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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 !

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    $\begingroup$ First things first: there is not one universally accepted definition of "principal" argument. For some it will lie in $(-\pi,\pi] $, for others in $[0, 2\pi)$. $\endgroup$ – Arnaud Mortier May 29 '18 at 22:50
  • $\begingroup$ We know that argument represents the angle in the complex plane. For any z , Arg is bounded by $0 \leq arg(z) < 2 \pi$ $\endgroup$ – Clark Makmur May 29 '18 at 22:51
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    $\begingroup$ I think it simply means $\;\arg z^n\equiv n \arg z \mod 2\pi$, and the value of $k$ is such that $n\arg z +2\pi k$ lies in the interval we've chosen for the principal argument. $\endgroup$ – Bernard May 29 '18 at 22:53
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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).

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  • $\begingroup$ I think you want to remove "except for $ z=0 $". $\endgroup$ – Arnaud Mortier May 31 '18 at 12:38
  • $\begingroup$ @ArnaudMortier: well , I put that to keep it valid for $n$ in all $\mathbb Z$. But your comment is accepted in making the statement more precise. $\endgroup$ – G Cab May 31 '18 at 14:10
  • $\begingroup$ Right. But then you can't say that a function is entire if it is not holomorphic on the whole of $\Bbb C $. $\endgroup$ – Arnaud Mortier May 31 '18 at 14:29
  • $\begingroup$ @ArnaudMortier: I tried an additional step towards precision (which in complex field is always necessary, you are right! thanks so much). Is that correct now ? or , since you are more expert , could you please amend it correctly or put as your own answer? $\endgroup$ – G Cab May 31 '18 at 15:47

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