Write the complex number 3 - 4i in polar form. Please show all work

Question

complex number

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 = 3 - 4i.

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

Here a = 3 > 0 and b = - 4.

So, first find the absolute value of r .

r = | z | = √(a ² + b ²)

            = √[ (3)² + (- 4)² ]

            = √[ 9 + 16 ]

            = √[ 25 ]

            = 5.

Now find the argument θ.

Since a = 3 > 0, use the formula θ = tan^{- 1}(b / a).

θ = tan^{- 1}[ - 4/(3) ]

θ ≅ 53.13^o

Note that here θ is measured in degrees.

Therefore, the polar form of 3 - 4i is about 5[ cos(53.13^o) + i sin(53.13^o) ].

Comments

The value of θ = tan^{- 1}[ - 4/(3) ] = - tan^{- 1}[4/(3)] ≅ - 53.13^o.

Therefore, the polar form of 3 - 4i is about 5[ cos(- 53.13^o) + i sin(- 53.13^o) ] = 5[ cos(53.13^o) - i sin(53.13^o) ].