Let $z=-1-i$, find the principal value ? Here $x=-1,y=-1$ therefore $\arg(z)=\tan \alpha=|\frac{y}{x}|=|\frac{-1}{-1}|$

Therefore, $\alpha =\tan^{-1}=\frac{\pi}{4}$ which lies between $0$ and $\frac{\pi}{2}$. Since the point $(-1,-1)$ lie in third quadrant :

Therefore $\arg(z)=\pi+\alpha=\pi+\frac{\pi}{4}=\frac{5\pi}{4}$

Why principal value lies in range $-\pi<\alpha \leq \alpha$?

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    $\begingroup$ Your last question's inequalities don't seem to make much sense... And other question: what is "the principal value"? The one with an angle in $\,[0,2\pi)\,$ , or perhaps in $\,[-\pi, \pi)\,$ ...? $\endgroup$ – DonAntonio Mar 23 '13 at 15:59

Denote principal value of a complex number $z$ by Arg $z$. The principal argument Arz $z$ satisfies this inequality $-\pi<Arg Z\leq\pi$. The idea behind this inequality is to make the principal argument unique as you may know that the argument itself can take on infinitely many values. That is $arg z$=Arg$z$+$2\pi k$ where $k\in\mathbb{Z}$. And the reason why we don't include $-\pi$ is that $\pi$ and $-\pi$ are the same exact position there violating the uniqueness we wish to preserve.


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