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I am seeking a succinct mathematical definition of the "principal argument" of a complex number.

Principal argument of a complex number, denoted $\operatorname{Arg}(z)$ means the argument of a complex number $\arg(z)$ within the range $(-\pi, \pi]$.

How do I represent this mathematically?

I've tried using $\operatorname{Arg}(z) = \arg(z) \in (-\pi, \pi]$, but this doesn't make sense.

There is also a very complicated one on Wikipedia, $\operatorname{Arg}(z) = \{\arg(z) - 2\pi k \mid k \in \mathbb{Z} \wedge -\pi < \operatorname{Arg}(z) \leq \pi \}$

There definition makes it seem as if there can be multiple values of $\operatorname{Arg}(z)$.

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  • $\begingroup$ what's the mathematical definition of $arg(z)$ ? $\endgroup$
    – mercio
    Commented Sep 20, 2018 at 20:05
  • $\begingroup$ Let there be a circle on a Cartesian plane with center ($0,0$). Let ($a, b$) be a point on the circle arg(a,b) is the number of radians of the circular arc between ($a, b$) and ($\sqrt{a^2+b^2}, 0$) $\endgroup$ Commented Feb 14 at 1:59

1 Answer 1

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For each complex number $z\in\mathbb C\setminus\{0\}$ there is a unique $r>0$ and $\theta\in(-\pi,\pi]$ such that $z=r\,\exp(i\theta)$, where $\exp\colon\mathbb C\to\mathbb C$ is the complex exponential function. You can now define the principal argument as $\operatorname{Arg}(z) = \theta$. The multivalued argument assigns to $z$ all possible $\theta\in\mathbb R$ such that $z=r\,\exp(i\theta)$ which amounts to $\operatorname{Arg}(z)+2\pi n$ where $n$ ranges over all integers.


The definition of the principal argument in terms of the multivalued argument as stated in the Wikipedia article you referenced is circular and doesn't make any sense. It could be corrected as

$\operatorname{Arg}(z)$ is the unique element $\theta\in\operatorname{arg}(z)$ such that $\theta\in(-\pi,\pi]$.

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  • $\begingroup$ If we have $Arg(z^5)$ where $z = 1 + i$ for example, what happens to our principal argument? Does it remain unchanged or does it become $-5\pi< \theta < 5\pi$ $\endgroup$
    – Oofy2000
    Commented May 19 at 0:51

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