One of the hints on KhanAcademy appears to skip how they found the inverse tangent as a fraction of pi over a number.
Every time I type this into a calculator it comes up only as a decimal, and I'm not sure how you could find it with pi inside the fraction unless you already memorized the number to 9 digits or something.
The main problem:
Express $z_1 = 3\sqrt{3} - 9i$ in polar form.
Express your answer in exact terms, using radians, where you rangle is between 0 and $2\pi$ radians, inclusive.
The part that appears to skip how to find the answer as a fraction with pi involved:
Finding $\theta$
Using the formula, we have:
$\theta = \tan^{-1}(\frac{b}{a})$
$\theta = \tan^{-1}(\frac{-9}{3\sqrt{3}})$
$\theta = -\frac{\pi}{3}$
I am assuming you have the requisite knowledge of radians and the unit circle in this answer.
Your question just seems to be a question on why $\arctan({-\sqrt{3}}) = -\frac{\pi}{3}$. The definition of $\arctan$ is just the inverse of the tangent function. If you know how to take the tangent of an angle, then you know the arctangent. If I want the value of $\arctan(x)$, then the arctangent function just asks "What ANGLE do I need to make such that the tangent of that angle is $x$?"
In your case, you want $\arctan({-\sqrt{3}}).$ On the unit circle, what ANGLE do I need to make from the axis such that the tangent of that angle is $-\sqrt{3}$? The answer? $-\frac{\pi}{3}$. And you can verify that $\tan(\frac{-\pi}{3}) = -\sqrt{3}$.
This is assuming you have the requisite knowledge of the unit circle to deduce that $\tan({\frac{-\pi}{3}}) = -\sqrt{3},$ though. Let me know if you have any questions.