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.

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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