Find all cube roots of the complex number. Leave answers in trigonometric form.
-$4 - 4 i \sqrt { 3 }$
A) $\sqrt [ 3 ] { 4 } \operatorname { cis } 80 ^ { \circ } , \sqrt [ 3 ] { 4 } \operatorname { cis } 200 ^ { \circ } , \sqrt [ 3 ] { 4 } \operatorname { cis } 320 ^ { \circ }$
B) $\sqrt { 8 } \operatorname { cis } 100 ^ { \circ } , \sqrt { 8 } \operatorname { cis } 220 ^ { \circ } , \sqrt { 8 } \operatorname { cis } 340 ^ { \circ }$
C) $\sqrt [ 3 ] { 8 } \operatorname { cis } 100 ^ { \circ } , \sqrt [ 3 ] { 8 } \operatorname { cis } 220 ^ { \circ } , \sqrt [ 3 ] { 8 } \operatorname { cis } 340 ^ { \circ }$
D) $\sqrt { 4 } \operatorname { cis } 60 ^ { \circ } , \sqrt { 4 } \operatorname { cis } 180 ^ { \circ } , \sqrt { 4 } \operatorname { cis } 300 ^ { \circ }$

Question

Quizplus · Accepted Answer

First, convert $4 - 4i\sqrt{3}$ to polar form. The magnitude is $\sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$. The angle $\theta$ can be found from $\tan(\theta) = \frac{-4\sqrt{3}}{4} = -\sqrt{3}$, which gives $\theta = 300^\circ$ (or $-60^\circ$ in standard position). The cube roots will have magnitudes of $\sqrt[3]{8}$ and angles of $\frac{300^\circ + 360^\circ k}{3}$, where $k = 0, 1, 2$. Calculating these gives angles of $100^\circ$, $220^\circ$, and $340^\circ$.

Find All Cube Roots of the Complex Number 4−4i34 - 4 i \sqrt { 3 }4−4i3​

Find All Cube Roots of the Complex Number $4 - 4 i \sqrt { 3 }$