$$P(x) = (x-1)(x-\omega_{15})(x-\omega_{15}^2)...(x-\omega_{15}^{14})$$

Where $\omega$ is a root of unity of order 15 and $\omega_{15}= e^{2\pi i/15}$

I need to simplify this.

The solution is using fast fourier transform but I am not able to simplify it. The only lead I have is that 2 polynomials of same degree which have the same coefficient in front of the largest power and have same zeros are equal.

Any ideas on how to approach this problem?

  • $\begingroup$ I took the liberty to rename "W" to $\omega$ which is quite a standard way to denote roots of unity. $\endgroup$ – Zubzub Apr 8 '17 at 15:40
  • $\begingroup$ Thanks! Not very familiar with editing as of now. $\endgroup$ – John Keaton Apr 8 '17 at 15:41
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    $\begingroup$ It is probably better to say that $\omega$ is a fifteenth root of unit, i.e. $\omega^{15} = 1$ where $\omega = e^{2\pi i /15}$. Then the equation $P(x) = 0$ is satisfied by the integer powers of $\omega$, and there are fifteen distinct roots. $\endgroup$ – hardmath Apr 8 '17 at 15:47
  • $\begingroup$ @hardmath, the key point is that $\omega$ is a primitive root of unit. $\endgroup$ – lhf Apr 8 '17 at 16:16
  • 1
    $\begingroup$ @dxiv Thanks for the help. Solved it. $\endgroup$ – John Keaton Apr 9 '17 at 1:23

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