How to calculate $\theta$ when we know $\tan \theta$. Hej
I'm having difficulties calculating the angle given the tangent.
Example:
In a homework assignement I'm to express a complex variable $z = \sqrt{3} -i$ in polar form. I know how to solve this except for when I get to calculating the angle $\theta$.
I know that $\tan \theta = -\frac{1}{\sqrt{3}}$  but I do not know how to continue and compute the angle from that.
 A: You shouldn't use the tangent for this kind of problems; compute
$$
|z|=\sqrt{z\bar{z}}=\sqrt{(\sqrt{3}-i)(\sqrt{3}+i)}=
\sqrt{3+1}=2
$$
Then you have $z=|z|u$, where
$$
u=\frac{\sqrt{3}}{2}-i\frac{1}{2}
$$
and you need an angle $\theta$ such that
$$
\cos\theta=\frac{\sqrt{3}}{2},\quad\sin\theta=-\frac{1}{2}.
$$
Since the sine is negative and the cosine is positive, you see that you can take
$$
\theta=-\frac{\pi}{6}
$$
(the pair of values is well known). If you need an angle in the interval $[0,2\pi)$, just take
$$
-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}.
$$
A: $$\tan\theta=\frac{-1}{\sqrt3}=\pm\frac{\frac12}{\frac{\sqrt3}2}\implies\theta=\arctan\frac{-1}{\sqrt3}=\ldots$$
Remember though that
$$\sin\frac\pi6=\frac12\;,\;\;\cos\frac\pi6=\frac{\sqrt3}2$$
and since the minus sign belongs to the imaginary part (i.e., to the sine), it must be that
$$\theta=-\frac\pi6+2k\pi\;,\;\;k\in\Bbb Z$$
A: You can find the reference angle by disregarding the sign , you still need to figure out which quadrant you are in (which is easy) so you can add or subtract the reference angle accordingly
$$ \theta_R = tan^{-1} \frac{1}{ \sqrt{3}} $$
$$ \theta_R = \frac{ \pi}{6} $$
We are in the 4th quadrant so ,
$$ \theta = 2 \pi - \frac{ \pi}{6} = \frac{ 11 \pi}{6} $$
Now , if you need more working angles just add integer multiples of  $ 2 \pi $
