$\arg(r_+) = 0$ and $\arg(r_-) = \pi$

Question

$r_+$ = positive real number

$r_-$ = negative real number

I am confused $\arg(r_+) = 0$ and $\arg(r_-) = \pi$

for $\arg(r-) = \pi$ :

for example, $\arg(-1)$ can be represented as $-1 + 0i $

$\cos(\arg(-1)) = \frac{-1}{1} = -1$; this equals to $\pi$

$\sin(\arg(-1)) = \frac{0}{1} = 0$ this equals to $0$ or $\pi$

$\arctan({\arg(-1)}) = 0$

so we care about real number is $(-1)$ so its cosine and is $\pi$, correct?

for next one is where I have confusing of $\arg(r_+) = 0$

for example $\arg(1) = 0$

$\cos(\arg(1)) = \frac{1}{1} = 1$ , $\sin(\arg(1)) = \frac{0}{1} = 0$ , $\arctan({\arg(1)}) = 0 $, so the answer is zero. So we chose value of $\arctan$?

now let's do example of $\arg(\lvert2+i\rvert)$ and $r = \sqrt{5}$ , so automatically is zero? This is where I am confused. I hope from the start to now is correctly shown so far?

Answer 1

You seem to confuse $\arg(|2+i|)$ and $\arg(2+i)$. The first is indeed the argument of a positive real number and is $0$.

For every nonzero $z$, $\arg(|z|)=\arg(|re^{i\theta}|)=\arg(r)=\arg(re^{i0})=\arg(e^{i0})=0$, while $\arg(z)=\arg(re^{i\theta})=\arg(e^{i\theta})=\theta$.