calculate the modulus argument of

Question

-1+√3i

Answer 1

The polar form of a complex number z = a + bi is z = r (cos θ + i sin θ), where r = | z | = √(a² + b²), a = r cos θ, and b = r sin θ, and θ = tan^{- 1}(b / a) for a > 0 or θ = tan^{- 1}(b / a) + π or θ = tan^{- 1}(b / a) + 180^o for a < 0.

The complex number is z = - 1 + i√3.

The polar form of a complex number z = a + bi is z = r (cos θ + i sin θ).

Here a = - 1 < 0 and b = √3.

So, first find the absolute value of r .

r = | z | = √(a² + b²)

            = √[ (-1)² + (√3)² ]

            = √[ 1 + 3 ]

            = √[ 4 ]

            = 2.

Now find the argument θ.

Since a = - 1 < 0, use the formula θ = tan^{- 1}(b / a) + 180^o.

θ = tan^{- 1}[ √3/(- 1) ] + 180^o

θ = - 60 + 180^o

θ = 120^o

Note that here θ is measured in degrees.

Therefore, the polar form of - 1 + √3i is about 2[ cos(120^o) + i sin(120^o) ].