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I know that, given $n\in\mathbb{N}$, for every $z\in\mathbb{C}$, $z\neq0$, there are exactly $n$ distinct $n$th roots of $z$.

To prove this: given $z\neq0$, a $n$th root of $z$ is a complex number $w$ such that $w^n=z$. By posing $$ w = r(\cos\theta+i\sin\theta), $$ we have $w^n=r^n(\cos n\theta+i\sin n\theta)$. So, in order to have $w^n=z$, it must be $$ r = |z|,\qquad n\theta=\phi+2k\pi,\quad k\in\mathbb{Z}, $$ where $\phi$ is the argument of $z$, because two complex numbers are equal if and only if their modulus coincide and their arguments are the same up to a multiple of $2\pi$. Is it right?

My question: Why in some book, in order to have $w^n=z$, is $n\theta=\operatorname{Arg}z+2k\pi$, where $\operatorname{Arg}z$ is the principal value of the argument of $z$? Should $\phi$ be the argument (and not the principal value of the argument) of $z$?

Thank You

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There is no such thing as the argument: every complex number has infinitely many arguments. If you say that you are working with the principal value of the argument, then there is no ambiguity (although, in fact, any argument will do).

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  • $\begingroup$ +1. The point being you have to choose one argument, and the principal value is usually the easiest one to choose $\endgroup$ – Henry Apr 4 '18 at 8:52

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