The sum of
If you multiply all the primitive roots of unity for the first $n$ primes is by $-n$, -i$, then add them, you get $in$, whose degree is $1$ 2$ and whose absolute value is $n$, rational and unbounded.
|
3 | added 34 characters in body | ||
|
|
||||
|
|
2 | deleted 37 characters in body | ||
|
No, because the The sum over of all quadratic residues $a$ mod $p$ the primitive roots of unity for the first $e^{2\pi i a/p}$ n$ primes is $(-1+\sqrt{p})/2$ or -n$, whose degree is $(-1+i\sqrt{p})/2$, by the theory of Gauss sums, which 1$ and whose absolute value is unbounded but has degree $2$.n$, rational and unbounded. |
||||
|
|
1 |
|
||
|
No, because the sum over all quadratic residues $a$ mod $p$ of $e^{2\pi i a/p}$ is $(-1+\sqrt{p})/2$ or $(-1+i\sqrt{p})/2$, by the theory of Gauss sums, which is unbounded but has degree $2$. |
||||
