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Detail solution

Given the equation

$$z^{4} = 1$$

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:

$$\sqrt[4]{\left(1 z + 0\right)^{4}} = 1$$

$$\sqrt[4]{\left(1 z + 0\right)^{4}} = -1$$

or

$$z = 1$$

$$z = -1$$

We get the answer: z = 1

We get the answer: z = -1

or

$$z_{1} = -1$$

$$z_{2} = 1$$

All other 2 root(s) is the complex numbers.

do replacement:

$$w = z$$

then the equation will be the:

$$w^{4} = 1$$

Any complex number can presented so:

$$w = r e^{i p}$$

substitute to the equation

$$r^{4} e^{4 i p} = 1$$

where

$$r = 1$$

- 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 w

Consequently, the solution will be for w:

$$w_{1} = -1$$

$$w_{2} = 1$$

$$w_{3} = - i$$

$$w_{4} = i$$

do backward replacement

$$w = z$$

$$z = w$$

The final answer:

$$z_{1} = -1$$

$$z_{2} = 1$$

$$z_{3} = - i$$

$$z_{4} = i$$

$$z^{4} = 1$$

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:

$$\sqrt[4]{\left(1 z + 0\right)^{4}} = 1$$

$$\sqrt[4]{\left(1 z + 0\right)^{4}} = -1$$

or

$$z = 1$$

$$z = -1$$

We get the answer: z = 1

We get the answer: z = -1

or

$$z_{1} = -1$$

$$z_{2} = 1$$

All other 2 root(s) is the complex numbers.

do replacement:

$$w = z$$

then the equation will be the:

$$w^{4} = 1$$

Any complex number can presented so:

$$w = r e^{i p}$$

substitute to the equation

$$r^{4} e^{4 i p} = 1$$

where

$$r = 1$$

- 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 w

Consequently, the solution will be for w:

$$w_{1} = -1$$

$$w_{2} = 1$$

$$w_{3} = - i$$

$$w_{4} = i$$

do backward replacement

$$w = z$$

$$z = w$$

The final answer:

$$z_{1} = -1$$

$$z_{2} = 1$$

$$z_{3} = - i$$

$$z_{4} = i$$

Sum and product of roots
[src]

sum

-1 + 1 + -I + I

$$\left(-1\right) + \left(1\right) + \left(- i\right) + \left(i\right)$$

=

0

$$0$$

product

-1 * 1 * -I * I

$$\left(-1\right) * \left(1\right) * \left(- i\right) * \left(i\right)$$

=

-1

$$-1$$

Rapid solution
[src]

z_1 = -1

$$z_{1} = -1$$

z_2 = 1

$$z_{2} = 1$$

z_3 = -I

$$z_{3} = - i$$

z_4 = I

$$z_{4} = i$$

The graph