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$r_+$ = 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?

Kenta S
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EM4
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  • It is hard to comprehend your question(the cosine and sine part), how do you define $\arg(z)$? – UmbQbify Jul 24 '20 at 23:15
  • arg(-1) can be arg(-1+0i) , then I found r = $\sqrt{x^2 + y^2)$ which is 1. – EM4 Jul 24 '20 at 23:27
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    Yeah, I get that, argument is simply defined as arctan of imaginary part over real part of $\tan^{-1}\left( \frac{y} x\right)$ so why do you need sine and cosine? – UmbQbify Jul 24 '20 at 23:31
  • I believe I went the long route I guess, but the arctan will give you zero and arg(-1) is pi...and I couldn't understand it that route. Even with your the arg(1) and afterwards. – EM4 Jul 24 '20 at 23:35
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    Okay, first of all, you know that $\frac{\sin^{-1}x}{\cos^{-1}x}\ne \tan^{-1}x$, right? Secondly, range of $\tan^{-1}x$ is $\left[-\frac{\pi}2,\frac{\pi}2\right]$ so it cannot give a value of $\pi$ (because it is outside of that interval). Note, $\tan^{-1}(-1)=-\frac{\pi}4$ but $\arg(-1+i)= \frac{3\pi}4$ – UmbQbify Jul 24 '20 at 23:41
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    Argument is closely related to rotation. By convention, anticlockwise rotation is taken as postive and clockwise as negative. Thus, the all the angles in $[-\pi,\pi]$ cover the whole Argand plane. That's why the principal values of argument lie in between $-\pi$ to $\pi$. I can't comment on why is it so because I'm not totally sure (because it depends on the sign of real and imaginary ) – UmbQbify Jul 24 '20 at 23:45
  • A much simpler question is: What does the OP think what the meaning is of Argument of a complex number? No trig needed for this answer. – imranfat Jul 25 '20 at 01:10
  • huh? I am just more confused why Arg of positive real is 0 and Arg negative real is pi. Is it a point in the graph? – EM4 Jul 25 '20 at 01:17
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    Just look at the Argand diagram. The argument is the amount of anti-clockwise rotation from the positive real axis needed to reach the number. So, if your number is on the positive real axis then no rotation is needed and the argument is $0$. However, if the number is on the negative real axis then you need $180^\circ$ but in, this context, radians would always be used so $\pi$. Similarly, the argument of $i$ is $90^\circ$ or $\frac{\pi}{2}$ and the argument of $-i$ is $270^\circ$ or $\frac{3 \pi}{2}$ or $- \frac{\pi}{2}$. Pick a rule to choose one of these. – badjohn Jul 25 '20 at 08:53

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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 $\arg(z)=\arg(re^{i\theta})=\arg(e^{i\theta})=\theta$.

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    I figured it out, used this understanding and drawing it out. This is what I did:

    I basically arg(|2+|) = arg(√5) it lies on the 1st quadrant in the Re axis, and there is no rotation and the arg is 0 for positive real number.

    and for the negative Re it lies on second quadrant (0 + pi) and it will be pi.

     – EM4 Jul 25 '20 at 12:03