Mister Exam
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How to use it?
- Equation:
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Equation z^8-1=0
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Identical expressions
- z^ eight - one = zero
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Similar expressions
- z^8+1=0
z^8-1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{8} - 1 = 0$$
Because equation degree is equal to = 8 - contains the even number 8 in the numerator, then
the equation has two real roots.
Get the root 8-th degree of the equation sides:
We get:
$$\sqrt[8]{\left(1 z + 0\right)^{8}} = 1$$
$$\sqrt[8]{\left(1 z + 0\right)^{8}} = -1$$
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 6 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{8} = 1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{8} e^{8 i p} = 1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{8 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(8 p \right)} + \cos{\left(8 p \right)} = 1$$
so
$$\cos{\left(8 p \right)} = 1$$
and
$$\sin{\left(8 p \right)} = 0$$
then
$$p = \frac{\pi N}{4}$$
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:
$$w_{1} = -1$$
$$w_{2} = 1$$
$$w_{3} = - i$$
$$w_{4} = i$$
$$w_{5} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{6} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$w_{7} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{8} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = -1$$
$$z_{2} = 1$$
$$z_{3} = - i$$
$$z_{4} = i$$
$$z_{5} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{6} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$z_{7} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{8} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$z^{8} - 1 = 0$$
Because equation degree is equal to = 8 - contains the even number 8 in the numerator, then
the equation has two real roots.
Get the root 8-th degree of the equation sides:
We get:
$$\sqrt[8]{\left(1 z + 0\right)^{8}} = 1$$
$$\sqrt[8]{\left(1 z + 0\right)^{8}} = -1$$
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 6 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{8} = 1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{8} e^{8 i p} = 1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{8 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(8 p \right)} + \cos{\left(8 p \right)} = 1$$
so
$$\cos{\left(8 p \right)} = 1$$
and
$$\sin{\left(8 p \right)} = 0$$
then
$$p = \frac{\pi N}{4}$$
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:
$$w_{1} = -1$$
$$w_{2} = 1$$
$$w_{3} = - i$$
$$w_{4} = i$$
$$w_{5} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{6} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$w_{7} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{8} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = -1$$
$$z_{2} = 1$$
$$z_{3} = - i$$
$$z_{4} = i$$
$$z_{5} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{6} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$z_{7} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{8} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
Rapid solution
[src]
z_1 = -1
$$z_{1} = -1$$
z_2 = 1
$$z_{2} = 1$$
z_3 = -I
$$z_{3} = - i$$
z_4 = I
$$z_{4} = i$$
___ ___
\/ 2 I*\/ 2
z_5 = - ----- - -------
2 2
$$z_{5} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
___ ___
\/ 2 I*\/ 2
z_6 = - ----- + -------
2 2
$$z_{6} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
___ ___
\/ 2 I*\/ 2
z_7 = ----- - -------
2 2
$$z_{7} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
___ ___
\/ 2 I*\/ 2
z_8 = ----- + -------
2 2
$$z_{8} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ ___ ___ ___ ___
\/ 2 I*\/ 2 \/ 2 I*\/ 2 \/ 2 I*\/ 2 \/ 2 I*\/ 2
-1 + 1 + -I + I + - ----- - ------- + - ----- + ------- + ----- - ------- + ----- + -------
2 2 2 2 2 2 2 2
$$\left(-1\right) + \left(1\right) + \left(- i\right) + \left(i\right) + \left(- \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) + \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right) + \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) + \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right)$$
=
0
$$0$$
product
___ ___ ___ ___ ___ ___ ___ ___
\/ 2 I*\/ 2 \/ 2 I*\/ 2 \/ 2 I*\/ 2 \/ 2 I*\/ 2
-1 * 1 * -I * I * - ----- - ------- * - ----- + ------- * ----- - ------- * ----- + -------
2 2 2 2 2 2 2 2
$$\left(-1\right) * \left(1\right) * \left(- i\right) * \left(i\right) * \left(- \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) * \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right) * \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) * \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right)$$
=
-1
$$-1$$
Numerical answer
[src]
z1 = 1.0
z2 = -0.707106781186548 - 0.707106781186548*i
z3 = -0.707106781186548 + 0.707106781186548*i
z4 = 0.707106781186548 - 0.707106781186548*i
z5 = 0.707106781186548 + 0.707106781186548*i
z6 = 1.0*i
z7 = -1.0
z8 = -1.0*i
z8 = -1.0*i
The graph