Complex numbers equation: $z^4 = -16$ Probably dumb question, but I would like to ask it anyway.
I was to solve this equation:
$z^4 = -16$
At first glance, the way to solve it would be (as any other equation):
$z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$
$z^2 = 4i \lor z^2 = -4i$
$z = 2\sqrt{i} \lor z = -2\sqrt{i} \lor z = 2i\sqrt{i} \lor z = -2i\sqrt{i}$
However, in suggested answers, there are $z_1, z_2, z_3$ and $z_4$ found using trigonometric formulas. Is there something particulary wrong in my solution and the only proper one uses trigonometric form of complex number?

We have
$$z^4 + 16 = (z^2 + 4)^2 - 8z^2 = (z^2 - 2\sqrt{2} z + 4)(z^2 + 2\sqrt{2} z + 4).$$
Anotherway $$z^2 - 2\sqrt{2} z + 4 = (z-\sqrt{2})^2 + 2 = (z-\sqrt{2})^2 -(\sqrt{2} i)^2 =  (z-\sqrt{2} + \sqrt{2} i)(z-\sqrt{2} + \sqrt{2} i)).$$
The same for $z^2 + 2\sqrt{2} z + 4$
