Mister Exam
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How to use it?
- Equation:
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Equation 8x-15=-12+5x
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Equation 141*10^(-6)*x*3*sqrt(2)=sqrt((1-19881*10^(-12)*x^2)^2+9)
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- Express {x} in terms of y where:
- -20*x-4*y=-14
- 17*x+9*y=0
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Identical expressions
- z^ four + eighty-one = zero
- z to the power of 4 plus 81 equally 0
- z to the power of four plus eighty minus one equally zero
- z4+81=0
- z⁴+81=0
- z^4+81=O
-
Similar expressions
- z^4-81=0
z^4+81=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{4} + 81 = 0$$
Because equation degree is equal to = 4 and the free term = -81 < 0,
so the real solutions of the equation d'not exist
All other 4 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{4} = -81$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -81$$
where
$$r = 3$$
- 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
$$p = \frac{\pi N}{2} + \frac{\pi}{4}$$
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:
$$w_{1} = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$w_{2} = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
$$w_{3} = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$w_{4} = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$z_{2} = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
$$z_{3} = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$z_{4} = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
$$z^{4} + 81 = 0$$
Because equation degree is equal to = 4 and the free term = -81 < 0,
so the real solutions of the equation d'not exist
All other 4 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{4} = -81$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -81$$
where
$$r = 3$$
- 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
$$p = \frac{\pi N}{2} + \frac{\pi}{4}$$
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:
$$w_{1} = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$w_{2} = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
$$w_{3} = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$w_{4} = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$z_{2} = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
$$z_{3} = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
$$z_{4} = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ ___ ___ ___ ___
3*\/ 2 3*I*\/ 2 3*\/ 2 3*I*\/ 2 3*\/ 2 3*I*\/ 2 3*\/ 2 3*I*\/ 2
- ------- - --------- + - ------- + --------- + ------- - --------- + ------- + ---------
2 2 2 2 2 2 2 2
$$\left(\left(\frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}\right) + \left(\left(- \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}\right) + \left(- \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}\right)\right)\right) + \left(\frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}\right)$$
=
0
$$0$$
product
/ ___ ___\ / ___ ___\ / ___ ___\ / ___ ___\ | 3*\/ 2 3*I*\/ 2 | | 3*\/ 2 3*I*\/ 2 | |3*\/ 2 3*I*\/ 2 | |3*\/ 2 3*I*\/ 2 | |- ------- - ---------|*|- ------- + ---------|*|------- - ---------|*|------- + ---------| \ 2 2 / \ 2 2 / \ 2 2 / \ 2 2 /
$$\left(- \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}\right) \left(- \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}\right) \left(\frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}\right) \left(\frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}\right)$$
=
81
$$81$$
81
Rapid solution
[src]
___ ___
3*\/ 2 3*I*\/ 2
z1 = - ------- - ---------
2 2
$$z_{1} = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
___ ___
3*\/ 2 3*I*\/ 2
z2 = - ------- + ---------
2 2
$$z_{2} = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
___ ___
3*\/ 2 3*I*\/ 2
z3 = ------- - ---------
2 2
$$z_{3} = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2} i}{2}$$
___ ___
3*\/ 2 3*I*\/ 2
z4 = ------- + ---------
2 2
$$z_{4} = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2} i}{2}$$
z4 = 3*sqrt(2)/2 + 3*sqrt(2)*i/2
Numerical answer
[src]
z1 = 2.12132034355964 - 2.12132034355964*i
z2 = 2.12132034355964 + 2.12132034355964*i
z3 = -2.12132034355964 - 2.12132034355964*i
z4 = -2.12132034355964 + 2.12132034355964*i
z4 = -2.12132034355964 + 2.12132034355964*i