5. Complex Numbers and Quadratic Equations Mathematics class 11 in English Medium ncert book solutions 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 = 2^x

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