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

Mathematics Asked on January 5, 2022

$$r_+$$ = 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 $$arg(lvert2+irvert)$$ 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?

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 $$z$$, $$arg(|z|)=arg(|re^{itheta}|)=arg(r)=arg(re^{i0})=arg(e^{i0})=0$$, while $$arg(z)=arg(re^{itheta})=arg(e^{itheta})=theta$$.

Answered by user65203 on January 5, 2022

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