Reading a Russian book "Lectures on Mathematical Analysis" by Arkhipov, Sadovnichy and Chubarikov, in the section "Integral Sums Method" I encountered an equality, which I cannot prove. Here it is: $$ \prod_{k=1}^{n}|(\alpha-e^{ix_k})\cdot(\alpha-e^{-ix_k})|=|\alpha^{2n}-1|, x_k={\pi k\over n} $$ Could somebody, please, explain, how to prove this. I think it has something to do with root of unity.

  • $\begingroup$ Look at the roots of the polynomial $z^{2n}-1$ $\endgroup$
    – shdp
    Mar 14, 2017 at 19:44
  • $\begingroup$ I see now. It's simple, but complex numbers is my weak part so far, I studied them long time ago. I hope to read again a book on complex analysis by Privalov in years to come. $\endgroup$ Mar 14, 2017 at 21:11

1 Answer 1


Hint: Consider the roots of $f(x)=x^{2n}-1$. What is $f(\alpha)$?

  • $\begingroup$ You're welcome! Glad to help. $\endgroup$ Mar 14, 2017 at 21:12
  • $\begingroup$ I've noticed, however, that left hand side has a double root -1, while right hand side has roots 1 and -1 instead, I think that there's mistake in the book, and right hand side should contain multiplier $$ {|\alpha+1|\over |\alpha-1|} $$ Although the final answer to the problem in the book remains correct. $\endgroup$ Mar 14, 2017 at 22:10
  • $\begingroup$ Yes, you are correct indeed. $\endgroup$ Mar 14, 2017 at 22:22

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