Using complex solutions in a factorisation

Question

I'm working through an assignment, and have become stuck understanding the question...

In part (a) I am asked to solve the equation: $z^5 = -1$

I have done this, so I now have a set of solutions: $z_0, z_1, z_2, z_3, z_4$ (one $\in \mathbb{R}$ the rest $\in \mathbb{C}$)

My confusion arises in part (b):

Let $z_0, z_1, z_2, z_3, z_4$ be the solutions that you found in part (a). Use the factorisation $$z^5+1=(z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)$$ to determine the complex number that is obtained by multiplying together all the solutions of the equation $z^5=-1$.

I can happily multiply the solutions $z_0, z_1, z_2, z_3, z_4$, but I just have no idea what the relevance of the factorisation is. I'm missing something but I have no idea what, I'm guessing I have forgotten something fundamental, or my cognitive leaps are mere cognitive hops.

Sorry of any gaps/vaguery in the above, I don't want to be in a position of being accused of publishing any answers to the problem online. And please - I don't want the answer, I just want a nudge in the direction the question is pointing.

Answer 1

By expanding the expression and regrouping the terms on $z^5,z^4,\ldots$ we find:

\begin{align}&(z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)\\&=z^5-(z_0+\cdots+z_4)z^4+\cdots+(-z_0)\times(-z_1)\times\cdots\times(-z_4)\\&=z^5+1\end{align} so by identification we get

$$(-z_0)\times(-z_1)\times\cdots\times(-z_4)=1$$ Can you take it from here?

Answer 2

HINT:

As $\displaystyle(-1)^5=-1,(z+1)|(z^5+1)$

By actual division or using Geometric Series formula, $\displaystyle\frac{z^5+1}{z+1}=z^4-z^3+z^2-z+1 $

Now use the idea of Equation with high exponents or Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$