Question

Determine which regions contain cube roots of -1. Check all that apply.
on real axis
on imaginary axis
quadrant 1
quadrant 2
quadrant 3
quadrant 4

Answer 1

Answer:

A:on real axis C:quadrant 1 F:quadrant 4

Step-by-step explanation:

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Answer 2

Cube roots of -1 lies in the region given in the best options are  A. on real axis, B. quadrant 1, F. quadrant 4.

What is cube roots of -1?

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 $(-1)^\frac{1}{3}$.

Step (1): Write the given number in polar form

x = $(cos 0 - i sin 0)^\frac{1}{3}$

Step (2): Add 2kπ to the argument

x = $(cos 2k\pi - i sin 2k\pi )^\frac{1}{3}$

Step (3): Apply $De$ $Moivre's$  theorem = $(cos \alpha + i sin\alpha) = (cos n\alpha + i sin\alpha )$

x =  $\frac{cos2k\pi }{3} - i sin \frac{2k\pi }{3}$

Step (4): Put k = 0,1,2 .... (n-1)                   [sin (-θ) = -sin θ, cos(-θ) = cos θ]

The three roots are

cos 0 - $i$ sin 0, cos $\frac{2\pi }{3}$ - $i$ sin $\frac{2\pi }{3}$, cos $\frac{4\pi }{3}$ - $i$ sin $\frac{4\pi }{3}$

-1, $-\frac{1}{2} - i\frac{\sqrt{3} }{2}$, $\frac{-1}{2} + i \frac{\sqrt{3} }{2}$ are the three roots of cube roots of -1.

Clearly 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.

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