This entry was written by Lawrence Sun.

**Problem:** Let be positive integers. Prove that the sum of roots of unity, whose sum is not , has modulus at least .

Inspired by a result of Gerald Myerson.

**Solution:**

Consider the field where denotes a primitive root of unity. Now, for each denote the automorphism which maps to . If we let be the set of residues modulo relatively prime to we have

which is stated in this article : https://www.dropbox.com/s/nlfn8ldodz4uvgb/Cyclotomic%20Polynomials.pdf. Now, let

be our sum of roots of unity. Clearly as for each .

Now, I claim:

The proof of this is simple by the Fundamental Theorem of Galois Theory or the Fundamental Theorem of Symmetric Polynomials and is thus left to the reader. As a sketch of the elementary second proof, simply let:

And then note that the product is a symmetric function in for each , and is thus an integer polynomial in terms of the symmetric sums of those numbers. But the symmetric sums are the coefficients of , so the result follows.

Now we prove the product is nonzero. Suppose for some . Take the inverse in the automorphism group of , call it . But then , absurd since an automorphism fixes the rationals. Thus the product has modulus as it is a nonzero integer.

We now show that each , . The proof of this is simple; simply remark that:

Therefore it follows:

as desired.

**Extensions:**

You can find much more sophisticated methods to find much stronger bounds on this mathoverflow thread : http://mathoverflow.net/questions/46068/how-small-can-a-sum-of-a-few-roots-of-unity-be

Excellent. @lawrence, nice article!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!