Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Complex Numbers and Sets.
Complex Numbers and Sets – AIME I, 1990
The sets A={z:\(z^{18}=1\)} and B={w:\(w^{48}=1\)} are both sets of complex roots with unity, the set C={zw: \(z \in A and w \in B\)} is also a set of complex roots of unity. How many distinct elements are in C?.
- is 107
- is 144
- is 840
- cannot be determined from the given information
Key Concepts
Integers
Complex Numbers
Sets
Check the Answer
Answer: is 144.
AIME I, 1990, Question 10
Complex Numbers from A to Z by Titu Andreescue
Try with Hints
18th and 48th roots of 1 found by de Moivre’s Theorem
=\(cis(\frac{2k_1\pi}{18})\) and \(cis(\frac{2k_2\pi}{48})\)
where \(k_1\), \(K_2\) are integers from 0 to 17 and 0 to 47 and \(cis \theta = cos \theta +i sin \theta\)
zw= \(cis(\frac{k_1\pi}{9}+\frac{k_2\pi}{24})=cis(\frac{8k_1\pi+3k_2\pi}{72})\)
and since the trigonometric functions are periodic every period \({2\pi}\)
or, at (72)(2)=144 distinct elements in C.
Other useful links
- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA