solving ecuations in ℂ

[ana.agatha.black]

I would like help with this if any of you has time:
If ε is a solution of x²+ x + 1 = 0 calculate (aε²+bε)(bε²+aε)
I know this particular equation (x²+ x + 1 = 0) was kind of special and ε had certain properties like ε³=... but i can't seem to remember any of them.
this is all I have done so far: from x²+ x + 1 = 0 x_1,2= ( -1 +/- i√3) / 2 . I really have no idea how to begin solving this.
I don't need this solved step by step but any help pointing me in the right direction would be truly appreciated.

 

[Deleted member 4993]

I would like help with this if any of you has time:
If ε is a solution of x²+ x + 1 = 0 calculate (aε²+bε)(bε²+aε)
I know this particular equation (x²+ x + 1 = 0) was kind of special and ε had certain properties like ε³= 1

... but i can't seem to remember any of them.
this is all I have done so far: from x²+ x + 1 = 0 x_1,2= ( -1 +/- i√3) / 2 . I really have no idea how to begin solving this.
I don't need this solved step by step but any help pointing me in the right direction would be truly appreciated.

.

 

[DrPhil]

I would like help with this if any of you has time:
If ε is a solution of x²+ x + 1 = 0 calculate (aε²+bε)(bε²+aε)

I know this particular equation (x²+ x + 1 = 0) was kind of special and ε had certain properties like ε³=... but i can't seem to remember any of them.
this is all I have done so far: from x²+ x + 1 = 0...... x_1,2= ( -1 ± i√3) / 2 .

I really have no idea how to begin solving this.
I don't need this solved step by step but any help pointing me in the right direction would be truly appreciated.

One way or another, knowing ε² is going to be helpful.

\(\displaystyle \displaystyle \epsilon_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm i\)

\(\displaystyle \displaystyle \epsilon_{1,2}^2 = \frac{1}{4} \mp \frac{\sqrt{3}}{2} \mathrm i - \frac{3}{4} = -\frac{1}{2} \mp \frac{\sqrt{3}}{2} \mathrm i = \epsilon_{1,2}^*\)

That looks interesting.

 

[Deleted member 4993]

I would like help with this if any of you has time:
If ε is a solution of x²+ x + 1 = 0 calculate (aε²+bε)(bε²+aε)
I know this particular equation (x²+ x + 1 = 0) was kind of special and ε had certain properties like ε³=... but i can't seem to remember any of them.
this is all I have done so far: from x²+ x + 1 = 0 x_1,2= ( -1 +/- i√3) / 2 . I really have no idea how to begin solving this.
I don't need this solved step by step but any help pointing me in the right direction would be truly appreciated.

ε³ = 1 and ε² = ε^*

then

(aε²+bε)(bε²+aε) = abε⁴ + a²ε³ + b²ε³ + abε² ............................ Corrected

= abε + a²+ b² + abε^* ............................ Corrected

= ab(ε +ε^*) + a² + b² = ab(-1) + a² + b² = a² + b² - ab ............................ Corrected

 

[Romsek]

ε³ = 1 and ε² = ε^*

I wondered where this came from and discovered it's due to the fact that the roots of 1+z+z^2 are a conjugate pair spaced by 2pi/3 radians on the unit circle.

This lead me to wonder about higher order polynomials with all coefficients equal to 1.

Consider 1 + z + z^2 + ... + z^N-1 = 0

This is a geometric series, and for z != 1 is equal to (1 - z^N)/(1 - z)

When z=1, this polynomial = N+1 and thus 1 is not a root of that polynomial, so

(1 - z^N)/(1-z) = 0, (1 - z^N) = 0, 1 = z^N

Thus the N-1 roots of this polynomial are the N roots of z^N=1 except for z=1

The powers of these roots generate a finite group and in the case of N=3, i.e this problem, that group is ε, ε^*, and 1

This must be what led the original poster to consider that this polynomial had special properties.

 

[HallsofIvy]

A rather obvious point is that if \(\displaystyle x^2+ x+ 1= 0\), then, multiplying both sides by x, \(\displaystyle x^3+ x^2+ x= 0\) and, then, adding 1 to both sides, \(\displaystyle x^3+ (x^2+ x+ 1)= 1\) so that, as Subhotosh Khan said, \(\displaystyle x^3= 1\). From that, \(\displaystyle x^4= x\).

Now, for any a and b, by direct multiplication, \(\displaystyle (ax^2+ bx)(bx^2+ ax)= abx^4+ 2(a^2+ b^2)x^3+ abx^2\).

For \(\displaystyle \epsilon\), since \(\displaystyle \epsilon^2+ \epsilon+ 1= 0\), so that \(\displaystyle \epsilon^2= -\epsilon- 1\), \(\displaystyle \epsilon^3= 1\), and \(\displaystyle \epsilon^4= \epsilon\), that is \(\displaystyle ab\epsilon+ 2(a^2+ b^2)+ ab(-\epsilon- 1)= 2a^2- ab+ 2b^2\).

 

[Deleted member 4993]

A rather obvious point is that if \(\displaystyle x^2+ x+ 1= 0\), then, multiplying both sides by x, \(\displaystyle x^3+ x^2+ x= 0\) and, then, adding 1 to both sides, \(\displaystyle x^3+ (x^2+ x+ 1)= 1\) so that, as Subhotosh Khan said, \(\displaystyle x^3= 1\). From that, \(\displaystyle x^4= x\).

Now, for any a and b, by direct multiplication, \(\displaystyle (ax^2+ bx)(bx^2+ ax)= abx^4+ \)2\(\displaystyle (a^2+ b^2)x^3+ abx^2\).

For \(\displaystyle \epsilon\), since \(\displaystyle \epsilon^2+ \epsilon+ 1= 0\), so that \(\displaystyle \epsilon^2= -\epsilon- 1\), \(\displaystyle \epsilon^3= 1\), and \(\displaystyle \epsilon^4= \epsilon\), that is \(\displaystyle ab\epsilon+ 2(a^2+ b^2)+ ab(-\epsilon- 1)= 2a^2- ab+ 2b^2\).

How did you get that 2?