0

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?

Xander Henderson
  • 29,772
Samar Imam Zaidi
  • 8,800
  • 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 $\arg$. – Xander Henderson Sep 03 '17 at 15:00
  • $arg(z)<0$ this is the only condition we gave – Samar Imam Zaidi Sep 03 '17 at 15:05
  • $\alpha$ represent the angle representation w.r.t Co-ordinate in terms of principal values – Samar Imam Zaidi Sep 03 '17 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. – Gribouillis Sep 03 '17 at 15:16
  • @Gribouillis arg(z)<0 in 2&4 Quadrant, tan value – Samar Imam Zaidi Sep 03 '17 at 15:19
  • What definition of $\arg(z)$ do you have ? – Gribouillis Sep 03 '17 at 15:21
  • The framer of this question has provided only the info presented in question rest is my assumption, the heading is complete question, my assumption can be wrong but question is correct. With respect to arg(z) i am confused whether we need negative tan value or negative angle value – Samar Imam Zaidi Sep 03 '17 at 15:32
  • Negative argument means negative angle. The argument of a complex number is its angle with the $x$-axis. Problems arise because an angle is only defined modulo $2\pi$. – Gribouillis Sep 03 '17 at 15:38
  • $https://math.stackexchange.com/questions/174154/how-to-get-principal-argument-of-complex-number-from-complex-plane$ may be this will solve the problem arg(z)<0 $\in$ (-π/2,0) considerint principal value – Samar Imam Zaidi Sep 03 '17 at 15:43

1 Answers1

1

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 $z$.

Cye Waldman
  • 7,524