Find the trigonometric form of $- 6 - 6 \sqrt { 3 } i$ .
A) $12 \left( \cos \frac { 4 \pi } { 3 } + i \sin \frac { 4 \pi } { 3 } \right)$
B) $6 \left( \cos \frac { 4 \pi } { 3 } + i \sin \frac { 4 \pi } { 3 } \right)$
C) $12 \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$
D) $12 \left( \cos \frac { 7 \pi } { 6 } + i \sin \frac { 7 \pi } { 6 } \right)$
E) $6 \sqrt { 2 } \left( \cos \frac { 7 \pi } { 6 } + i \sin \frac { 7 \pi } { 6 } \right)$

Question

Quizplus · Accepted Answer

The magnitude of the complex number is $r = \sqrt{(-6)^2 + (-6\sqrt{3})^2} = 12$. The angle $\theta$ in radians, found from the positive x-axis (real axis), corresponds to $\frac{4\pi}{3}$ for the point $-6 - 6\sqrt{3}i$ in the complex plane. Thus, the trigonometric form is $12(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3})$.

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Find the Trigonometric Form Of $- 6 - 6 \sqrt { 3 } i$