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
Best Answer
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 $z^5-1$, $$\begin{align*} Az^5 &= z^5 &&\implies&A &= 1\\ -Aa_0 &= -1 &&\implies& pqrst &= a_0 = 1 \end{align*}$$