# $\arg (z) < 0$ , then $\arg (- z) - \arg (z)$ =

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 $B=\arg(z)=\pi -\alpha$

$$A-B=-\pi.$$

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?

• 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 $\arg$. Sep 3, 2017 at 15:00
• $arg(z)<0$ this is the only condition we gave Sep 3, 2017 at 15:05
• $\alpha$ represent the angle representation w.r.t Co-ordinate in terms of principal values Sep 3, 2017 at 15:09
• 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. Sep 3, 2017 at 15:16
• @Gribouillis arg(z)<0 in 2&4 Quadrant, tan value Sep 3, 2017 at 15:19

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

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

Now,

$$-z=|z|e^{i(\theta\pm\pi)}=|z|e^{i\tilde\theta}$$

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

Then

$$\arg(-z)-\arg(z)=\tilde\theta-\theta=\pm\pi$$

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

• But answer is π and not -π, i cross checked it z lies in IV Quadant Sep 3, 2017 at 15:47
• 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. Sep 3, 2017 at 15:51
• Because arg(z) $\in (-π,π]$ & tan z values is (-π/2,π/2) then Quadrant IV is the location Sep 3, 2017 at 15:58
• 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 Sep 3, 2017 at 16:04