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?

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

  • $\begingroup$ We'll can I just choose $ \pi$ or $-\pi$ ? Because using that definition you provided, I got the same answer, by trying both values. $\endgroup$ – user367640 Sep 9 '16 at 23:23
  • $\begingroup$ @user367640 en.wikipedia.org/wiki/… says it can't be $-\pi$. $\endgroup$ – Simply Beautiful Art Sep 9 '16 at 23:43
  • $\begingroup$ @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. $\endgroup$ – Mariuslp Sep 10 '16 at 0:00

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