x^4+4=0 equation

Given the equation
$$x^{4} + 4 = 0$$
Because equation degree is equal to = 4 and the free term = -4 < 0,
so the real solutions of the equation d'not exist

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

The final answer:
$$x_{1} = -1 - i$$
$$x_{2} = -1 + i$$
$$x_{3} = 1 - i$$
$$x_{4} = 1 + i$$