A certain troublesome property of the complex argument function

Question

When writing $$\arg(z^n) = n\arg(z) + 2πk$$ and letting $\arg$ denote the principal complex argument of $z$. Is $k$ generally an integer or is it that $0\lt k\lt n$ or $k=[\frac{1}{2}-\frac{n}{2\pi}\arg(z)]$ as some books suggest? Obviously, I don't understand any of this and would appreciate if someone explained this tricky situation.. Thanks in advance !

Answer 1

If $n$ is a non-negative integer, then $z^n=z \cdot z \cdot \; \cdots$ is well defined (analytic and entire) on all $\mathbb C$.
If $n$ is a negative integer, then $z^{n}=(1/z)^{|n|}$ and it is meromorphic, with only a pole of order $|n|$ at $z=0$.

In both cases,the argument (apart the $i2\pi$) is also well defined to be $\{n\arg(z)/(2\pi)\}(2\pi)=(n\arg(z))\mod{(2\pi)}$ where the brackets indicate the fractional part. That as much as $\arg(z)$ is defined.

The above if you define $0\le \arg(z) <2\pi$.
If instead, as rightly indicated in a comment, the definition is $-\pi < \arg(z) \le \pi$ (which is that adopted in all major CAS nowadays), in any case you shall reduce $n\arg(z)$ to fall therein.

If $n$ is instead rational, then it comes that you have to choice the branch: the example of $z^{1/2}=\pm \sqrt{z}$ is well known and I will not continue further (you can find a more authoritative explanation here).