Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Complex Numbers and prime.

## Complex Numbers and primes – AIME 2012

The complex numbers z and w satisfy \(z^{13} = w\) \(w^{11} = z\) and the imaginary part of z is \(\sin{\frac{m\pi}{n}}\), for relatively prime positive integers m and n with m<n. Find n.

- is 107
- is 71
- is 840
- cannot be determined from the given information

**Key Concepts**

Complex Numbers

Algebra

Number Theory

## Check the Answer

But try the problem first…

Answer: is 71.

AIME I, 2012, Question 6

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

First hint

Taking both given equations \((z^{13})^{11} = z\) gives \(z^{143} = z\) Then \(z^{142} = 1\)

Second Hint

Then by De Moivre’s theorem, imaginary part of z will be of the form \(\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}}\) where \(k \in {1, 2, upto 70}\)

Final Step

71 is prime and n = 71.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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