Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation -4x=15-3(3x-5)
-
Equation 9x−9(x+1)=−2x
-
Equation -25-x=-17
-
Equation z^4+16=0
- Express {x} in terms of y where:
- -11*x-15*y=14
- 16*x-3*y=17
- -6*x-12*y=9
- -5*x-2*y=-14
- Graphing y =:
- z^4+16
-
Identical expressions
- z^ four + sixteen = zero
- z to the power of 4 plus 16 equally 0
- z to the power of four plus sixteen equally zero
- z4+16=0
- z⁴+16=0
- z^4+16=O
-
Similar expressions
- z^4-16=0
z^4+16=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$z^{4} + 16 = 0$$
Because equation degree is equal to = 4 and the free term = -16 < 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:
$$w^{4} = -16$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -16$$
where
$$r = 2$$
- 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:
$$w_{1} = - \sqrt{2} - \sqrt{2} i$$
$$w_{2} = - \sqrt{2} + \sqrt{2} i$$
$$w_{3} = \sqrt{2} - \sqrt{2} i$$
$$w_{4} = \sqrt{2} + \sqrt{2} i$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - \sqrt{2} - \sqrt{2} i$$
$$z_{2} = - \sqrt{2} + \sqrt{2} i$$
$$z_{3} = \sqrt{2} - \sqrt{2} i$$
$$z_{4} = \sqrt{2} + \sqrt{2} i$$
$$z^{4} + 16 = 0$$
Because equation degree is equal to = 4 and the free term = -16 < 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:
$$w^{4} = -16$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = -16$$
where
$$r = 2$$
- 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:
$$w_{1} = - \sqrt{2} - \sqrt{2} i$$
$$w_{2} = - \sqrt{2} + \sqrt{2} i$$
$$w_{3} = \sqrt{2} - \sqrt{2} i$$
$$w_{4} = \sqrt{2} + \sqrt{2} i$$
do backward replacement
$$w = z$$
$$z = w$$
The final answer:
$$z_{1} = - \sqrt{2} - \sqrt{2} i$$
$$z_{2} = - \sqrt{2} + \sqrt{2} i$$
$$z_{3} = \sqrt{2} - \sqrt{2} i$$
$$z_{4} = \sqrt{2} + \sqrt{2} i$$
Rapid solution
[src]
___ ___ z1 = - \/ 2 - I*\/ 2
$$z_{1} = - \sqrt{2} - \sqrt{2} i$$
___ ___ z2 = - \/ 2 + I*\/ 2
$$z_{2} = - \sqrt{2} + \sqrt{2} i$$
___ ___ z3 = \/ 2 - I*\/ 2
$$z_{3} = \sqrt{2} - \sqrt{2} i$$
___ ___ z4 = \/ 2 + I*\/ 2
$$z_{4} = \sqrt{2} + \sqrt{2} i$$
z4 = sqrt(2) + sqrt(2)*i
Sum and product of roots
[src]
sum
___ ___ ___ ___ ___ ___ ___ ___ - \/ 2 - I*\/ 2 + - \/ 2 + I*\/ 2 + \/ 2 - I*\/ 2 + \/ 2 + I*\/ 2
$$\left(\left(\sqrt{2} - \sqrt{2} i\right) + \left(\left(- \sqrt{2} - \sqrt{2} i\right) + \left(- \sqrt{2} + \sqrt{2} i\right)\right)\right) + \left(\sqrt{2} + \sqrt{2} i\right)$$
=
0
$$0$$
product
/ ___ ___\ / ___ ___\ / ___ ___\ / ___ ___\ \- \/ 2 - I*\/ 2 /*\- \/ 2 + I*\/ 2 /*\\/ 2 - I*\/ 2 /*\\/ 2 + I*\/ 2 /
$$\left(- \sqrt{2} - \sqrt{2} i\right) \left(- \sqrt{2} + \sqrt{2} i\right) \left(\sqrt{2} - \sqrt{2} i\right) \left(\sqrt{2} + \sqrt{2} i\right)$$
=
16
$$16$$
16
Numerical answer
[src]
z1 = -1.4142135623731 - 1.4142135623731*i
z2 = 1.4142135623731 - 1.4142135623731*i
z3 = -1.4142135623731 + 1.4142135623731*i
z4 = 1.4142135623731 + 1.4142135623731*i
z4 = 1.4142135623731 + 1.4142135623731*i
The graph