(z+sqrt(2))^4=-16 equation

Given the equation
$$\left(z + \sqrt{2}\right)^{4} = -16$$
Because equation degree is equal to = 4 and the free term = -16 < 0,
so the real solutions of the equation d'not exist

All other 4 root(s) is the complex numbers.
do replacement:
$$w = z + \sqrt{2}$$
then the equation will be the:
$$w^{4} = -16$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -16$$
where
$$r = 2$$
- 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 w
Consequently, the solution will be for w:
$$w_{1} = - \sqrt{2} - \sqrt{2} i$$
$$w_{2} = - \sqrt{2} + \sqrt{2} i$$
$$w_{3} = \sqrt{2} - \sqrt{2} i$$
$$w_{4} = \sqrt{2} + \sqrt{2} i$$
do backward replacement
$$w = z + \sqrt{2}$$
$$z = w - \sqrt{2}$$

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