TITLE: Question on the $arg$ function
QUESTION [2 upvotes]: We recently came across $arg$ in our lectures:
$\arg\colon \mathbb{C}\setminus\{0\} \to \mathopen] -\pi, \pi]$,
$ z \to \arg(z)$, so $z=|z|e^{i{\arg(z)}}$
First question: why is the function bounded by $\mathopen]-\pi,\pi]$, surely it could also be bounded by $\mathopen]0,2\pi]$ or even by $[0,2\pi\mathclose[$. I do not understand the significance of its bound as it is periodical in any case?
Second question: I've had a look at the following problem: Solve $z^3=-i$, where $z \in \mathbb{C}$. 
Since $\arg(-i) \in \mathopen]-\pi,\pi]$ we are required to say $-i=e^{-i\frac{\pi}{2}+2ki\pi}=z^3$, $k\in \mathbb{Z}$. 
Therefore, $z_{k}=e^{\frac{-i\frac{\pi}{2}+2ki\pi}{3}}$, $k=0,1,2$
Using this we get:
$z_{0}=e^{\frac{-i\pi}{6}}$, $\arg(z_{0})=\frac{-\pi}{6}$, which lies in $\mathopen]-\pi,\pi]$
$z_{1}=e^{i\frac{\pi}{2}}$, which is correct as arg also lies in the range
Now the problem arrives, as:
$z_{2}= e^{i\frac{7}{6}\pi}$ and $\arg(z_{2})=\frac{7}{6}\pi$, which does not lie in the range $\mathopen]-\pi,\pi]$. Does this mean $z_{2}$ is not a solution to the problem? Or, is $z_{2}$ still as solution, as it is periodical and $\arg(z_{2})=\arg(z_{2})+2(-1)\pi=-\frac{5}{6}\pi \in \mathopen]-\pi,\pi]$. Am I missing something, or just leading myself astray? Thank you for your help.

REPLY [1 votes]: The convention that $\arg z$ has range $]-\pi,\pi]$ corresponds to the convention that the cut to form the Riemann surface for $\sqrt{z}$ is along the negative real axis; and the convention that the cut to form the Riemann surfsce for $\log z$ is along the negative real axis.
With these matching conventions, we have
$$
\sqrt{z} = \sqrt{|z|}\;e^{i(\arg z)/2}
\\
\log z = \log |z| + i \arg z
$$
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