Write the complex number z=(1-sqrt3i)^8

Question

 in polar form z=r(cosθ+i sinθ Where r=? θ=? Find what r= and θ= (show work please!

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).

if a > 0, then θ = tan^{- 1}(b / a), and

if a < 0, then θ = tan^{- 1}(b / a) + π or θ = tan^{- 1}(b / a) + 180^o.

 

The complex number is z = (1 - i√3)⁸ .

Let, w = 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 | = √(a ² + b ²)

            = √[ 1² + (- √3)² ]

            = √[1 + 3]

            = √4

            = 2.

Now find the argument θ.

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

θ = tan^{- 1}[ (- √3)/1 ]

θ = tan^{- 1}(- √3)

θ = 120⁰

θ = 120 * π/180 radians

θ = 2π/3 radians.

The polar form of w is about  2[ cos (2π/3) + i sin (2π/3) ].

By euler's equation, we obtain the polar form as : z = re ^{i θ}.

⇒ w = 2e^{i (2π/3)} .

z = w⁸ = [ 2e^{i (2π/3)} ]⁸

 z = 2⁸ * e^{i (16π/3)}

 z = 256e^{i (16π/3)} .

Compare it with, polar form as : z = re ^{i θ} .

r = 256 and θ = 16π/3.