modulus Proofs: Special attention is given to properties like ∣z1​z2​∣=∣z1​∣∣z2​∣ and z1​+z2​​=zˉ1​+zˉ2​, ensuring students understand the underlying theory.
  • Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts and follow the NCERT syllabus. The solutions use correct mathematical formulas, symbols, and steps.
  • Simple and Easy-to-Follow Solutions: Each concept is explained in simple language with clear steps. This helps students understand complex numbers and quadratic equations without confusion.
  • Coverage of NCERT Exercises and Examples: The solutions cover the exercises and examples from NCERT Class 11 Maths Chapter 4 – Complex Numbers and Quadratic Equations. Students can use them to understand concepts, check their answers, and revise the chapter.

Table of Contents

  • 1.0Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations: Key Concepts
  • 2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 4
  • 2.1EXERCISE - 4.1
  • 2.1.1Find the multiplicative inverse of the complex numbers given in Q. 11 to 13.
  • 2.2MISCELLANEOUS EXERCISE
  • 3.0Class 11 Maths NCERT Solutions – Chapter-wise Links
  • 4.0Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations: Exercise-wise Questions and Topics
  • 5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 4 (Complex Numbers and Quadratic Equations)