Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 2^2-x-1=x^2-5*x-(-1-x^2)
-
Equation cos(x)=(1/2)
-
Equation (x+2)^2=(1-x)^2
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Equation z^4+1=0
- Express {x} in terms of y where:
- 4*x-15*y=19
- 16*x-17*y=-9
- 7*x-2*y=20
- -7*x-7*y=-14
- Derivative of:
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z^4+1
-
Identical expressions
- z^ four + one = zero
- z to the power of 4 plus 1 equally 0
- z to the power of four plus one equally zero
- z4+1=0
- z⁴+1=0
- z^4+1=O
-
Similar expressions
- z^4-1=0
z^4+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{4} + 1 = 0$$
Because equation degree is equal to = 4 and the free term = -1 < 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:
$$w^{4} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -1$$
where
$$r = 1$$
- 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:
$$w_{1} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$w_{3} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$z_{3} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$z^{4} + 1 = 0$$
Because equation degree is equal to = 4 and the free term = -1 < 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:
$$w^{4} = -1$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -1$$
where
$$r = 1$$
- 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:
$$w_{1} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$w_{3} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$w_{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
$$z_{3} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
$$z_{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ ___ ___ ___ ___
\/ 2 I*\/ 2 \/ 2 I*\/ 2 \/ 2 I*\/ 2 \/ 2 I*\/ 2
- ----- - ------- + - ----- + ------- + ----- - ------- + ----- + -------
2 2 2 2 2 2 2 2
$$\left(\left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) + \left(\left(- \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) + \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right)\right)\right) + \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right)$$
=
0
$$0$$
product
/ ___ ___\ / ___ ___\ / ___ ___\ / ___ ___\ | \/ 2 I*\/ 2 | | \/ 2 I*\/ 2 | |\/ 2 I*\/ 2 | |\/ 2 I*\/ 2 | |- ----- - -------|*|- ----- + -------|*|----- - -------|*|----- + -------| \ 2 2 / \ 2 2 / \ 2 2 / \ 2 2 /
$$\left(- \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right) \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}\right) \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}\right)$$
=
1
$$1$$
1
Rapid solution
[src]
___ ___
\/ 2 I*\/ 2
z1 = - ----- - -------
2 2
$$z_{1} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
___ ___
\/ 2 I*\/ 2
z2 = - ----- + -------
2 2
$$z_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
___ ___
\/ 2 I*\/ 2
z3 = ----- - -------
2 2
$$z_{3} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2} i}{2}$$
___ ___
\/ 2 I*\/ 2
z4 = ----- + -------
2 2
$$z_{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2} i}{2}$$
z4 = sqrt(2)/2 + sqrt(2)*i/2
Numerical answer
[src]
z1 = -0.707106781186548 - 0.707106781186548*i
z2 = 0.707106781186548 + 0.707106781186548*i
z3 = -0.707106781186548 + 0.707106781186548*i
z4 = 0.707106781186548 - 0.707106781186548*i
z4 = 0.707106781186548 - 0.707106781186548*i
The graph