Write the complex number in polar form with $0 \leq \theta< 2 \pi$. Write your answer as a simplified fraction, if necessary. Give an exact answer.
Part: 0 / 2
Part 1 of 2
(a) $4-4 \sqrt{3} i$
The polar form is

Step 1 ：Given the complex number $4-4 \sqrt{3} i$, we need to convert it into polar form.

Step 2 ：The polar form of a complex number is given by $r(\cos \theta + i \sin \theta)$ where $r$ is the magnitude of the complex number and $\theta$ is the angle it makes with the positive real axis.

Step 3 ：The magnitude $r$ is given by $\sqrt{a^2 + b^2}$ where $a$ and $b$ are the real and imaginary parts of the complex number respectively.

Step 4 ：For the complex number $4-4 \sqrt{3} i$, $a = 4$ and $b = -4 \sqrt{3}$. So, $r = \sqrt{4^2 + (-4 \sqrt{3})^2} = 8$.

Step 5 ：The angle $\theta$ is given by $\arctan(\frac{b}{a})$ if $a > 0$, $\arctan(\frac{b}{a}) + \pi$ if $a < 0$ and $b > 0$, and $\arctan(\frac{b}{a}) - \pi$ if $a < 0$ and $b < 0$.

Step 6 ：Here, $\theta = \arctan(\frac{-4 \sqrt{3}}{4}) = -\frac{\pi}{3}$. Since we want $0 \leq \theta < 2 \pi$, we add $2 \pi$ to $\theta$ to get $\theta = \frac{5 \pi}{3}$.

Step 7 ：So, the polar form of the complex number $4-4 \sqrt{3} i$ is $8(\cos \frac{5 \pi}{3} + i \sin \frac{5 \pi}{3})$.

Step 8 ：Final Answer: The polar form of the complex number $4-4 \sqrt{3} i$ is \(\boxed{8(\cos \frac{5 \pi}{3} + i \sin \frac{5 \pi}{3})}\).