NCERT Solutions for Class 11 – Complete Chapter-wise Study Material
5. Complex Numbers and Quadratic Equations is one of the most important chapters in the Class 11 Mathematics English NCERT Solutions curriculum. This chapter plays a significant role in helping students build a strong conceptual foundation while preparing for school examinations, class tests, unit tests, half-yearly examinations, annual examinations, and CBSE board assessments. The chapter has been carefully designed according to the latest NCERT syllabus, making it an essential part of every student's study plan.
The 5. Complex Numbers and Quadratic Equations - Class 11 Mathematics English NCERT Solutions available on ATP Education explain every question in a simple, accurate, and step-by-step manner. Each answer is prepared according to the latest CBSE guidelines so that students can understand the concepts clearly without confusion. Whether you are completing your homework, revising before examinations, or strengthening your understanding of the subject, these solutions provide reliable academic support throughout your learning journey.
One of the biggest advantages of studying 5. Complex Numbers and Quadratic Equations is that it helps students understand important concepts, definitions, examples, and textbook exercises in an organized way. Instead of memorizing answers, students learn how to develop logical thinking, improve analytical skills, and write well-structured answers in examinations. This chapter also helps improve problem-solving ability and encourages conceptual learning, which is essential for scoring higher marks in school and competitive examinations.
Our Class 11 Mathematics NCERT Solutions cover all textbook questions, important exercise questions, and chapter-wise explanations in English Medium. Every solution is written in easy-to-understand language, allowing students to revise the chapter quickly before examinations. Regular practice of these solutions improves confidence, strengthens subject knowledge, and reduces examination stress.
Students preparing for school assessments should carefully study 5. Complex Numbers and Quadratic Equations because questions from this chapter are frequently asked in objective questions, short answer questions, long answer questions, competency-based questions, and case-study questions. Understanding the concepts explained in this chapter also helps students connect related topics from other chapters, making overall learning more effective and meaningful.
At ATP Education, we continuously update our Class 11 Mathematics English NCERT Solutions according to the latest NCERT textbooks and CBSE curriculum. Students can confidently use these chapter-wise solutions for daily study, homework assistance, quick revision, examination preparation, and self-learning. By studying 5. Complex Numbers and Quadratic Equations thoroughly and practising every question regularly, students can strengthen their concepts, improve writing skills, and achieve better academic performance in both school and board examinations.
5. Complex Numbers and Quadratic Equations - Class 11 Mathematics English NCERT Solutions
5. Complex Numbers and Quadratic Equations
Exercise 5.1
Exercise 5.1
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
5. Complex Numbers and Quadratic Equations
Exercise 5.2
Exercise 5.2
5. Complex Numbers and Quadratic Equations
Exercise 5.3
Exercise 5.3
5. Complex Numbers and Quadratic Equations
Miscellaneous Exercise on Chapter - 5
Miscellaneous Exercise on Chapter 5
Q2. For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Solution:
Let z1 = a + ib, z2 = c + id
Re z1 = a, Re z2 = c, Im z1 = b, Im z2 = d ….. (1)
z1z2 = (a + ib) (c + id)
= ac + iad + ibc + bd i2
= ac + iad + ibc + bd (-1)
= ac + iad + ibc - bd
= ac - bd + i(ad + bc)
Comparing real and imaginary part we obtain,
Re(z1z2) = ac - bd, Im(z1z2) = ad + bc
Now we take real part
⇒ Re(z1z2) = ac - bd
⇒ Re(z1z2) = Re z1 Re z2 – Im z1 Im z2 [using (1) ]
Hence, proved
Multiplying numerator and denominator by 28 + 10i
On multiplying numerator and denominator by (2 – i), we get
On comparing real and imaginary part we obtain
Q14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Solution:
Let z = (x – iy) (3 + 5i)
= 3x + 5xi - 3yi - 5yi2
= 3x + 5xi - 3yi + 5y
= 3x + 5y + 5xi - 3yi
= (3x + 5y) + (5x - 3y)i
Comparing both sides and equating real and imaginary parts, we get
3x + 5y = – 6 …… (1)
5x - 3y = 24 …… (2)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them,
(x + iy)3 = u + iv
⇒ x3 + 3 . x2 . iy + 3 . x . (iy)2 + (iy)3 = u + iv
⇒ x3 + 3x2 y i + 3 x y2 i2 + y3i3 = u + iv
⇒ x3 + 3x2 y i - 3 x y2 - i y3 = u + iv
⇒ x3 - 3 x y2 + 3x2 y i - iy3 = u + iv
⇒( x3 - 3 x y2) + i(3x2 y - y3) = u + iv
On equating both sides real and imaginary parts, we get;
u = x3 - 3 x y2, …… (1)
v = 3x2 y - y3, ……(2)
Q18. Find the number of non-zero integral solutions of the equation |1 - i|x = 2x
Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0.
Q19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a2 + b2) (c2 + d2) (e2 + f 2) (g2 + h2) = A2 + B2
Solution :
(a + ib) (c + id) (e + if) (g + ih) = A + iB ……….. (1) given
Replacing i by (-i) we have
(a - ib) (c - id) (e - if) (g - ih) = A - iB ……….. (2)
Multiplying (1) and (2)
(a + ib) (a - ib) (c + id) (c - id) (e + if) (e - if) (g + ih) (g - ih) = (A + iB) (A - iB)
(a2 + b2) (c2 + d2) (e2 + f 2) (g2 + h2) = A2 + B2 [ ∵ (x + iy) (x - iy) = x2 + y2]
Hence, proved
Or m = 4k
Hence, least positive integer is 1.
Therefore, the least positive integral value of m is 4 × 1 = 4
Mathematics
Class 11 (English Medium)
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