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???
Best Answer
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.