Let ω=e^iπ/4 and P=|(1-ω)(1-ω^2(1-ω^3|^2. Find the value of P.

Since $\omega$ is a primitive 8th root of unity, $x^7+x^6+\dots+1=\prod_{k=1}^{7}(x-\omega^k)$, and putting $x=1$ gives $\prod_{k=1}^{7}(1-\omega^k)=8$. Pair the factors $k$ and $8-k$: because $\omega^{8-k}=\overline{\omega^{k}}$, we get $(1-\omega^k)(1-\omega^{8-k})=|1-\omega^k|^2$. Hence $8=|1-\omega|^2|1-\omega^2|^2|1-\omega^3|^2\,(1-\omega^4)$. Now $\omega^4=-1$, so $1-\omega^4=2$ and the product of the three squared moduli is $8/2=4$. But $P=|1-\omega|^2|1-\omega^2|^2|1-\omega^3|^2$ exactly, so $P=4$. A numerical check of the three chords of the unit circle agrees. The key idea is that the full product over all non-trivial roots is a known constant, and the conjugate pairing converts it into squared moduli, so no trigonometric evaluation of individual chord lengths is needed. Students who forget the unpaired factor at $k=4$ would wrongly obtain $8$ instead of $4$.