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.
 A: 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 $$\frac{z_k^5-1}{z_k-1}=z_k^4+z_k^3+z_k^2+z_k^1+z_k ^0$$

A: Well, those are $4$ of the five fifth roots of unity; you're just missing $k=5$ or $z_5=1$; because of the fundamental theorem of algebra we can factorize $x^5-1$ as $(x-z_1)\cdots(x-z_5)$, but we don't want $z_5$ which is $1$, so we divide $x^5-1$ by $(x-1)$ which gets us exactly the polynomial we wanted. If you don't know how did I know that those were the roots of $x^5-1$ I recommend you to search about De Moivre's theorem and apply it to calculate each $z_i^5$
A: Your "high school" approach is valid: 
$(x-\cos\dfrac{2\pi}5-i\sin\dfrac{2\pi}5)(x-\cos\dfrac{4\pi}5-i\sin\dfrac{4\pi}5)$$(x-\cos\dfrac{6\pi}5-i\sin\dfrac{6\pi}5)(x-\cos\dfrac{8\pi}5-i\sin\dfrac{8\pi}5)$
$=(x-\cos\dfrac{2\pi}5-i\sin\dfrac{2\pi}5)(x-\cos\dfrac{4\pi}5-i\sin\dfrac{4\pi}5)$
$(x-\cos\dfrac{4\pi}5+i\sin\dfrac{4\pi}5)(x-\cos\dfrac{2\pi}5+i\sin\dfrac{2\pi}5)$
Combine conjugate pairs:
$$(x-\cos\dfrac{2\pi}5-i\sin\dfrac{2\pi}5)(x-\cos\dfrac{2\pi}5+i\sin\dfrac{2\pi}5)$$
$$=x^2+\cos^2\dfrac{2\pi}5+\sin^2\dfrac{2\pi}5-2x\cos\dfrac{2\pi}5=x^2-2x\cos\dfrac{2\pi}5+1$$
and
$$(x-\cos\dfrac{4\pi}5-i\sin\dfrac{4\pi}5)(x-\cos\dfrac{4\pi}5+i\sin\dfrac{4\pi}5)$$
$$=x^2+\cos^2\dfrac{4\pi}5+\sin^2\dfrac{4\pi}5-2x\cos\dfrac{4\pi}5=x^2-2x\cos\dfrac{4\pi}5+1.$$
Now multiply to get $$ (x^2-2x\cos\dfrac{2\pi}5+1)(x^2-2x\cos\dfrac{4\pi}5+1)$$
$$=(x^2+1)^2+4x^2\cos\dfrac{2\pi}5\cos\dfrac{4\pi}5-2x(x^2+1)(\cos\dfrac{2\pi}5+\cos\dfrac{4\pi}5)$$
$$=x^4+2x^2+1-x^2+x(x^2+1)=x^4+x^3+x^2+x+1$$
A: 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.
