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What is the value of $\text{arg}(z) + \text{arg}(\bar{z})$ for a complex number?

According to me the answer should be $0$, by basic $\tan^{-1}$ logic and also since $$arg(z) + \text{arg}(\bar{z}) = \text{arg}(z\bar{z}) = \text{arg}(x^2 + y^2) = 0$$

Considering that $z = x + iy$, therefore I wanted to clarify that is it always $0$ or can it be $2\pi$ aswell? If so in which case and how?

Sebastiano
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marks_404
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    Depends on your interval of "principal values of arguments". – xbh Aug 16 '21 at 16:10
  • $\arg(z)$ is a multi-valued function, you have to specify first which value is chosen. – Martin R Aug 16 '21 at 16:12
  • @xbh oh now I understand why there was an option of "Can't say" since the question would be baseless without mentioning the principal argument. – marks_404 Aug 16 '21 at 16:38

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The argument of a complex number is only defined modulo $2\pi$ so the arguments $0$ and $2\pi$ are essentially the same.

Sometimes texts use the convention that arguments must be in the interval $[0,2\pi)$. In that case $2\pi$ is not the argument.

Ethan Bolker
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