# Range of principal argument [duplicate]

For a polar representation of a complex number, why the principal argument is to be considered in $(-\pi, \pi]$ or how can it arise?

I read some books where they just maintained it, but they don't give any reason. Also I read in Wikipedia page where they maintain: Some authors define the range of the principal value as being in the closed-open interval $[0, 2\pi)$, but why? Further I search in MSE and get two same problems(links are 1, 2). But i am not satisfied with such comments only. I expect for a detailed explanation. My special focus is how did it assumed. Any help is highly appreciated.

## marked as duplicate by Xander Henderson, Brian Borchers, user99914, Sahiba Arora, NamasteFeb 6 '18 at 17:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Those 'comments' are the answers. You need to explain why you are not satisfied. – user525761 Feb 6 '18 at 14:04
• @treeleaf can you briefly explain it? – SAHEB PAL Feb 6 '18 at 14:08

## 1 Answer

The reason that makes it feel natural for me is that the value of the argument in $(-\pi,\pi)$ corresponds to the shortest distance you have to walk, from the origin at $1$, to reach a given point on the circle, where the sign tells you when the shortest path involves walking backwards.

The fact that you choose $\pi$ for the principal argument of $-1$, instead of $-\pi$, is a matter of choice.

Note: this is assuming walking along a unit circle, so that the angle actually corresponds to the arc-length.