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Let $a \in \mathbb{C}$. It might as well be on the unit circle, so $a=e^{i \theta}$. I'm interesting in finding, for $n \ge 2$, how many $n$-th roots, $\omega_i$ (with $ 0 \le i \le n-1$), have $\Im (\omega _i) \ge0$. Intuitively, the answer should be roughly a half, since they are placed evenly on the unit circle in increments of $2\pi/n$ starting from $a$, but I'm having trouble showing this.

Spine Feast
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  • For third roots of unity, $2/3$ have non-negative imaginary part. – Dietrich Burde Sep 09 '13 at 09:53
  • But for example with third roots of, say $e^{-i \pi/100}$, only one of the three has non-neg. imaginary part. So (at least for odd $n$) the answer depends somehow on $\theta$ (conjecture: it's either $(n-1)/2$ or $(n+1)/2$), and I'd like to find this dependence. – Spine Feast Sep 09 '13 at 09:58

1 Answers1

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The roots of such an $a$ are the $n$-th roots of unity after a rotation (the rotation depends on $a$). The roots of unity are easy to draw on the plane, they form the vertices of an $n$-gon with $1$ as one of the vertices. $n$ is even iff $-1$ is also an $n$-th root of unity. Thus, if $n$ is even then there are $n-2$ remaining roots, half of them lie in the upper half plane, so you get a total of $\frac{n-2}{2}$ roots with positive imaginary part. If $n$ is odd then there are $n-1$ remaining roots, half of which lie in the upper half plane, so you get a total of $\frac{n-1}{2}$ roots with positive imaginary part.

This sums up the situation for roots of unity. Now just figure out what may happen if you rotate these roots. It's still an $n$-gon of course so the analysis is very similar. In any case, it is most certainly the case that about half of the roots have positive imaginary part. (You can say much more than about half. You can say precisely how many.)

Ittay Weiss
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  • By 'positive' you mean here 'nonnegative', right? – Spine Feast Sep 09 '13 at 10:03
  • oh, I answered the question for positive rather than non-negative. Sorry about that. The adaptation to my argument is easy to make. – Ittay Weiss Sep 09 '13 at 10:04
  • It only makes a difference in the case of even $n$, making the amount $(n+2)/2$, right? – Spine Feast Sep 09 '13 at 10:07
  • It would seem that in the case of even $n$, we always 'lose' one when rotating by any angle which isn't a multiple of $2 \pi /n$. Is this right? So the number of roots with non-neg. im. part is either $1+ n/2$ (when $\theta = 2k \pi /n$), or $n/2$ otherwise. – Spine Feast Sep 09 '13 at 10:16
  • You are on your way to answering your question. Check your answer for small values of $n$ by drawing some equilateral triangles, squares, and a regular pentagon and you'll see if you are correct or not. – Ittay Weiss Sep 09 '13 at 10:18
  • From what I've drawn it looks about right, but maybe I'm missing something. In the case of odd $n$, the number of roots it at most the number of roots of unity, which is $(n+1)/2$. But one can be lost if we rotate by an angle between $\phi /2 $ and $\phi$, where $\phi = 2\pi /n$. So the number of roots is $(n-1)/2$ if $ \phi /2 \le \theta \le \phi$,modulo $\phi$ and $(n+1)/2$ otherwise. – Spine Feast Sep 09 '13 at 10:23