<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] Well, with
$z_k = e^{2ki\pi / 5}, \; 1 \le k \le 4, \tag 1$
we have
$z_k^5 = (e^{2ki\pi / 5})^5 = e^{2ki\pi} = 1, \; 1 \le k \le 4; \tag 2$
that is, each $z_k$ satisfies
$x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1); \tag 3$
but with
$z_k \ne 1, \; 1 \le k \le 4, \tag 4$
(3) yields
$0 = z_k^5 - 1 = (z_k - 1)(z_k^4 + z_k^3 + z_k^2 + z_k + 1) \Longrightarrow z_k^4 + z_k^3 + z_k^2 + z_k + 1 = 0, \; 1 \le k \le 4; \tag 5$
that is, each $z_k$ satisfies
$x^4 + x^3 + x^2 + x + 1 = 0. \tag 6$
Nota Bene:  We observe that this little result generalizes to the case 
$z_k = e^{2k i\pi / p}, \; p \in \Bbb P, \; 1 \le k \le p - 1, \tag 7$
for then we have
$z_k^p = (e^{2ki \pi / p})^p = e^{2ki \pi} = 1, \; 1 \le k \le p - 1; \tag 8$
that is, the $z_k$ satisfy the polynomial
$x^p - 1 = (x - 1)(x^{p - 1} + x^{p - 2} + \ldots + x + 1) = (x - 1) \displaystyle \sum_0^{p - 1} x^j; \tag 9$
thus,
$(z_k - 1)\displaystyle \sum_0^{p - 1} z_k^j = 0, \; 1 \le k \le p - 1; \tag{10}$
from (7) we see that
$z_k - 1 \ne 0, \; 1 \le k \le p - 1, \tag{11}$
and hence (10) yields
$\displaystyle \sum_0^{p - 1} z_k^j = 0, \; 1 \le k \le p - 1; \tag{12}$
that is, the $z_k$ all satisfy the polynomial
$x^{p - 1} + x^{p - 2} + \ldots + x + 1 = \displaystyle \sum_0^{p - 1} x^j. \tag{13}$
End of Note.
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