I know how to convert a complex number from rectangular form to polar form, but I can't understand when should I add or subtract $\pi$ from/to arctan in different quadrants.
Can some one explain this for all quadrants? (maybe except for I)
Converting rectangular form complex numbers to polar form in different quadrants
complex numbers
Best Answer
Normally, I'm not supposed to answer, because the OP hasn't shown any work. But this is a different situation, so if I get downvoted, okay. The way that I was taught ("An Introduction To Complex Function Theory" : Bruce Palka) you are supposed to use the sine and cosine functions.
Let $z = (x + iy)$, where $x$ and $y$ are not both $= 0$.
Let $r = +\sqrt{x^2 + y^2}$.
Then, within a modulus of $(2\pi)$, there will be a unique angle $\theta$, such that
$\cos(\theta) = \frac{x}{r}$ and $\sin(\theta) = \frac{y}{r}.$
Then, $(x + iy) = z = r[\cos(\theta) + i\sin(\theta)] = re^{i\theta}.$
Therefore, you never have to concern yourself with the tangent function, or wondering which quadrant $\theta$ is in. The quadrant is determined by whether (for example) $\frac{x}{r}$ and/or $\frac{y}{r}$ is positive.
Addendum
Responding to the OP's comment/question. $z = -3 + i\sqrt{3}$.
$r^2 = (-3)^2 + \left(\sqrt{3}\right)^2 = 9 + 3 = 12 \implies $
$r = 2\sqrt{3}.$
Therefore, you are looking for $\theta$ such that
$\cos(\theta) = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$ and
$\sin(\theta) = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$.
Looking at the sign of $\cos(\theta)$ and $\sin(\theta)$, you immediately deduce that $\theta$ is in the 2nd quadrant. Then, for this particular problem, you can simply draw the unit circle, identify the corresponding angle in the first quadrant, and then infer the value of $\theta$ in the 2nd quadrant.
For example, you know that $\cos(\pi/6) = \frac{\sqrt{3}}{2}.$
Therefore, you can infer, by symmetry, that $\theta$ must be $(\pi - \pi/6) = (5\pi/6)$.
You could have instead focused on the fact that $\sin(\pi/6) = (1/2)$, and reached the same inference. Using symmetry around a drawing of the unit circle is a good way of stretching your intuition to easily handle situations like this.
Anyway, once you determine that $(\theta = 5\pi/6)$ and $(r = 2\sqrt{3})$, you know that
$-3 + i\sqrt{3} = 2\sqrt{3}[\cos(5\pi/6) + i\sin(5\pi/6)].$
For a more systematic way of handling situations like this, first see this wikipedia article.
Then, you could (for example) use the appropriate formula to deduce that $\sin(\pi - \theta) = \sin(\theta).$
Then, you can easily attack a situation like this sytematically by first learning (not memorizing) the sine and cosine functions for such special angles as $\{0, \pi/6, \pi/4, \pi/3, \pi/2\}$, and the corresponding angles in each of the other 4 quadrants.
Actually, I used the verb learn to indicate that once you've got the 1st quadrant down, you should be able to infer the trig functions that correspond to the special angles in the other quadrants, simply by looking at the unit circle, and making the intuitive leaps, around ideas involving symmetry.