Why is principle value of argument generally taken as $(-\pi , \pi]$?

Question

I have just started reading about the Modulus and Argument of Complex Numbers. In the definition, it is said that:

If z is not equal to 0 and $-\pi < \theta \le \pi$ , then $\theta$ is the principal argument of z, written $\theta = \arg(z)$.

My question is about the interval: why do we take $-\pi < \theta \le \pi$, which, if I understand it correctly starts at $180$ degrees, namely $-\pi$ (the most left point on the $x$ axis of the zero circle), and moves counter clockwise until it reaches $180$ degrees again? Why don't we use, say, $[0,2\pi)$?

Not much of an advantage, but in this principal value range, if $\arg(z) = t$ then $\arg(z^*)$ is $-t$. — jonsno, Sep 03 '17 at 12:08
@Shaun Have I incorrectly used the MathJax's symbols? I did my best :) Please, show me what I did wrong. Thank you very much! — Vitale, Sep 03 '17 at 12:09
@samjoe What does asterisk refer to? Sorry, I am not not that good yet. ) — Vitale, Sep 03 '17 at 12:11
Note that $-\pi<\Theta\le\pi$ because $\Theta=\pi$ is possible. — Shaun, Sep 03 '17 at 12:12

score 2 · Answer 1 · answered Sep 03 '17 at 12:06

2

It's just a matter of convention.

answered Sep 03 '17 at 12:06

Shaun

44,997

Thank you. What does this convention refer too, and what meaning lies behind it? – Vitale Sep 03 '17 at 12:07
It's somewhat arbitrary. – Shaun Sep 03 '17 at 12:08
3

I wouldn't say that it is arbitrary. This convention is related to the principal determination of the complex log function which, rather often, is defined by a cut along the negative $x$ axis. – Jean Marie Sep 03 '17 at 16:51

Why is principle value of argument generally taken as $(-\pi , \pi]$?

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