Product of complex solutions via factorisation I'm wondering if someone could help me out. 
I am asked to solve the equation: $z^6 =−1$ in part (a) of a question.
I have done this and so I now have a set of solutions: $z_0,z_1,z_2,z_3,z_4.$   
I'm lost in part (b):
Let $z_0, z_1, z_2, z_3, z_4, z_5$  be the solutions that you found in part (a). Use the factorization 
$z_6 +1 = (z−z_0)(z−z_1)(z−z_2)(z−z_3)(z−z_4)$ 
to determine the complex number that is obtained by multiplying together all the solutions of the equation $z^6 =−1$.
What I don't understand is why do I have to use the factorisation...? Can't I just multiple the separate solutions???
 A: You can multiply all of the $6$ complex solutions you got, but that might be tedious. 
Instead, since $z^6+1 = (z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4)(z-z_5)$, we can plug in $z = 0$ to get $0^6+1 = (0-z_0)(0-z_1)(0-z_2)(0-z_3)(0-z_4)(0-z_5)$ $= (-1)^6z_0z_1z_2z_3z_4z_5$. 
This gives you the answer much faster than multiplying out $6$ complex numbers.
A: $$z^6=-1\Longleftrightarrow$$
$$z^6=e^{\left(\pi+2\pi k\right)i}\Longleftrightarrow$$
$$z=e^{\frac{1}{6}\left(\pi+2\pi k\right)i}\Longleftrightarrow$$
$$z=e^{\left(\frac{\pi}{6}+\frac{\pi k}{3}\right)i}$$
with $k \in \mathbb{Z}$ and $k$ goes from $0-5$

So the $6$ solutions are:
$$z_0=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 0}{3}\right)i}=e^{\frac{\pi}{6}i}$$
$$z_1=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 1}{3}\right)i}=i$$
$$z_2=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 2}{3}\right)i}=e^{\frac{5\pi}{6}i}$$
$$z_3=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 3}{3}\right)i}=e^{-\frac{5\pi}{6}i}$$
$$z_4=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 4}{3}\right)i}=-i$$
$$z_5=e^{\left(\frac{\pi}{6}+\frac{\pi \cdot 5}{3}\right)i}=e^{-\frac{\pi}{6}i}$$
