Class 11 Maths NCERT Solutions for Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2

Class 11 Maths NCERT Solutions for Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.2

Complex Numbers and Quadratic Equations Exercise 5.2 Solutions

1. Find the modulus and the argument of the complex number z = -1 - i√3 

Solution

z = -1- i√3 
Let r cos θ = -1 and r sin θ = -√3 
On squaring and adding, we obtain
(r cos θ)2 + (r sin θ)2 = (-1)2 + (-√3)2 
⇒ r2 (cos2 θ + sin2 θ) = 1 + 3 
⇒  r2 = 4  [cos2 θ + sin2 θ = 1]
⇒ r = √4 = 2  [Conventionally, r > 0]
∴ Modulus = 2 
∴ 2 cos θ = -1 and 2sin θ = - √3 
⇒ cos θ = -1/2  and sin θ = -√3/2 
Since both the values of sin θ and cos θ are negative and sin θ and cos θ are negative in III quadrant, 
Argument = -(π - π/3) = -2π/3 
Thus, the modulus and argument of the complex number -1 - √3 i are 2 and -2π/3 respectively.


2. Find the modulus and the argument of the complex number z = -√3 + i 

Solution

z = -√3 + i 
Let r cos θ = -√3 and r sin θ = 1 
On squaring and adding, we obtain 
r2 cos2 θ + r2 sin2 θ = (-√3)2 + 12 
⇒ r2 = 3 + 1 = 4  [cos2 θ + sin2 θ = 1] 
⇒ r = √4 = 2  [Conventionally, r > 0]
∴ Modulus = 2 
∴ 2 cos θ = -√3 and 2sin θ = 1 
⇒ cosθ = -√3/2 and sinθ = 1/2 
∴ θ = π - π/6 = 5π/6  [As θ lies in the II quadrant] 
Thus, the modulus and argument of the complex number -√3 + i are 2 and 5π/6 respectively.


3. Convert the given complex number in polar form : 1 - i 

Solution

1 - i 
Let r cos θ = 1 and r sin θ = -1 
On squaring and adding, we obtain 
r2 cos2 θ + r2 sin2 θ = 12 + (-1)2 
⇒ r2 (cos2 θ + sin2 θ) = 1 + 1 
⇒ r2 = 2 
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cos θ = 1 and √2 sin θ = -1 
⇒ cos θ = 1/√2  and sinθ = -1/√2
∴ θ = -π/4  [As θ lies in the IV quadrant] 
∴ 1 - i = r cos θ + i r sinθ = √2 cos (-π/4)  + i√2sin(-π/4) = √2[cos (-π/4) + i sin(-π/4) ] This is the required polar form.


4. Convert the given complex number in polar form : -1 +i 

Solution

-1+ i 
Let r cos θ = -1 and r sin θ = 1 
On squaring and adding, we obtain 
r2 cos2 θ + r2 sin2 θ = (-1)2 + 12 
⇒ r2 (cos2 θ + sin2 θ) = 1 + 1 
⇒ r2 = 2
⇒ r = √2  [Conventionally, r > 0] 
∴ √2 cos θ = -1 and √2sin θ = 1 
⇒ cos θ = -1/√2 and sinθ = 1/√2 
∴ θ = π - π/4 = 3π/4  [As θ lies in the II quadrant] 
It can be written,  
∴ -1 + i = r cos θ + i r sinθ 

This is the required polar form.


5. Convert the given complex number in polar form : -1 - i . 

Solution

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


6. Convert the given complex number in polar form: -3 

Solution

-3 
Let r cos c = -3 and r sin θ = 0 
On squaring and adding we obtain  
r2 cos2 θ + r2 sin2 θ = (-3)2 
⇒ r2 (cos2 θ + sin2 θ) = 9 
⇒ r2 = 9 
⇒ r = √9 = 3  [Conventionally, r > 0] 
∴ 3 cosθ = -3 and 3 sinθ = 0 
 ⇒ cos θ= -1 and sinθ = 0 
∴ θ = π 
∴ -3 = r cos θ + i r sin θ = 3 cos π + B sinπ = 3(cos π + isin π)
This is the required polar form.


7. Convert the given complex number in polar form : √3 + i 

Solution

√3 + i 
let r cos θ = √3 and r sin θ = 1 
On squaring and adding, we obtain 
r2 cos2 θ + r2 sin2 θ = (√3)2 + 12 
⇒ r2 (cos2 θ + sin2 θ) = 3 + 1 
⇒ r2 = 4 
⇒ r = √4 = 2   [Conventionally, r > 0] 
∴ 2 cosθ = √3 and 2sinθ = 1 
 ⇒ cos θ = √3/2 and sinθ = 1/2
∴ θ = π/6  [As θ lies n the I quadrant] 
∴ √3 + i = r cos θ + i r sinθ = 2 cos (π/6 ) + i 2 sin (π/6) = 2[cos(π/6) + i sin(π/6)]
This is the required polar form.


8. Convert the given complex number in polar form : i

Solution

i
Let r cosθ = 0 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2 θ = 02 + 12 
⇒ r2 (cos2 θ + sin2 θ) = 1
⇒ r2 = 1
⇒ r = √1 = 1  [Conventionally, r > 0]
∴ cosθ = 0 and sinθ = 1
∴ θ = π/2
∴ i = r cosθ + i r sin θ = cos (π/2) + i sin (π/2) 
This is the required polar form.

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