Does the $\lim_{z \to -1} \sqrt{|z|} \,e^{i\operatorname{Arg}(z)/2}=i$? I am asked to show whether or not the $\lim_{z \to -1}f(z)$ where $f(z)=\sqrt{|z|} \,e^{i\operatorname{Arg}(z)/2}$ exists.
Simply plugging in $-1$ into the limit, we get, $$\lim_{z \to -1} \sqrt{\left|-1\right|}\: e^{i\operatorname{Arg}(-1)/2}\\=\lim_{z \to -1} e^{i\operatorname{Arg}(-1)/2}$$
The major problem I am facing is figuring out what the $\operatorname{Arg}(-1)$ is going to be. After doing some research, I came across a Wolfram Alpha article which said the $\operatorname{arg}(-1)=\pi$. I am assuming this holds true for the principal argument as well. Solving, we get, $$\lim_{z \to -1} e^{i(\pi/2)}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i$$
Although, I need some help in proving why the $\operatorname{Arg}(-1)=\pi$. Unfortunately, I was not able to find much information about it online, so I would appreciate some assistance.
 A: *

*" I came across a Wolfram Alpha article which said the
$\arg(−1)=\pi$. I am assuming this holds true for the principal
argument as well. "
$\pi$ is the principal argument of $−1$, i.e., Arg$(-1)=\pi.$
But in general, $\arg(−1)=(2k+1)\pi,$ where $k\in\mathbb{Z}.$


*" I have always thought about finding $\arg(z)$ by applying the
$\tan^{−1}\left(\frac{y}{x}\right).$ "
This formula doesn't generally work, since
$\tan^{-1}\left(\frac{y}{x}\right) \in
   \left(-\frac{\pi}2,\frac{\pi}2\right),$ which doesn't even span the
principal range $\left(-\pi, \pi\right]$ of $\arg(z)$:
$$\displaystyle\text{Arg}(-1-i)=-\frac34\pi \\\neq \frac{\pi}4
   =\tan^{−1}\left(\frac{-1}{-1}\right).$$
The given limit $$\lim_{z\to-1} \sqrt{|z|} \exp\left(i\frac{\text{Arg}(z)}2\right)$$ doesn't exist because $\displaystyle\lim_{z\to-1}\frac{\text{Arg}(z)}2$ (and in fact, $\displaystyle\lim_{z\to-1}\frac{\arg(z)}2$) has two non-overlapping representations $\displaystyle\pm\frac{\pi}2$ on the Argand diagram, as hinted by F. Tomas in the comments.
