SOLUTION: How many complex roots does the equation below have? x 6 + x 3 + 1 = 0

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Question 1063890: How many complex roots does the equation below have?
x 6 + x 3 + 1 = 0

Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E6%2Bx%5E3%2B1=0

Let  x3 = u
Then x6 = u2

x%5E6%2Bx%5E3%2B1=0


The discriminant is 

b%5E2-4ac%22%22=%22%221%5E2-4%281%29%281%29%22%22=%22%221-4-3

which is negative u has two complex roots.

So there are two complex values x3.

There are 3 complex cube roots of each of those two complex
roots for u.  Therefore, 

there are 6 complex roots.  <--answer

Edwin



Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
Every polynomial of degree "n" has exactly "n" complex roots.

It is the Main Algebra Theorem firstly discovered and proved by famous German mathematician Gauss in 19-th century.

So, the given equation has exactly 6 complex roots in Complex domain.

Regarding your equation 

x%5E6+%2B+x%5E3+%2B+1 = 0,             (0)

we can solve it explicitly in this way.

First, introduce new variable y = x%5E3.
In terms of this new equations, the original equation takes the form

y%5E2+%2B+y+%2B+1 = 0.

It is a quadratic equation. You can solve it by applying the quadratic formula

y%5B1%2C2%5D = %28-1+%2B-+sqrt%281%5E2+-+4%2A1%2A1%29%29%2F2 = %28-1+%2B-+sqrt%28-3%29%29%2F2 = %28-1+%2B-+i%2Asqrt%283%29%29%2F2.

So, one root is y%5B1%5D = %28-1+%2Bi%2Asqrt%283%29%29%2F2. The other root is y%5B2%5D = %28-1+-i%2Asqrt%283%29%29%2F2.

You can write these complex numbers (complex roots) in a trigonometric (polar) form:

y%5B1%5D = %28-1+%2Bi%2Asqrt%283%29%29%2F2 = %28-1%2F2%29+%2B+i%2A%28sqrt%283%29%2F2%29 = cos%282pi%2F3%29%2Bi%2Asin%282pi%2F3%29.  The modulus is sqrt%28%28-1%2F2%29%5E2%2B%28sqrt%283%29%2F2%29%5E2%29 = sqrt%281%2F4+%2B3%2F4%29 = sqrt%281%29 = 1  and the argument is 2pi%2F3.

y%5B2%5D = %28-1+-i%2Asqrt%283%29%29%2F2 = %28-1%2F2%29+-+i%2A%28sqrt%283%29%2F2%29 = cos%284pi%2F3%29%2Bi%2Asin%284pi%2F3%29.  The modulus is sqrt%28%28-1%2F2%29%5E2%2B%28-sqrt%283%29%2F2%29%5E2%29 = sqrt%281%2F4+%2B3%2F4%29 = sqrt%281%29 = 1  and the argument is 4pi%2F3.


Next, since y%5B1%5D = x%5E3, then x = root%283%2Cy%5B1%5D%29 = root%283%2C+%28cos%282pi%2F3%29%2Bi%2Asin%282pi%2F3%29%29%29.


    Now, applying the de Moivre's formula, you have THREE values for x, that all are the cubic roots of the complex number cos%282pi%2F3%29%2Bi%2Asin%282pi%2F3%29:

        x%5B1%5D = root%283%2Cy%5B1%5D%29 = root%283%2C+%28cos%282pi%2F3%29%2Bi%2Asin%282pi%2F3%29%29%29 = cos%282pi%2F9%29+%2B+i%2Asin%282pi%2F9%29,            (1)

        x%5B2%5D = root%283%2Cy%5B1%5D%29 = root%283%2C+%28cos%282pi%2F3%29%2Bi%2Asin%282pi%2F3%29%29%29 = cos%282pi%2F9+%2B+2pi%2F3%29+%2B+i%2Asin%282pi%2F9+%2B+2pi%2F3%29,     (2)

        x%5B3%5D = root%283%2Cy%5B1%5D%29 = root%283%2C+%28cos%282pi%2F3%29%2Bi%2Asin%282pi%2F3%29%29%29 = cos%282pi%2F9+%2B+4pi%2F3%29+%2B+i%2Asin%282pi%2F9+%2B+4pi%2F3%29.     (3)


By the same way as the root y%5B1%5D produces the three solutions x%5B1%5D, x%5B2%5D and x%5B3%5D, the root  y%5B2%5D produces three other solutions x%5B4%5D, x%5B5%5D and x%5B6%5D.

Namely, since y%5B2%5D = x%5E3, you have x = root%283%2Cy%5B2%5D%29 = root%283%2C+%28cos%284pi%2F3%29%2Bi%2Asin%284pi%2F3%29%29%29.

    Now, applying the de Moivre's formula again, you have THREE other values for x, that all are the cubic roots of the complex number cos%284pi%2F3%29%2Bi%2Asin%284pi%2F3%29:

        x%5B4%5D = root%283%2Cy%5B2%5D%29 = root%283%2C+%28cos%284pi%2F3%29%2Bi%2Asin%284pi%2F3%29%29%29 = cos%284pi%2F9%29+%2B+i%2Asin%284pi%2F9%29,            (4)

        x%5B5%5D = root%283%2Cy%5B2%5D%29 = root%283%2C+%28cos%284pi%2F3%29%2Bi%2Asin%284pi%2F3%29%29%29 = cos%284pi%2F9+%2B+2pi%2F3%29+%2B+i%2Asin%284pi%2F9+%2B+2pi%2F3%29,     (5)

        x%5B6%5D = root%283%2Cy%5B2%5D%29 = root%283%2C+%28cos%284pi%2F3%29%2Bi%2Asin%284pi%2F3%29%29%29 = cos%284pi%2F9+%2B+4pi%2F3%29+%2B+i%2Asin%284pi%2F9+%2B+4pi%2F3%29.     (6)

These complex numbers, x%5B1%5D, x%5B2%5D, x%5B3%5D, x%5B4%5D, x%5B5%5D and x%5B6%5D, are all different and are all the roots of the original equation (0).

Probably, this explanation is too complicated for the high school students.

But it is how the university students who study "High Algebra" or/and "Abstract Algebra" solve the equations like (0) in complex domain.

Expressions (1) - (6) could be simplified further and written in rectangular coordinates.
I will not go into these details here.

The complex numbers (1) - (6) lie on the unit circle of the complex domain.


There is a bunch of my lessons on complex numbers
    - Complex numbers and arithmetical operations on them
    - Complex plane
    - Addition and subtraction of complex numbers in complex plane
    - Multiplication and division of complex numbers in complex plane
    - Raising a complex number to an integer power
    - How to take a root of a complex number
    - Solution of the quadratic equation with real coefficients on complex domain
    - How to take a square root of a complex number
    - Solution of the quadratic equation with complex coefficients on complex domain
in this site.

Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook.