Find all cube roots of the complex number. Leave answers in trigonometric form.
-$64 \left( \cos 237 ^ { \circ } + \mathrm { i } \sin 237 ^ { \circ } \right)$
A) $4 \operatorname { cis } 79 ^ { \circ } , 4 \operatorname { cis } 139 ^ { \circ } , 4 \operatorname { cis } 19 ^ { \circ }$
B) $4 \operatorname { cis } 79 ^ { \circ } , 4 \operatorname { cis } 199 ^ { \circ } , 4 \operatorname { cis } 319 ^ { \circ }$
C) $4 \operatorname { cis } 169 ^ { \circ } , 4 \operatorname { cis } 229 ^ { \circ } , 4 \operatorname { cis } 109 ^ { \circ }$
D) $4 \operatorname { cis } 79 ^ { \circ } , 4 \operatorname { cis } 199 ^ { \circ } , 4 \operatorname { cis } 199 ^ { \circ }$

Question

Quizplus · Accepted Answer

To find the cube roots of a complex number in trigonometric form, $r(\cos \theta + i\sin \theta)$, we use the formula for the nth roots: $\sqrt[n]{r}(\cos \frac{\theta + 2k\pi}{n} + i\sin \frac{\theta + 2k\pi}{n})$, where $k = 0, 1, ..., n-1$. For cube roots ($n=3$), and given $r = 64$ and $\theta = 237^\circ$, we find the magnitude of the roots as $\sqrt[3]{64} = 4$, and the angles as $\frac{237^\circ + 2k\pi}{3}$ for $k = 0, 1, 2$. Calculating these gives angles of $79^\circ$, $199^\circ$, and $319^\circ$, corresponding to choice B.

Find All Cube Roots of the Complex Number 64(cos⁡237∘+isin⁡237∘)64 \left( \cos 237 ^ { \circ } + \mathrm { i } \sin 237 ^ { \circ } \right)64(cos237∘+isin237∘)

Find All Cube Roots of the Complex Number $64 \left( \cos 237 ^ { \circ } + \mathrm { i } \sin 237 ^ { \circ } \right)$