Find all cube roots of the complex number. Leave answers in trigonometric form.
-$$- 6 \sqrt { 3 } - 6 i$$
A) $\sqrt { 6 } \operatorname { cis } 70 ^ { \circ } , \sqrt { 6 } \operatorname { cis } 190 ^ { \circ } , \sqrt { 6 } \operatorname { cis } 310 ^ { \circ }$
B) $\sqrt [ 3 ] { 12 } \operatorname { cis } 70 ^ { \circ } , \sqrt [ 3 ] { 12 } \operatorname { cis } 190 ^ { \circ } , \sqrt [ 3 ] { 12 } \operatorname { cis } 310 ^ { \circ }$
C) $\sqrt [ 3 ] { 6 } \operatorname { cis } 70 ^ { \circ } , \sqrt [ 3 ] { 6 } \operatorname { cis } 190 ^ { \circ } , \sqrt [ 3 ] { 6 } \operatorname { cis } 310 ^ { \circ }$
D) $\sqrt [ 3 ] { 12 } \operatorname { cis } 80 ^ { \circ } , \sqrt [ 3 ] { 6 }$ cis $200 ^ { \circ } , \sqrt [ 3 ] { 12 }$ cis $320 ^ { \circ }$

Question

Quizplus · Accepted Answer

First, convert the complex number to polar form. The magnitude is $\sqrt{(-6\sqrt{3})^2 + (-6)^2} = \sqrt{108 + 36} = \sqrt{144} = 12$. The angle $\theta$ can be found using $\tan^{-1}\left(\frac{-6}{-6\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$, but since the number is in the third quadrant, the actual angle is $180^\circ + 30^\circ = 210^\circ$. The polar form is $12\operatorname{cis}210^\circ$.To find the cube roots, divide the angle by 3 and take the cube root of the magnitude: $\sqrt[3]{12}$ and $210^\circ / 3 = 70^\circ$. The other two roots are found by adding $120^\circ$ and $240^\circ$ to the initial angle due to the symmetry of the roots in the complex plane. Thus, the cube roots are $\sqrt[3]{12}\operatorname{cis}70^\circ$, $\sqrt[3]{12}\operatorname{cis}(70^\circ + 120^\circ) = \sqrt[3]{12}\operatorname{cis}190^\circ$, and $\sqrt[3]{12}\operatorname{cis}(70^\circ + 240^\circ) = \sqrt[3]{12}\operatorname{cis}310^\circ$.

Find All Cube Roots of the Complex Number −63−6i- 6 \sqrt { 3 } - 6 i−63​−6i

Find All Cube Roots of the Complex Number $- 6 \sqrt { 3 } - 6 i$