# Principal argument without imaginary components

this is a pretty simple question. I have some minor confusion.

If I had complex number: $$z = -2 + 0j$$

The prinicpal argument would then be:

$$Arg(z) = \pi, -\pi$$ correct?

And if I also had $$z = 2 +0j$$

The prinicpal argument would then be:

$$Arg(z) = 0$$ correct?

• Principal = main (single) argument $0\le\theta\le2\pi$ But in 'principal', you are correct. – Simply Beautiful Art Sep 9 '16 at 23:12
• Ok, thank you for the response. – user367640 Sep 9 '16 at 23:14
• Oh, according to Joe, its $\theta\in[-\pi,\pi]$. But same thing basically – Simply Beautiful Art Sep 9 '16 at 23:17
• But for -2, the value could be either $\pi$ or $-\pi$? Aren't I just choosing different routes to get there? – user367640 Sep 9 '16 at 23:20
• Yes, I guess that would be right. Or you could modify it so $\theta\in(-\pi,\pi]$ instead. – Simply Beautiful Art Sep 9 '16 at 23:22

The principal argument of a complex nonzero number $z$ is defined as the unique $\theta\in]-\pi,\pi]$ such that $$z=|z|e^{i\theta}\;\;.$$ The principal argument is a function so it can't be multivalued.
• We'll can I just choose $\pi$ or $-\pi$ ? Because using that definition you provided, I got the same answer, by trying both values. – user367640 Sep 9 '16 at 23:23
• @user367640 en.wikipedia.org/wiki/… says it can't be $-\pi$. – Simply Beautiful Art Sep 9 '16 at 23:43
• @user367640: the very principle of principal argument is that there is only one possible value. By convention, the principal argument lies in $(-\pi, \pi]$, so $-\pi$ cannot be a possibility. But this is only a matter of convention, really. If you say that your principal range is $[-2\pi, 0)$, then the only possibility is $-\pi$. But with the classical convention, this is not possible. – Mariuslp Sep 10 '16 at 0:00