Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x-3)(x^2-2x+1)=3(x-1)
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Equation z^6+1=0
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Equation (x-4)/(x-6)=2
- Equation 3*80+(n-3)*160
- Express {x} in terms of y where:
- 12*x+11*y=20
- 13*x-18*y=-16
- -18*x-3*y=2
- 5*x+20*y=-4
- Graphing y =:
- z^6+1
-
Identical expressions
- z^ six + one = zero
- z to the power of 6 plus 1 equally 0
- z to the power of six plus one equally zero
- z6+1=0
- z⁶+1=0
- z^6+1=O
-
Similar expressions
- z^6-1=0
z^6+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{6} + 1 = 0$$
Because equation degree is equal to = 6 and the free term = -1 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{6} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
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:
$$w_{1} = - i$$
$$w_{2} = i$$
$$w_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$w_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - i$$
$$z_{2} = i$$
$$z_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$z_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$z^{6} + 1 = 0$$
Because equation degree is equal to = 6 and the free term = -1 < 0,
so the real solutions of the equation d'not exist
All other 6 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{6} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = -1$$
so
$$\cos{\left(6 p \right)} = -1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3} + \frac{\pi}{6}$$
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:
$$w_{1} = - i$$
$$w_{2} = i$$
$$w_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$w_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$w_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - i$$
$$z_{2} = i$$
$$z_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
$$z_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
$$z_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
Sum and product of roots
[src]
sum
___ ___ ___ ___
I \/ 3 I \/ 3 \/ 3 I I \/ 3
-I + I + - - - ----- + - - ----- + ----- - - + - + -----
2 2 2 2 2 2 2 2
$$\left(\left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right) + \left(\left(\left(- \frac{\sqrt{3}}{2} - \frac{i}{2}\right) + \left(- i + i\right)\right) + \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right)\right)\right) + \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)$$
=
0
$$0$$
product
/ ___\ / ___\ / ___ \ / ___\
| I \/ 3 | |I \/ 3 | |\/ 3 I| |I \/ 3 |
-I*I*|- - - -----|*|- - -----|*|----- - -|*|- + -----|
\ 2 2 / \2 2 / \ 2 2/ \2 2 /
$$- i i \left(- \frac{\sqrt{3}}{2} - \frac{i}{2}\right) \left(- \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right) \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)$$
=
1
$$1$$
1
Rapid solution
[src]
z1 = -I
$$z_{1} = - i$$
z2 = I
$$z_{2} = i$$
___
I \/ 3
z3 = - - - -----
2 2
$$z_{3} = - \frac{\sqrt{3}}{2} - \frac{i}{2}$$
___
I \/ 3
z4 = - - -----
2 2
$$z_{4} = - \frac{\sqrt{3}}{2} + \frac{i}{2}$$
___
\/ 3 I
z5 = ----- - -
2 2
$$z_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}$$
___
I \/ 3
z6 = - + -----
2 2
$$z_{6} = \frac{\sqrt{3}}{2} + \frac{i}{2}$$
z6 = sqrt(3)/2 + i/2
Numerical answer
[src]
z1 = -0.866025403784439 - 0.5*i
z2 = 1.0*i
z3 = -1.0*i
z4 = 0.866025403784439 - 0.5*i
z5 = -0.866025403784439 + 0.5*i
z6 = 0.866025403784439 + 0.5*i
z6 = 0.866025403784439 + 0.5*i
The graph