If $\arg (z) < 0$ , then $\arg (- z) - \arg (z)$ = ? Concept used

Principal Value

  1. Quad I = $\alpha$
  2. Quad II = $\pi -\alpha$
  3. Quad III = $-\pi+\alpha$
  4. Quad IV = $-\alpha$

Case 1: As $\arg(z) <0$, let us presume $z$ lies in QUADRANT II, therefore $-z$ lies in QUADRANT IV, therefore $A=\arg(-z) = -\alpha$ and $B=\arg(z)=\pi -\alpha$


Case 2: As $\arg(z) <0$, let us presume $z$ lies in QUADRANT IV therefore $-z$ lies in QUADRANT II , therefore$ A=\arg(-z)=\pi -\alpha$ and $B=\arg(z)=-\alpha$,

$$ A-B=\pi$$

Where I am making mistake?

  • $\begingroup$ What is $\alpha$? How are you defining $\arg$? that is, is $\arg$ the principal argument, or just some branch of the argument? Also \arg will produce a properly formatted $\arg$. $\endgroup$
    – Xander Henderson
    Sep 3, 2017 at 15:00
  • $\begingroup$ $arg(z)<0$ this is the only condition we gave $\endgroup$ Sep 3, 2017 at 15:05
  • $\begingroup$ $\alpha$ represent the angle representation w.r.t Co-ordinate in terms of principal values $\endgroup$ Sep 3, 2017 at 15:09
  • $\begingroup$ It seems the formulas are correct if $\alpha = \tan^{-1}\left|\frac{y}{x}\right| \in [0, \frac{\pi}{2})$, where $z = x+i y$. Note that $\arg(z)<0$ occurs only when $z$ is in the 3rd and 4th quadrants. $\endgroup$ Sep 3, 2017 at 15:16
  • $\begingroup$ @Gribouillis arg(z)<0 in 2&4 Quadrant, tan value $\endgroup$ Sep 3, 2017 at 15:19

1 Answer 1


It doesn't matter that $\arg z<0$. Consider

$$ z=|z|e^{i\theta}\\ $$



where $\tilde\theta=\arg(-z)$



This has been verified numerically for random values of $z$.

  • $\begingroup$ But answer is π and not -π, i cross checked it z lies in IV Quadant $\endgroup$ Sep 3, 2017 at 15:47
  • $\begingroup$ That is correct, when $z$ is in the $4^{th}$ quadrant the result is $+\pi$. I have resolved the problem to $\pm\pi$. The rest is easy. $\endgroup$ Sep 3, 2017 at 15:51
  • $\begingroup$ Because arg(z) $\in (-π,π]$ & tan z values is (-π/2,π/2) then Quadrant IV is the location $\endgroup$ Sep 3, 2017 at 15:58
  • $\begingroup$ Excellent concept but one problem you cannot multiply by -π because arg (z) $\in$ (-π,π] it is open at -π but closed at π $\endgroup$ Sep 3, 2017 at 16:04

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