How to know that $(z-z_0^3)(z-z_0^5)(z-z_0^7) = \sum_{k=0}^3 z^k z^{3-k}_0$

Question

How to know that $$(z-z_0^3)(z-z_0^5)(z-z_0^7) = \sum_{k=0}^3 z^k z^{3-k}_0$$

with $z_0$ a root of $z^4+1$.

I can check that it is true, but is there a way to tell, by seeing the LHS expression, that it can be in the RHS form ?

Answer 1

I'm not sure if that is what you are after, but:

If $\omega$ is a primitive $2k$-th root of unity, then the solutions of $z^k+1 = 0$ are $\omega^{2n+1},\, 0 \leqslant n < k$, so

$$z^k + 1 = \prod_{n=0}^{k-1} (z-\omega^{2n+1}),$$

and hence

$$\begin{align} \prod_{n=1}^{k-1} (z-\omega^{2n+1}) &= \frac{z^k+1}{z-\omega}\\ &= \frac{z^k - \omega^k}{z-\omega}\\ &= \sum_{m=0}^{k-1} z^{k-1-m}\omega^{m}\\ &= \sum_{m=0}^{k-1} z^m\omega^{k-1-m}. \end{align}$$

So the "trick" is to recognise that $z_0^3,z_0^5,z_0^7$ are just the fourth roots of $-1$ that are different from $z_0$.