# Is there always a primitive m-th root of unity with imaginary part bigger than 1/2

Let $m$ be a positive integer.

I need the existence of a primitive $m$-th root of unity $\zeta_m$ such that its imaginary part is strictly greater than $1/2$.

We can write $\zeta_m = \exp(2\pi i a/m)$ for some $a$ coprime to $m$.

The condition above boils down to $\sin(2 \pi a /m ) > 1/2$. This just means that $$\frac{m}{12} < a < \frac{5m}{12}.$$ So I'm looking for the existence of an integer $a$ coprime to $m$ such that $$\frac{m}{12} < a < \frac{5m}{12}.$$

Is this always possible?

Probably I need that $m> 12$. For $m=12$, there is no such $a$. It's ok if it doesn't work for a finite number of $m$.

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By Bertrand's Postulate, which is a theorem, there is a prime between $k$ and $2k$ if $k>2$. –  André Nicolas Oct 13 '11 at 21:23
@André: That's not quite enough, since it might divide $m$. –  joriki Oct 13 '11 at 21:28
@joriki: I was giving a start. –  André Nicolas Oct 13 '11 at 21:31

By Bertrand's postulate, there is always a prime number between $\lceil m/12\rceil$ and $2\lceil m/12\rceil$ and one between $2\lceil m/12\rceil$ and $4\lceil m/12\rceil$. For sufficiently large $m$ we have $4\lceil m/12\rceil\lt5m/12$. For sufficiently large $m$, these primes cannot both divide $m$, so at least one of them is coprime to $m$.
Beautiful. Note that $4[m/12] < 4(m/12 + 1) < 5m/12$ iff $m>12$. –  Rayleigh Oct 13 '11 at 21:32
How big should I take $m$ though such that these primes won't both divide $m$? –  Rayleigh Oct 13 '11 at 21:32
Well their product is at least $2 \lceil m/12\rceil^2$, so as long as this exceds $m$, you know for sure that they can't both divide $m$. –  N. S. Oct 13 '11 at 21:43
Also, one more comment: There is a theorem by Chebashev which sais that there exists $N$ so that for all $n>N$ there are $k$ primes between $N$ and $2N$. This also follows from the Prime Number Theorem. Then using Joriki's idea with $k=2$, you get a shorter proof that there are only finitely many numbers for which your statement can fail, but unfortunatelly you don't get an estimate on how big this "finitely many" is. The above solution is nicer. –  N. S. Oct 13 '11 at 22:29