The complex fourth roots of 5-5sqrt3i

First convert the complex number to polar form

|5 - 5sqrt3i| =  25sqrt10

argument =  arctan - sqrt3  = -pi/3

so  5 - 5sqrt3i  =  25sqrt10(cos(-pi/3) + i sin(-pi/3))

the angle is in the 4th quadrant so we could write it as 2pi-pi/3 =  5p/3

= 25sqrt10(cos 5pi/3 + i sin 5pi/3)

Now if r(cosx + isin x) is a 5th root of 5-5sqrt3i  then

r^5(cos x + i sinx)^5 = 25sqrt10(cos 5pi/3 + i sin 5pi/3)

r^5 = 25sqrt10 and cos5x + i sin 5x =  cos 5pi/3 + i sin 5pi/3

i have to go urgently  so i have to leave it to you to finish this

Just tell koda to adjust the number of shapes himself.

One statement could be 12/4=3