Write the complex number in trigonometric form r(c $$r ( \cos \theta + i \sin \theta )$$ ), with $$\theta$$ in the interval [0 $$\left[ 0 ^ { \circ } , 360 ^ { \circ } \right)$$
-$- 4 - 4 i \sqrt { 3 }$
A) $8 \left( \cos 290 ^ { \circ } + i \sin 290 ^ { \circ } \right)$
B) $4 \left( \cos 240 ^ { \circ } + i \sin 240 ^ { \circ } \right)$
C) $8 \left( \cos 240 ^ { \circ } + i \sin 240 ^ { \circ } \right)$
D) $8 \left( \cos 180 ^ { \circ } + i \sin 180 ^ { \circ } \right)$

Question

Quizplus · Accepted Answer

The magnitude $r$ of the complex number $-4 - 4i\sqrt{3}$ is calculated using the formula $r = \sqrt{a^2 + b^2}$, where $a = -4$ and $b = -4\sqrt{3}$. Thus, $r = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$. The angle $\theta$ is found by using the formula $\tan(\theta) = \frac{b}{a}$, which gives $\tan(\theta) = \frac{-4\sqrt{3}}{-4} = \sqrt{3}$. This corresponds to an angle of $60^\circ$, but since the complex number is in the third quadrant (both real and imaginary parts are negative), we add $180^\circ$ to find the correct angle in the specified interval, resulting in $240^\circ$. Therefore, the trigonometric form is $8(\cos 240^\circ + i\sin 240^\circ)$.

Write the Complex Number in Trigonometric Form R(c r(cos⁡θ+isin⁡θ)r ( \cos \theta + i \sin \theta )r(cosθ+isinθ)

Write the Complex Number in Trigonometric Form R(c $r ( \cos \theta + i \sin \theta )$