How to calculate $\theta$ when we know $\tan \theta$.

Question

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.

Answer 1

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 $$ -\frac{\pi}{6}+2\pi=\frac{11\pi}{6}. $$

Answer 2

$$\tan\theta=\frac{-1}{\sqrt3}=\pm\frac{\frac12}{\frac{\sqrt3}2}\implies\theta=\arctan\frac{-1}{\sqrt3}=\ldots$$

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

$$\theta=-\frac\pi6+2k\pi\;,\;\;k\in\Bbb Z$$

Answer 3

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 $