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

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

Solution

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

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

Solution

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

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

Solution

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

This is the required polar form.

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

Solution

-1-i
Let r cos Î¸ = -1 and r sin Î¸ = -1
On squaring and adding, we obtain
r2 cos2 Î¸ + r2 sin2 Î¸ = (-1)2  + (-1)2
⇒ r2 (cos2 Î¸ + sin2 Î¸) = 1 + 1
⇒ r2 = 2
⇒ r = √2  [Conventionally, r > 0]
∴ √2 cos Î¸ = -1 and √2sin Î¸ = -1
⇒ cos Î¸ = -1/√2 and sinÎ¸ = -1/√2
∴ Î¸ = -(Ï€ - Ï€/4) = -3Ï€/4  [As Î¸ lies in the III quadrant]
∴ -1-i = r cos Î¸ + i r sin Î¸ = √2 cos (-3Ï€/4) + i√2 sin(-3Ï€/4) = √2[cos(-3Ï€/4) + i sin(-3Ï€/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 Î¸ = 0
On squaring and adding we obtain
r2 cos2 Î¸ + r2 sin2 Î¸ = (-3)2
⇒ r2 (cos2 Î¸ + sin2 Î¸) = 9
⇒ r2 = 9
⇒ r = √9 = 3  [Conventionally, r > 0]
∴ 3 cosÎ¸ = -3 and 3 sinÎ¸ = 0
⇒ cos Î¸= -1 and sinÎ¸ = 0
∴ Î¸ = Ï€
∴ -3 = r cos Î¸ + i r sin Î¸ = 3 cos Ï€ + B sinÏ€ = 3(cos Ï€ + isin Ï€)
This is the required polar form.

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

Solution

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

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

Solution

i
Let r cosÎ¸ = 0 and r sin Î¸ = 1
On squaring and adding, we obtain
r2 cos2 Î¸ + r2 sin2 Î¸ = 02 + 12
⇒ r2 (cos2 Î¸ + sin2 Î¸) = 1
⇒ r2 = 1
⇒ r = √1 = 1  [Conventionally, r > 0]
∴ cosÎ¸ = 0 and sinÎ¸ = 1
∴ Î¸ = Ï€/2
∴ i = r cosÎ¸ + i r sin Î¸ = cos (Ï€/2) + i sin (Ï€/2)
This is the required polar form.