Problem:
The sets A={z:z18=1} and B={w:w48=1} are both sets of complex roots of unity. The set C={zw:zβA and wβB} is also a set of complex roots of unity. How many distinct elements are in C?
Solution:
First observe that if zβA and wβB, 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(1442Οβk)+isin(1442Οβk)=cis(1442Οβk)=[cis(1442Οβ)]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=3β
18β48 and use this in the following:
cis(1442Οβ)=cis(8642Οββ
6)=cis(8642Οβ(3β
18β48))=cis(482Οβ3)cis(182Οβ(β1))
By another application of DeMoivre's formula, we now have
x=[cis(482Οβ3)cis(182Οβ(β1))]k=cis(482Οβ3k)cis(182Οβ(βk))
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 144β elements.
The problems on this page are the property of the MAA's American Mathematics Competitions