5. Complex Numbers and Quadratic Equations Mathematics Exercise - -5 class 11 Maths in English - CBSE Study
NCERT Solutions for Class 11 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 5. Complex Numbers and Quadratic Equations with detailed explanations and step-by-step answers for better exam preparation. Each Miscellaneous Exercise on Chapter - 5 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 11 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 11 English Medium Mathematics All Chapters:
5. Complex Numbers and Quadratic Equations
4. 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
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