# Complex Numbers and Quadratic Equations – Class XI – Exercise 5.2

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

Solution:

z = -1 – i√3

Let r cosθ = -1 and r sinθ = -√3

On squaring and adding, we get

(r cosθ)2 + ( r sinθ)2 = (-1)2 + (-√3)2

r2(cos2θ + sin2θ) = 1 + 3

r2 = 4

r = √4  = 2 [r>0]

Modulus = 2

2 cosθ = -1 and 2sinθ = -√3

cosθ = ­-1/2 and sinθ = –√3/2

Since both the values of sinθ and cosθ are negative and sinθ and cosθ are negative in III quadrant

= -(π – π/3) = –/3

Thus the modulus and argument of the complex number -1 – √3i are 2 and –/3 respectively.

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

Solution:

z = – √3 + i

Let r cos θ = – √3 and r sinθ = 1

On squaring and adding, we obtain

r3cos2 θ + r2 sin2 θ = (-√3)2 + 12

r2 = 3 + 1 = 4

r = √4 = 2 [r>0]

Modulus = 2

2 cosθ = – √3 and 2sin θ = 1

cos θ = –√3/2 and sin θ = 1/2

θ = π – π/6 = /6         [As θlies in II quadrant]

Thus the modulus and argument of the complex number -√3 + I are 2 and /6 respectively.

1. Convert the given complex number in polar form 1 – i

Solution:

1 – i

Let r cos θ = 1 and r sinθ = -1

On squaring and adding , we get

r2cos2 θ + r2 sin2 θ = 12 + (-1)2

r2 (cos2 θ + sin2 θ) = 1 + 1

r2 = 2

r = √2 [r>0]

√2 cosθ = 1 and √2sinθ = –1/√2

Therefore, θ = – π/4 [As θ lies in IV quadrant]

1 – i = r cosθ + i rsin θ = √2[cos(-π/4)+i sin(-π/4)]

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

Solution:

-1 + i

Let r cos θ = -1 and r sinθ = 1

On squaring and adding , we get

r2cos2 θ + r2 sin2 θ = (-1)2 + (1)2

r2 (cos2 θ + sin2 θ) = 1 + 1

r2 = 2

r = √2 [r>0]

√2 cosθ = -1 and √2sinθ = 1

cos θ = – 1/√2 and sinθ = 1/√2

θ = π – π/4 = /4 [θ lies in II quadrant]

It can be written as

-1 + I = r cosθ + I rsinθ = √2cos(/4) + i √2 sin(-π/4) = √2[cos(/4) + i sin(/4)]

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

Solution:

-1 – i

Let r cos θ = -1 and r sinθ = -1

On squaring and adding, we get

r2cos2 θ + r2 sin2 θ = (-1)2 + (-1)2

r2 (cos2 θ + sin2 θ) = 1 + 1

r2 = 2

r = √2 [r>0]

√2 cosθ = -1 and √2sinθ = 1

cos θ = – 1/√2 and sinθ = –1/√2

θ = π – π/4 = –/4 [Asθ lies in III quadrant]

It can be written as

-1 + I = r cosθ + I rsinθ = √2cos(-/4) + i √2 sin(-π/4) = √2[cos(-/4) + i sin(-/4)]

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

Solution:

-3

Let r cos θ = -3 and r sinθ = 0

On squaring and adding , we get

r2cos2 θ + r2 sin2 θ = (-3)2

r2 (cos2 θ + sin2 θ) = 9

r2 = 9

r = 3 [r>0]

√2 cosθ = -3 and √2sinθ = 0

cos θ = -1 and sinθ = 0

θ = π

It can be written as

-3 = r cosθ + I rsinθ = 3cosπ + i sinπ = 3[cosπ + i sinπ]

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

Solution:

√3 + i

Let r cos θ = √3 and r sinθ = 1

On squaring and adding , we get

r2cos2 θ + r2 sin2 θ = (√3)2 + (1)2

r2 (cos2 θ + sin2 θ) = 3 + 1

r2 = 4

r = 2 [r>0]

2 cosθ = √3 and 2sinθ = 1

cos θ = – √3/2 and sinθ = 1/2

θ = π/6 [As θlies in the I quadrant]

It can be written as

√3 + i = r cosθ + i rsinθ = 2cos(π/6) + i 2sin(π/6) = 2[cos(π/6) + i sin(π/6)]

1. Convert the given complex number in polar form i

Solution:

i

Let r cos θ = 0 and r sinθ = 1

On squaring and adding , we get

r2cos2 θ + r2 sin2 θ = (0)2 + (1)2

r2 (cos2 θ + sin2 θ) = 0 + 1

r2 = 1

r = √1 = 1 [r>0]

cosθ = 0 and sinθ = 1

θ = π/2

It can be written as

i = r cosθ + i rsinθ = cos(π/2) + i sin(π/2)