5. Complex Numbers and Quadratic Equations Mathematics class 11 exercise Miscellaneous Exercise on Chapter - 5
5. Complex Numbers and Quadratic Equations Mathematics class 11 exercise Miscellaneous Exercise on Chapter - 5 ncert book solution in english-medium
NCERT Books Subjects for class 11th Hindi Medium
Exercise 5.1
Exercise 5.1
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
Exercise 5.2
Exercise 5.2
Exercise 5.3
Exercise 5.3
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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Mathematics Chapter List
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle Of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability
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