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In fast fourier transform, what are the square and the square root of $\omega_{128}$, the primitive 128th root of 1?

Is it possible for you to help me on this one?

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  • $\begingroup$ Write the root out in exponential notation : it should then be easy to square it or find its square roots. $\endgroup$ – Joppy Oct 13 '17 at 23:51
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For the roots of unity $$w^{128}=1$$ we can use DeMoivre's Theorem $$\omega_k =(\cos \frac {2 k\pi}{n} + i \sin \frac {2 k\pi}{n}), k \in (0, 127).$$

EDIT: $\omega_0$ is equal to 1 (as 1 always a root of unity). The primitive root of $\omega_1$ (with $k=1$) is $$\omega_1 =(\cos \frac {\pi}{64} + i \sin \frac {\pi}{64}).$$ Thus $$(\omega_1)^2 =(\cos \frac {\pi}{32} + i \sin \frac {\pi}{32})$$ and $$\sqrt {\omega_1} =(\cos \frac {\pi}{128} + i \sin \frac {\pi}{128})$$

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  • $\begingroup$ i am not sure if w128=1 is correct, man. How about primitive 128th root of 1? $\endgroup$ – CS Help Oct 13 '17 at 15:38
  • $\begingroup$ It wasn't...good catch. See my edit above. $\endgroup$ – bjcolby15 Oct 13 '17 at 19:55

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