find the value $\prod_{k=0}^{9}(2+z^k)$ Let $$z=\cos\left( \frac{2\pi}{10}\right)+i\sin\left(\frac{2\pi}{10}\right)$$
Find the value
$$\prod_{k=0}^{9}(2+z^k)$$

Clearly, $z^k,0\le k\le9$ are the roots of $$z^{10}-1=0$$
Let $y=2+z^k\iff z^k=y-2\implies$  $$0=(y-2)^{10}-1=y^{10}-\binom{10}1y^9+\cdots+2^{10}-1$$
$$\implies\prod_{k=0}^9(2+z^k)=\dfrac{2^{10}-1}1$$
