How to Find the cuberoots of 4√3 + 4i?

Question

The question also stated "Do not evaluate " and leave the answers in cos  and i sin form which confused me.

Answer 1

Hi Youssef,

 

Find the magnitude or absolute value of this complex number.

It is equal to  m = sqrt((4√3)^2 + 4^2)

= sqrt(64)

= 8

tangent of the angle theta is b/a

tan theta = 4/4sqrt(3)

tan theta = 1/sqrt(3)

theta = 30 degrees

 

the complex number in polar form is 

8cis 30°

now take its cube root

(8cis 30°)^1/3

= 2 cis 10°

=2cis 130° =2cis 250°The solution for 10&deg is

=2(cos(10°) + isin(10°))

= 2(0.98 + i0.17)

= 1.96 + 0.34i

Answer 2

||4√3 + 4i|| = 8

θ = π/6

 

(4√3 + 4i)^1/3 = 2(cos θ + i sin θ)^1/3

 

Three roots are:

2(cos θ/3 + i sin θ/3)

2(cos (θ + 2π)/3 + i sin (θ + 2π)/3))

2(cos (θ + 4π)/3 + i sin (θ + 4π)/3))