NCERT Solutions For Class 11 Math Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise

Class 11 Maths NCERT Solutions for Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise

Complex Numbers and Quadratic Equations Miscellaneous Exercise Solutions

1. Evaluate : [i¹⁸ + (1/i)²⁵ ]³ .

Solution

= [-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

2. For any two complex numbersz₁ and z₂, prove that

Re (z₁z₂)= Re z₁ Re z₂ – Im z₁ Im z₂

Solution 

Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ 
∴ z₁z₂ = (x₁ + iy₁)(x₂ + iy₂)
= x₁ (x₂ + iy₂) + iy₁ (x₂ + iy₂) 
= x₁x₂ + ix₁y₂ + iy₁x₂ + i² y₁y₂ 
= x₁x₂ + ix₁y₂ + iy₁x₂ + y₁y₂ [i² = -1] 
= (x₁x₂ - y₁y₂) + i(x₁y₂ + y₁x₂) 
⇒ Re (z₁z₂) = x₁x₂ - y₁x₂ 
⇒ Re (z₁z₂) = Re z₁ Re z₂ - Im z₁ Im z₂ 
Hence, proved. 

3. Reduce [(1/(1-4i) - 2/(1 + i)][(3 - 4i)/(5 +i)] to the standard form. 

Solution

4. If x - iy = √(a - ib)/(c-id) prove that (x² + y²)²  = (a² + b²)/(c² + d²) 

Solution

5. Convert the following in the polar form : 
(i) (1 + 7i)/(2 - i)² ,
(ii) (1 + 3i)/(1 - 2i) 

Solution

Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r² (cos² θ + sin² θ) = 1 + 1
⇒ r² (cos² θ + sin² θ) = 2
⇒ r² = 2  [cos² θ + sin² θ = 1]
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cosθ = -1 and √2 sinθ = 1 
⇒ cos θ = -1/√2 and sin θ = 1/√2 
∴ θ = π - π/4 = 3π/4   [As θ lies in II quadrant] 
∴ z = r cos θ + i r sin θ 
= √2 cos (3π/4) + i √2sin (3π/4) = √2[ cos(3π/4) + i sin(3π/4) 

This is the required polar form.

(ii) Here, z = (1 + 3i)/(1 - 2i) 

Let r cos θ = –1 and r sin θ = 1

On squaring and adding, we obtain
r² (cos² θ + sin² θ) = 1 + 1
⇒r² (cos² θ + sin² θ) = 2
⇒ r² = 2 [cos² θ + sin² θ = 1]
⇒ r = √2   [Conventionally, r > 0] 
∴ √2 cosθ = -1 and √2sinθ = 1 
⇒ cos θ = -1/√2 and sinθ = 1/√2
∴ θ = π - π4 = 3π/4  [As θ lies in II quadrant] 
∴ z = r cos θ + i r sin θ 
= √2 cos 3π/4 + i√2 sin3π/4 = √2 [cos (3π/4 + i sin(3π/4)] 
This is the required polar form. 

6. Solve the equation 3x² - 4x + 20/3 = 0 

Solution

The given quadratic equation is 3x² - 4x + 20/3 = 0 
This equation can also be written as 9x² - 12x + 20 = 0 
On comparing this equation with ax² + bx + c = 0 , we obtain 
a = 9 b = -12 and c = 20 
Therefore, the discriminant of the given equation is 
D = b² - 4ac = (-12)² - 4× 9 × 20 = 144 - 720 = -576 
Therefore, the required solutions are 

7. Solve the equation x² - 2x + 3/2 = 0 

Solution

The given quadratic equation is x² - 2x + 3/2 = 0 

This equation can also be written as 2x² - 4x +3 = 0 
On comparing this equation with ax² + bx + c = 0 we obtain 
a = 2, b = -4 and c = 3 
Therefore, the discriminant of the given equation is
D = b² - 4ac = (-4)² - 4 × 2 × 3 = 16 - 24 = -8 
Therefore, the required solutions are  

8. Solve the equation 27x² - 10x + 1 = 0 

The given quadratic equation is 27x² - 10x + 1 = 0 
On comparing the given equation with ax² + bx + c = 0, we obtain 
a = 27, b = -10, and c = 1 
Therefore, the discriminant of the given equation is  
D = b² - 4ac = (-10)² - 4 × 27 × 1 = 100 - 108 = -8 
Therefore, the required solutions are  

9. Solve the equation 21x² - 28x + 10 = 0 

Solution

The given quadratic equation is 21x² - 28x + 10 = 0 
On comparing the given equation with ax² + bx + c = 0 , we obtain  
a = 21, b = -28 and c = 10 
therefore, the discriminant of the given equation is 
D = b² - 4ac = (-28)² - 4×21 ×10 = 784 - 840 = -56 

Therefore, the required solutions are 

10. If z₁ = 2 – i , z₂ = 1+ i , find |(z₁ + z₂ + 1)/(z₁ – z­₂ + 1)|

Solution

z₁ = 2 – I , z₂ = 1+ i , 

11. If (a + ib) = (x + 1)² /(2x² + 1) , prove that a² + b² = (x² + 1)² /(2x + 1)² 

Solution 

12. Let z₁ = 2 – i , z₂ = -2+ i . find
(i) Re(z₁ z₂ /z1) 
(ii) Im(1/z₁ z1)

Solution

z₁ = 2 – i , z₂ = -2+ i

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

13. Find the modulus and argument of the complex number (1 + 2i)/(1 - 3i).

Solution

Let z = (1 + 2i)/(1 - 3i) , then 

Let z = r cos θ = -1/2 and r sin θ = 1/2 
On squaring and adding, we obtain  

14. 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) 
z = 3x + 5xi - 3yi - 5yi² = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y) 
∴ z  = (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 = -6 ...(i) 
5x - 3y = 24 ...(ii) 
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain 

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

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

16. Find the modulus of [(1+i)/(1-i)] -[(1 -i)/(1 +i)]. 

Solution

17. If α and β are different complex numbers with |β| = 1, then find |(β - α)/(1 - α β)|. 

Solution

Let α = a + ib and β = x +iy
It is given that, |β| = 1 
∴ √(x² + y² ) = 1 
⇒ x²  + y²  = 1 ...(i)

18. Find the number of non-zero integral solutions of the equation |1-i|^x  =  2^x . 

Solution

|1-x|^x = 2^x 

⇒ 2^x/2 = 2^x 
⇒ x/2 = x 
⇒ x = 2x 
⇒ 2x - x = 0 
⇒ x = 0 
Thus, 0 is the only integral solution of the given equation. Therefore, the number of non- zero integral solutions of the given equation is 0. 

19. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that 
(a² + b²)(c² + d²)(e² + f² )(g² + h²) = A² + B². 

Solution

(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 + B|  [|z₁ z₂| z₁||z₂|] 

On squaring both sides, we obtain 
(a² + b² )(c² + d² )(e² + f² )(g² + h² ) = A² + B² . 
Hence, proved. 

20. If [(1 +i)/(1 - i)]^m = 1, then find the least positive integral value of m. 

Solution

∴ 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).