Problem:
The sets A={z:z18=1} and B={w:w48=1} are both sets of complex roots of unity. The set C={zw:z∈A and w∈B} is also a set of complex roots of unity. How many distinct elements are in C?
Solution:
First observe that if z∈A and w∈B, then
(zw)144=(z18)8(w48)3=1
This shows that the set C is contained in the set of 144th roots of unity. Next we show that any 144th root of unity is in C, thereby showing that C has 144 elements. Let x be a 144th root of unity. Then there is an integer k with
x=cos(1442πk)+isin(1442πk)=cis(1442πk)=[cis(1442π)]k
where the last equality follows by an application of DeMoivre's formula. We next express the greatest common divisor of 18 and 48 as 6=3⋅18−48 and use this in the following:
cis(1442π)=cis(8642π⋅6)=cis(8642π(3⋅18−48))=cis(482π3)cis(182π(−1))
By another application of DeMoivre's formula, we now have
x=[cis(482π3)cis(182π(−1))]k=cis(482π3k)cis(182π(−k))
which shows that x is a product of elements from A and B. Hence the set of 144th roots of unity is a subset of C. We may conclude that C is the set of 144th roots of unity, so C has 144 elements.
The problems on this page are the property of the MAA's American Mathematics Competitions