cube roots of -1 lies in the region given in the best options are A. on real axis, B. quadrant 1, F. quadrant 4.
<h3>What is cube roots of -1?</h3>
Cube roots of -1 mean "finding solution of the equation z³ = -1 since the equation is of order 3, the equation has 3 roots in complex field."
According to the question,
Cube roots of -1 can written as ![(-1)^\frac{1}{3}](https://tex.z-dn.net/?f=%28-1%29%5E%5Cfrac%7B1%7D%7B3%7D)
First, write the given number in polar form
![x = ( cos 0 - i sin 0)^\frac{1}{3}](https://tex.z-dn.net/?f=x%20%3D%20%28%20cos%200%20-%20i%20sin%200%29%5E%5Cfrac%7B1%7D%7B3%7D)
Add 2kπ to the argument
![x = (cos2k\pi - i sin 2k\pi )^\frac{1}{3}](https://tex.z-dn.net/?f=x%20%3D%20%28cos2k%5Cpi%20-%20i%20sin%202k%5Cpi%20%29%5E%5Cfrac%7B1%7D%7B3%7D)
Apply De Moivre's theorem
![(cos\alpha - i sin\alpha) = cosn\alpha + i sin\alpha](https://tex.z-dn.net/?f=%28cos%5Calpha%20%20-%20i%20sin%5Calpha%29%20%3D%20cosn%5Calpha%20%2B%20i%20sin%5Calpha)
x = ![\frac{2k\pi }{3} - i sin\frac{2k\pi }{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2k%5Cpi%20%7D%7B3%7D%20-%20i%20sin%5Cfrac%7B2k%5Cpi%20%7D%7B3%7D)
Put k = 0,1,2..., (n-1) [sin (-θ) = -sin θ, cos(-θ) = cos θ]
The three roots are
cos 0 - sin 0, cos
,
, cos
,
sin![\frac{4\pi }{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%5Cpi%20%7D%7B3%7D)
-1,
, are the three roots of cube roots of -1
It can be seen that the three roots lies in the regions of real axis, quadrant 1 and quadrant 4.
Hence, cube roots of -1 lies in the region given in the best options are A. on real axis, B. quadrant 1, F. quadrant 4.
Learn more about cube roots here:
brainly.com/question/310302
#SPJ1