Solutions of $z^5 = -1$ I have found the solutions to $z^5=-1$ but I have to use the following factorization to find the complex number produced when all solutions are multiplied. Each solution is denoted by $z_0-z_4$:
$$z^5 + 1 = (z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)$$
By my calculations, the complex number produced when $z_0-z_4$ are multiplied is simply $-1$, but I can't show it using the factorization.
Hoping someone can help me!
Thanks
 A: Expanding $f(x) = A(x-p)(x-q)(x-r)(x-s)(x-t)$ gives
$$f(x) = Ax^5 - Aa_4x^4 + Aa_3x^3-Aa_2x^2+Aa_1x^1-Aa_0$$
(mind the alternating sign), where 
$$\begin{align*}
a_4 &= p+q+r+s+t\\
a_3 &= pq + pr+ps+pt+qr+qs+qt+rs+rt+st\\
a_2 &= pqr + pqs + pqt + prs+prt+pst+qrs+qrt+qst+rst\\
a_1 &= pqrs+pqrt+pqst+prst+qrst\\
a_0 &= pqrst
\end{align*}$$
Comparing the coefficients of $f(z)$ and $z^5-1$,
$$\begin{align*}
Az^5 &= z^5 &&\implies&A &= 1\\
-Aa_0 &= -1 &&\implies& pqrst &= a_0 = 1
\end{align*}$$
A: Expanding out the factorization of $(x-p)(x-q)$ gives $x^2-(p+q)x+pq$. Notice that the last term is the product of the roots. In the general case with $n$ roots, the last term gives $(-1)^n$ times the product of the roots in just the same way. Your polynomial is $x^5+1=0$, so the product of the roots is $(-1)(1)=-1$. You'll also notice that the $n-1$ term is the sum of the roots. This too generalizes.
A: Um ... The factorization gives a polynomial.
$z^5 + 1 = (z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
We can set solve by setting $z = c$ and solving for:
$c^5 + 1 = (c-z_0)(c-z_1)(c-z_2)(c-z_3)(c-z_4)$
which could be really messy depending on what we choose for $c$.  But obviously if we choose $z = 0$ we get something simple:
$0^5 + 1 = (0-z_0)(0-z_1)(0-z_2)(0-z_3)(0-z_4)$
$1 = (-z_0)(-z_1)(-z_2)(-z_3)(-z_4) = -z_0z_1z_2z_3z_4$
$z_0z_1z_2z_3z_4 = -1$
