Determine the fourth roots of -16 Determine the fourth roots of -16 in the form $x +iy$ where $x$ and $y$ are not trigonometric functions.

I do not even know what they really want from me in this question.

My initial thought wass: $\sqrt[4]{-16} = 0 + 2i = 2i$

but that seems overly simplistic for a problem that counts as 8 points in an exam.
Your equation is $z^4=-16$. Notice that $-16=(4i)^2$, so
$$
z^4+16=(z^2-4i)(z^2+4i)
$$
and the equation reduces to solve $z^2-4i=0$ and $z^2+4i=0$, that is,  finding the square roots of $4i$ and $-4i$, which should be easy: write
$$
(x+iy)^2=4i
$$
and compare the real and imaginary parts. Do the same for $-4i$.
So let's use the formula: z=(fourth root of -16) times (cos x + i(sin (x)))
let k = 0, solution 1= 2(radical i) (cos (0 + 2(180)(0)/(4) + i(sin (0+2(180)(0)/(4))=
Solution 1: 2 (radical i)
let k = 1, solution 2= 2 (radical i) (cos (0 +2(180)(1)/(4) + i(sin(0+2(180)(1)/(4))
Solution 2: 2(i)(radical i)
let k = 2, solution 3= 2 (radical i) (cos (0+2(180)(2)/(4) + i(sin (0+2(180)(2)/(4))
Solution 3: -2(radical i)
let k = 3, solution 4= 2 (radical i) (cos (0+2(180)(3)/(4) + i(sin (0+2(180)(3)/(4))
Solution 4= -2(i)(radical i)
So the solutions are imaginary pairs (conjugates) and they are:
 2 (radical i), 2i(radical i), -2(radical i), and -2i(radical i). If you try them, they will all work. I solved it using De Moivre's Theorem.