<s>[INST] Polynomial equation for $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}$ I have $z_{k}=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5}, k=1,2,3,4$.
I need to find the polynomial equation for the roots $z_k(k=1,2,3,4)$
The right answer is $x^4+x^3+x^2+x+1=0$.I tried to replace k with 1,2,3,4 to find the roots and then to use $a(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but I didn't get too far.
 [/INST] Hint:
$$z_k=e^{\frac{2\pi i}{5}}$$

 Raising both sides to the 5th power yields $$z_k^5=e^{2\pi i}=1$$ or, $$z_k^5-1=0\tag{1}$$ The polynomial $(1)$ has 5 roots, one of them being $z_k=1$. But $z_k\neq 1$, because $k$ is not a multiple of $5$. Therefore $$\frac{z_k^5-1}{z_k-1}\tag{2}$$ is the polynomial we're after, which has 4 roots. $(2)$ can be simplified, by recognizing it as a geometric series $$\frac{z_k^5-1}{z_k-1}=z_k^4+z_k^3+z_k^2+z_k^1+z_k ^0$$

</s>