z^4=-8-8*(sqrt(3))*i equation

Given the equation
$$z^{4} = -8 - 8 \sqrt{3} i$$
Because equation degree is equal to = 4 and the free term = -8 - 8*i*sqrt(3) complex,
so the real solutions of the equation d'not exist

All other 4 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{4} = -8 - 8 \sqrt{3} i$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -8 - 8 \sqrt{3} i$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{4 i p} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
so
$$\cos{\left(4 p \right)} = - \frac{1}{2}$$
and
$$\sin{\left(4 p \right)} = - \frac{\sqrt{3}}{2}$$
then
$$p = \frac{\pi N}{2} + \frac{\pi}{12}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = -1 - \sqrt{3} i$$
$$w_{2} = 1 + \sqrt{3} i$$
$$w_{3} = - \sqrt{3} + i$$
$$w_{4} = \sqrt{3} - i$$
do backward replacement
$$w = z$$
$$z = w$$

The final answer:
$$z_{1} = -1 - \sqrt{3} i$$
$$z_{2} = 1 + \sqrt{3} i$$
$$z_{3} = - \sqrt{3} + i$$
$$z_{4} = \sqrt{3} - i$$