Question

Find the cube roots of -27.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let z = -27 = -27(cos $0^{\circ }$ + isin $0^{\circ }$)

Cube root z = $[-27(cos{0}^{\circ }+isin{0}^{\circ }){]}^{\frac{1}{3}}$

= - 3 $[cos\frac{2k\pi }{3}+isin\frac{2k\pi }{3}]$ where k = 0,1,2

When k = 0

z = -3[cos 0 + isin 0 ] = -3

When k = 1

z = - 3 $[cos\frac{2\pi }{3}+isin\frac{2\pi }{3}]$ = -3 $[\frac{-1}{2}+i\left(\frac{\sqrt{3}}{2}\right)]$

= $\frac{3}{2}$ - $\frac{i3\sqrt{3}}{2}$

When k = 2

z = - 3 $[cos\frac{4\pi }{3}+isin\frac{4\pi }{3}]$

= - 3 $[-\frac{1}{2}-i\left(\frac{\sqrt{3}}{2}\right)]$

= $\frac{3}{2}$ + $\frac{i3\sqrt{3}}{2}$

Cube roots are -3, $\frac{3}{2}$ - $\frac{3\sqrt{3}i}{2}$ , $\frac{3}{2}$ + $\frac{3\sqrt{3}i}{2}$