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I want to solve $z^n=1$ in the complex field.

We can express $z$ in standard form or in polar form. Which one is better? I guess that it is polar form as there are nice theorems about multiplication of complex numbers in polar form, so I will adopt this form.

Let's begin:

Using DeMoivre theorem:

$z^n = r^n[cos(nA)+isin(nA)]$

So, we want to solve $z^n = r^n[cos(nA)+isin(nA)]$ = 1

I am stuck here.

All I know, by some reasoning is that $isin(nA)$ must be $0$, and so, $nA$ must be either $0º$ or $180º$ (up to congruences), and, therefore, $cos(nA)$ must be either $1$ or $-1$. But what, about $r$? It could be either $1$ or $-1$?

I am having a big headache right now, so maybe there are typos here I did not perceive even after reading it all two times... Could somebody clarify this mess to me?

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  • $\begingroup$ In the real field: 1 is always a solution and -1 is sometimes (when n is even)? Is that it? What about the other solutions (since we are in the complex field)? $\endgroup$ – Viktor Kaspervich Feb 5 '18 at 10:25
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    $\begingroup$ actually, $r = |z| \ge 0$, so $r=1$ and $\cos(nA) =1$. $\endgroup$ – Quang Hoang Feb 5 '18 at 10:25
  • $\begingroup$ @Quang Hoang: Aaaahhha, that makes sense! Thanks! $\endgroup$ – Viktor Kaspervich Feb 5 '18 at 10:25
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Polar form is better here.

When do we have $r^n\bigl(\cos(n\theta)+i\sin(n\theta)\bigr)=1$. That's when $r=1$, $\cos(n\theta)=1$ and $\sin(n\theta)=0$. That happens when$$\theta\in\left\{0,\frac\pi{2n},\frac{2\pi}{2n},\ldots,\frac{(2n-1)\pi}{2n}\right\}$$Of course, you might keep going, but then you would not be getting new numbers.

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