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Let $n$ be any integer number, and let $w$ be a non-zero complex number. In polar form, let us suppose $w = r\exp(i\phi)$. The solutions of the equation $$ z^n = w $$ are therefore $$ \alpha, \quad \alpha\omega, \quad \alpha\omega^2, \quad \dots, \quad \alpha\omega^{n-1} $$ where $\omega$ is the $n$th complex root of unity with the smallest positive argument, and where $$ \alpha = \sqrt[n]{r}\exp\left(\frac{i\phi}{n}\right) $$ Using de Moivre's theorem, it is easy to prove this fact. In fact, writing $z = \rho \exp(i\theta)$ and substituting this into $z^n = w$ gives $$ \rho^n\exp(in\theta) = r\exp(i\phi) $$ and therefore $$ \rho = \sqrt[n]{r}, \qquad n\theta \equiv \phi \pmod{2\pi} $$ and the second of these equations gives $$ \theta = \frac{\phi}{n} + \dfrac{2k\pi}{n} \quad k = 0,1,2,\dots,(n-1) $$ so eventually $$ z = \sqrt[n]{r}\exp\left(\frac{i\phi}{n} + \frac{2i\pi}{n}\right) = \sqrt[n]{r}\exp\left(\frac{i\phi}{n}\right)\exp\left(\frac{2k\pi i}{n}\right) = \alpha\omega^k $$ The proof above is okay, but I was wondering whether there was a more 'algebraic' proof of this fact. After all, it is sort of considering $z^n = w$ as $z^n = w \times 1$ and extracting the $n$th root, one would get $z = \sqrt[n]{w}\times\sqrt[n]{1}$, but this is not formal. What I was looking for is whether there is a purely algebraic reason of this fact, that doesn't involve the polar form of a complex number. After all, the set $\mathbb{U} = \{ z \in \mathbb{C} : |z|=1\}$ is rich of structure.

Thank you for any suggestion you may give me.

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Given any two roots, $z_1,z_2$ you have $$\left(\frac{z_1}{z_2}\right)^n =\frac ww=1.$$ So, once you have one solution, the other $n$ can be expressed as the product of that solution and some solution to $y^n=1.$

You still need to find one root of $z^n=w,$ but I don’t think there is an algebraic solution there.

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Hint: If $z_1$ and $z_2$ are solutions of $z^n=w$, then $z_1/z_2$ is a solution of $z^n=1$ and so $z_1=z_2 \omega$, where $\omega$ is a $n$th root of unit.

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