x^4-81=0 equation

Given the equation
$$x^{4} - 81 = 0$$
Because equation degree is equal to = 4 - contains the even number 4 in the numerator, then
the equation has two real roots.
Get the root 4-th degree of the equation sides:
We get:
$$\sqrt[4]{x^{4}} = \sqrt[4]{81}$$
$$\sqrt[4]{x^{4}} = \left(-1\right) \sqrt[4]{81}$$
or
$$x = 3$$
$$x = -3$$
We get the answer: x = 3
We get the answer: x = -3
or
$$x_{1} = -3$$
$$x_{2} = 3$$

All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{4} = 81$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = 81$$
where
$$r = 3$$
- 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
$$p = \frac{\pi N}{2}$$
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:
$$z_{1} = -3$$
$$z_{2} = 3$$
$$z_{3} = - 3 i$$
$$z_{4} = 3 i$$
do backward replacement
$$z = x$$
$$x = z$$

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