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


But try the problem first…

Answer: is 144.

Source
Suggested Reading

AIME I, 1990, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

18th and 48th roots of 1 found by de Moivre’s Theorem

=\(cis(\frac{2k_1\pi}{18})\) and \(cis(\frac{2k_2\pi}{48})\)

Second Hint

where \(k_1\), \(K_2\) are integers from 0 to 17 and 0 to 47 and \(cis \theta = cos \theta +i sin \theta\)

zw= \(cis(\frac{k_1\pi}{9}+\frac{k_2\pi}{24})=cis(\frac{8k_1\pi+3k_2\pi}{72})\)

Final Step

and since the trigonometric functions are periodic every period \({2\pi}\)

or, at (72)(2)=144 distinct elements in C.

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