Are there only finitely many complex numbers $z$ such that $z^{634}=1$?
I think there are only finitely many; is that not so?
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3There are exactly $634$ unique complex solutions, two of which ($\pm 1$) are purely real. – Deepak Nov 26 '20 at 02:26
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2If $(re^{i\theta})^{634}=1$, then $r=1$ and $\theta=\dfrac{2\pi n}{634}$ – J. W. Tanner Nov 26 '20 at 02:32
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Check out de Moivre's theorem: https://en.wikipedia.org/wiki/De_Moivre%27s_formula – KingLogic Nov 26 '20 at 02:48
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Yes, there are only finitely many $634^{th}$ complex roots of unity.
This is easily seen with polar notation. Let $z=re^{i\theta}$ and $z^{634}=1$.
Then $r=1$ and $\theta=\dfrac{2\pi n}{634}$.
There are only finitely many values of $n$ (viz., $634$) such that $0\le\dfrac{2\pi n}{634}<2\pi$.
J. W. Tanner
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There are, however, countably infinite values of $\theta$ such that $e^{i\theta}=1$ without the restriction $0\leq\theta\leq2\pi$ – Graviton Nov 26 '20 at 02:49
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1@Graviton: you are correct, but they don't correspond to infinitely many complex numbers – J. W. Tanner Nov 26 '20 at 02:51