complex numbers
In my text book it says that the complex number z(not equal to 0) can be written in polar form as $z = r(\cos\theta + i \sin\theta)$, where r = mod z greater than 0 is the modulus and $\theta = \arg z$, with $-\pi \le \theta \le\pi$.
Why must the condition $-\pi < \theta \le \pi$ hold? Why not $0 < \theta \le 2 \pi$?
The problem is that points can be expressed in polar form in more than one way. Take a look at the diagram below:
There's your point (in blue). As you can see, it's in the second quadrant, and the angle $\theta = \frac{2\pi}{3}$ passes through it. So one could accurately say that the point's polar representation is $r = 1, \theta = \frac{2\pi}{3}$, i.e. $$z = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)$$
You'll notice, though, that $\theta = -\frac{\pi}{3}$ represents the same angle, but in the opposite direction. With this in mind, one could just as accurately say that the point's polar representation is $r = -1, \theta = -\frac{\pi}{3}$, i.e. $$z = -\cos\left(-\frac{\pi}{3}\right) - i\sin\left(-\frac{\pi}{3}\right)$$
Because $\cos\left(\frac{2\pi}{3}\right) = -\cos\left(-\frac{\pi}{3}\right)$ and $\sin\left(\frac{2\pi}{3}\right) = -\sin\left(-\frac{\pi}{3}\right)$, these two polar representations are equivalent. In other words, there's nothing wrong with concluding, as you did, that $\theta = -\frac{\pi}{3}$ You and the so-called "required answer" can reasonably disagree here and both be correct.
But you still did something wrong. If you choose to use $\theta = -\frac{\pi}{3}$, then you must also choose $r = -1$, yet you choose $r = 1$, based on your modulus computation. How can you avoid making this mistake in the future?
Simple: look at the quadrant in which the point lies. The point $-\frac{1}{2} + \frac{\sqrt{3}}{2}i$ has negative real part and positive imaginary part. Therefore it must be in the second quadrant. The angle $\theta = -\frac{\pi}{3}$ extends into the fourth quadrant, so if you intend to use that angle, then you must also make $r$ negative.
Hint: irrespective of $|z|$, you will get $\tan(\theta)=\frac{4}{3}$. So, taking the inverse will give you $\theta$ in radians.
Best Answer
Strictly speaking, the argument of a complex number is an element of the quotient group $\mathbf R/2\pi\mathbf Z$. Usually one takes a full set of representatives of this group – mainly $(-\pi,\pi]$ and $[0,2\pi)$, i.e. a set of real numbers such that any real number is congruent, modulo $2\pi\mathbf Z$, to exactly one number in the set.