Mister Exam
Other calculators
-
How to use it?
- Equation:
- Equation (x-4)^4-4*(x-4)^2-21=0
-
Equation 6x-19=-x-10
-
Equation x^2+4=5x
-
Equation cos(x)-sin(x)=1
- Express {x} in terms of y where:
- 18*x+12*y=18
- -6*x+12*y=11
- -2*x+20*y=-2
- 5*x+19*y=-3
- Graphing y =:
- z^3+1
- Derivative of:
-
z^3+1
-
Identical expressions
- z^ three + one = zero
- z cubed plus 1 equally 0
- z to the power of three plus one equally zero
- z3+1=0
- z³+1=0
- z to the power of 3+1=0
- z^3+1=O
-
Similar expressions
- z^3-1=0
z^3+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{3} + 1 = 0$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{z^{3}} = \sqrt[3]{-1}$$
or
$$z = \sqrt[3]{-1}$$
Expand brackets in the right part
We get the answer: z = (-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = -1$$
so
$$\cos{\left(3 p \right)} = -1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
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:
$$w_{1} = -1$$
$$w_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$w_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = -1$$
$$z_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$z_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
$$z^{3} + 1 = 0$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{z^{3}} = \sqrt[3]{-1}$$
or
$$z = \sqrt[3]{-1}$$
Expand brackets in the right part
z = -1^1/3
We get the answer: z = (-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -1$$
where
$$r = 1$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = -1$$
so
$$\cos{\left(3 p \right)} = -1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
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:
$$w_{1} = -1$$
$$w_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$w_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = -1$$
$$z_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$z_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 1$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = 1$$
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 1$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = 1$$
Rapid solution
[src]
z1 = -1
$$z_{1} = -1$$
___
1 I*\/ 3
z2 = - - -------
2 2
$$z_{2} = \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
___
1 I*\/ 3
z3 = - + -------
2 2
$$z_{3} = \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
z3 = 1/2 + sqrt(3)*i/2
Sum and product of roots
[src]
sum
___ ___
1 I*\/ 3 1 I*\/ 3
-1 + - - ------- + - + -------
2 2 2 2
$$\left(-1 + \left(\frac{1}{2} - \frac{\sqrt{3} i}{2}\right)\right) + \left(\frac{1}{2} + \frac{\sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ ___\ / ___\ |1 I*\/ 3 | |1 I*\/ 3 | -|- - -------|*|- + -------| \2 2 / \2 2 /
$$- (\frac{1}{2} - \frac{\sqrt{3} i}{2}) \left(\frac{1}{2} + \frac{\sqrt{3} i}{2}\right)$$
=
-1
$$-1$$
-1
Numerical answer
[src]
z1 = 0.5 - 0.866025403784439*i
z2 = 0.5 + 0.866025403784439*i
z3 = -1.0
z3 = -1.0
The graph