Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation -x^3+14*x^2-61*x+84=0
-
Equation asin(-1)-2=x
- Equation z^3=8*i
- Equation (x+3)(x-2)(x+9)=0
- Express {x} in terms of y where:
- -1*x+8*y=-11
- -9*x+19*y=-17
- 14*x-16*y=-20
- -7*x+15*y=0
- z^3
- Integral of d{x}:
- z^3
- Graphing y =:
- z^3
-
Identical expressions
- z^ three = eight *i
- z cubed equally 8 multiply by i
- z to the power of three equally eight multiply by i
- z3=8*i
- z³=8*i
- z to the power of 3=8*i
- z^3=8i
- z3=8i
-
Similar expressions
- (2sqrt(3)z^2-2sqrt(3)+8iz)/2=0
z^3=8*i equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{3} = 8 i$$
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:
$$\sqrt[3]{z^{3}} = \sqrt[3]{8 i}$$
or
$$z = 2 \sqrt[3]{i}$$
Expand brackets in the right part
We get the answer: z = 2*i^(1/3)
All other 3 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = 8 i$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 8 i$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = i$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = i$$
so
$$\cos{\left(3 p \right)} = 0$$
and
$$\sin{\left(3 p \right)} = 1$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{6}$$
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:
$$w_{1} = - 2 i$$
$$w_{2} = - \sqrt{3} + i$$
$$w_{3} = \sqrt{3} + i$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - 2 i$$
$$z_{2} = - \sqrt{3} + i$$
$$z_{3} = \sqrt{3} + i$$
$$z^{3} = 8 i$$
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:
$$\sqrt[3]{z^{3}} = \sqrt[3]{8 i}$$
or
$$z = 2 \sqrt[3]{i}$$
Expand brackets in the right part
z = 2*i^1/3
We get the answer: z = 2*i^(1/3)
All other 3 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = 8 i$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 8 i$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = i$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = i$$
so
$$\cos{\left(3 p \right)} = 0$$
and
$$\sin{\left(3 p \right)} = 1$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{6}$$
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:
$$w_{1} = - 2 i$$
$$w_{2} = - \sqrt{3} + i$$
$$w_{3} = \sqrt{3} + i$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - 2 i$$
$$z_{2} = - \sqrt{3} + i$$
$$z_{3} = \sqrt{3} + i$$
Vieta's Theorem
it is reduced cubic equation
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = - 8 i$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = - 8 i$$
$$p z^{2} + q z + v + z^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = - 8 i$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 0$$
$$z_{1} z_{2} z_{3} = - 8 i$$
The graph
Rapid solution
[src]
z1 = -2*I
$$z_{1} = - 2 i$$
___ z2 = I - \/ 3
$$z_{2} = - \sqrt{3} + i$$
___ z3 = I + \/ 3
$$z_{3} = \sqrt{3} + i$$
z3 = sqrt(3) + i
Sum and product of roots
[src]
sum
___ ___ -2*I + I - \/ 3 + I + \/ 3
$$\left(- 2 i + \left(- \sqrt{3} + i\right)\right) + \left(\sqrt{3} + i\right)$$
=
0
$$0$$
product
/ ___\ / ___\ -2*I*\I - \/ 3 /*\I + \/ 3 /
$$- 2 i \left(- \sqrt{3} + i\right) \left(\sqrt{3} + i\right)$$
=
8*I
$$8 i$$
8*i
Numerical answer
[src]
z1 = -1.73205080756888 + 1.0*i
z2 = -2.0*i
z3 = 1.73205080756888 + 1.0*i
z3 = 1.73205080756888 + 1.0*i