Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Complex Numbers and prime.
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.
Complex Numbers
Algebra
Number Theory
Answer: is 71.
AIME I, 2012, Question 6
Complex Numbers from A to Z by Titu Andreescue
Taking both given equations \((z^{13})^{11} = z\) gives \(z^{143} = z\) Then \(z^{142} = 1\)
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 \(k \in {1, 2, upto 70}\)
71 is prime and n = 71.
