# Finite product of complex functions

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.

• Look at the roots of the polynomial $z^{2n}-1$
– shdp
Mar 14, 2017 at 19:44
• 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. Mar 14, 2017 at 21:11

Hint: Consider the roots of $f(x)=x^{2n}-1$. What is $f(\alpha)$?
• 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. Mar 14, 2017 at 22:10