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) = –2π/3
Thus the modulus and argument of the complex number -1 – √3i are 2 and –2π/3 respectively.
- 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 = 5π/6 [As θlies in II quadrant]
Thus the modulus and argument of the complex number -√3 + I are 2 and 5π/6 respectively.
- 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)]
- 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 = 3π/4 [θ lies in II quadrant]
It can be written as
-1 + I = r cosθ + I rsinθ = √2cos(3π/4) + i √2 sin(-π/4) = √2[cos(3π/4) + i sin(3π/4)]
- 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 = –3π/4 [Asθ lies in III quadrant]
It can be written as
-1 + I = r cosθ + I rsinθ = √2cos(-3π/4) + i √2 sin(-π/4) = √2[cos(-3π/4) + i sin(-3π/4)]
- 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]
- 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)]
- 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)
