Adding theta to 360 ?

Question

The complex number given to me was 2-$\sqrt{3}$i.

I'm supposed to turn this into the polar form, and so I solved for r using r = $\sqrt{a^2+b^2}$, getting $\sqrt{7}$. When I solved for theta using tan$\theta$ = $\frac ba$ , I got tan$\theta$ = -$\sqrt{3}$ over 2. I know that I would get the inverse of that and then add it to 360. The degree would be around 319.1 degrees.

My question is why would I have to add additional degrees to it ? Another problem was similar to this but I added it to 180 instead but I didn't know why. Why do I add it and when is it a good time to use it ? How would I know which one is the appropriate one to use ?

Answer 1

We add the the additional $\pi$ or other values in order to adjust the angle to get the principle angle .

Take any non zero complex number , $z = x+iy $

$\arg(z) = \arctan\bigg(\frac {|y| }{|x| }\bigg)$

that gives you some angle , then find the principle angle by seeing the sign of the real and imaginary components of the complex number and referring to the diagram below.

Answer 2

Since your number is $\sqrt7\times\left(\frac2{\sqrt7}-\frac{\sqrt3}{\sqrt7}i\right)$, it is equal to $\sqrt7\bigl(\cos\theta+i\sin\theta\bigr)$ with $\theta=-\arccos\frac2{\sqrt7}$. The minus sign comes from the fact that $-\frac{\sqrt3}{\sqrt7}<0$. So, the set of all polar forms of your number is$$\left\{\sqrt7\bigl(\cos(\theta+2n\pi)+i\sin(\theta+2n\pi)\bigr)\,\middle|\,n\in\mathbb{Z}\right\}.$$