Finding ω¹⁰+ω⁵+3 where ω satisfies x²+x+1=0

[Asla]

complex number ??

Homework Statement

Let ω be the solution to the equation x²+x+1=0
Get the value of ω¹⁰+ω⁵+3=

Homework Equations

complex numbers?

The Attempt at a Solution

When I try solving the first equation I hit a complex number which is making me think I am wrong.
(x+1/2)²=-3/4
Again if the method is right, what is the relationship between the complex number and the later expression?

 

[NascentOxygen]

There are two solutions to the quadratic, both are complex numbers.

 

[TimeToShine]

Solve the equation and find the two complex solutions for x.

Now, you know that these values are equal to ω, apply De Moivre's theorem for complex numbers to the new expression.

 

[HallsofIvy]

You don't have to solve the equation- in fact, you don't have to use complex numbers at all.

From $$x^2+ x+ 1= 0$$ we get $$x^2= -(x+ 1)$$. $$x^{10}= (x^2)^5= -(x+ 1)^5= -x^5- 5x^4- 10x^3- 10x^2- 5x- 1$$ so that $$x^{10}+ x^5+ 3= -5x^4- 10x^3- 10x^2- 5x+ 2$$.

Now, continue using $$x^2= -(x+ 1)$$ to keep reducing the exponents until you have reduce it to a quadratic.

 

[Asla]

You don't have to solve the equation- in fact, you don't have to use complex numbers at all.

From $$x^2+ x+ 1= 0$$ we get $$x^2= -(x+ 1)$$. $$x^{10}= (x^2)^5= -(x+ 1)^5= -x^5- 5x^4- 10x^3- 10x^2- 5x- 1$$ so that $$x^{10}+ x^5+ 3= -5x^4- 10x^3- 10x^2- 5x+ 2$$.

Now, continue using $$x^2= -(x+ 1)$$ to keep reducing the exponents until you have reduce it to a quadratic.

I tried doing that but I really do not know how to go about the -10x³The cubic power keeps resurfacing?

 

[NascentOxygen]

Maybe try:

x^2 = -(x+1)

Therefore, x.x^2 = -x(x+1) = -x^2 - x

Now, substitute for x^2 the expression on the top line, and you have the equivalent for x^3.

 

[Asla]

Maybe try:

x^2 = -(x+1)

Therefore, x.x^2 = -x(x+1) = -x^2 - x

Now, substitute for x^2 the expression on the top line, and you have the equivalent for x^3.

Wow good insight now I have the quadratic equation and am stuck again.

 

[HallsofIvy]

Yes, you can reduce it to a quadratic. Now compare it to $$x^2+ x+ 1$$ which you know is 0.

 

[D H]

I tried doing that but I really do not know how to go about the -10x³The cubic power keeps resurfacing?

Determining the value of $x^3$ is critical. You know that $x^2 = -(x+1)$. Multiply both sides by x and simplify the right hand side.

Hall's approach is valid, but it's even easier if you use $x^5 = x^3 x^2$ and $x^{10} = (x^3)^3 x$.

 

[Asla]

Nice I got it.Thanks