138k views
3 votes
Which expression is a sixth root of -1 + i√3?

A.6rt2(cos(90°) + isin(90°))
B. 6rt2 (cos(60°) + isin(60°))
C. 6rt2 (cos(300°) + sin(300°))
D. 6rt2 (cos(20°) + isin(20°))
Pls help

Which expression is a sixth root of -1 + i√3? A.6rt2(cos(90°) + isin(90°)) B. 6rt-example-1
User Kurdtc
by
8.5k points

1 Answer

5 votes


-1 + i\sqrt3 lies in the second quadrant of the complex plane, so its argument is


\arg(-1 + i\sqrt3) = \pi - \tan^(-1)\left(-\sqrt3\right) = \frac{2\pi}3 = 120^\circ

Then any sixth root of
-1+i\sqrt3 will have an argument of


\frac{120^\circ + 360^\circ n}6

where
n\in\{0,1,2,3,4,5\}.

When
n=0, we have


\frac{120^\circ}6 = 20^\circ

so D is a sixth root.

The other sixth roots are separated by arguments of 60 degrees (80, 140, 200, 260, and 320).

User Hovestar
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories