Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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 $0 < \theta \le 2 \pi$?
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.
The argument of a non-zero complex number $z$ is a multi-valued function defined by $$\arg z = \theta + 2\pi n, \, \, \, n \in \mathbb{Z}$$
where $\theta \equiv \text{Arg }z $, the principal value of the argument, which by convention is taken to be $$-\pi < \theta \leq \pi$$
But this is just by convention, you could equally well have taken the principal value to be $$0 < \theta \leq 2\pi$$ and all your work with complex numbers would be just as valid.
We now have $$\arg z = \text{Arg }z + 2\pi n, \, \, \, \, n \in \mathbb{Z}$$ we can now define $$\text{Arg }z = \arg z + 2\pi \left\lfloor\frac{1}{2} - \frac{\arg z}{2\pi}\right\rfloor $$
