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I am trying to prove:

Let $z \neq 0$ be a complex number. Then there are exactly $n$ distinct complex numbers $x_1, \ldots, x_n$ so that $x_i^n = z$.

I know this can be done using the polar form, but am a bit unsure on the logic. Here is my attempt.

Consider $f(x) = x^n - z$ .By the fundamental theorem of algebra, there are $n$ solutions, which are $x \in \mathbb{C}$ such that $x^n - z = 0$, so $x^n = z$, so $x$ is a root of $z$. We now must show that these roots are distinct.

Given a root $t$ so $t^n = x$, write $$x = r_1 e^{i \theta_1}, \; t = r_2 e^{i \theta_2}, \; t^n = r_2^n e^{i (n\theta_2)}.$$ So $$r_2^n e^{i(n\theta_2)} = r_1 e^{i \theta_1}.$$ So $r_2^n = r_1$ and $n\theta_2 = \theta_1 + 2\pi k$. So $\theta_2 = \frac{\theta_1}{n} + \frac{2\pi}{n} k$. As $k$ ranges over $1 \leq k \leq n$, we get $n$ distinct roots.

The last step seems like a hand-wave. Is it on the right track?

  • Yes, that's the right idea. If you feel like the last step is handwavy, you should justify why your $n$ chosen values of $\theta$ yield $n$ distinct values of $e^{i\theta}$. – Karl Feb 09 '21 at 17:36
  • My only idea for how to use that was to use the identity $e^{i \theta} = r (\cos \theta + i \sin \theta)$. Is there a different way to do it? –  Feb 09 '21 at 17:43
  • That's the geometric reasoning I had in mind. $(\cos\alpha,\sin\alpha)$ and $(\cos\beta,\sin\beta)$ are points on the unit circle. The only way they can be the same point is if $\alpha$ and $\beta$ differ by a multiple of $2\pi$. – Karl Feb 09 '21 at 17:50
  • https://math.stackexchange.com/questions/192742/how-to-solve-x3-1/192743#192743 – lab bhattacharjee Feb 09 '21 at 18:13

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