Finding Solutions for Complex Numbers: A Case Study with ω10+ω5+3 = 0

AI Thread Summary
The discussion revolves around solving the equation ω10 + ω5 + 3 = 0, where ω is derived from the quadratic equation x^2 + x + 1 = 0, leading to complex solutions. Participants emphasize the importance of using De Moivre's theorem and reducing higher powers of x by substituting x^2 with -(x + 1). The conversation highlights the challenge of managing cubic terms and finding x^3, suggesting that multiplying by x can simplify the process. Ultimately, the goal is to reduce the expression to a quadratic form without necessarily solving the complex numbers directly. The thread concludes with participants successfully navigating the complexities of the problem.
Asla
Messages
35
Reaction score
0
complex number ??

Homework Statement


Let ω be the solution to the equation x2+x+1=0
Get the value of ω105+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?
 
Physics news on Phys.org
There are two solutions to the quadratic, both are complex numbers.
 
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.
 
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.
 
Last edited by a moderator:
HallsofIvy said:
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?
 
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. :smile:
 
NascentOxygen said:
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. :smile:
Wow good insight now I have the quadratic equation and am stuck again.
 
Yes, you can reduce it to a quadratic. Now compare it to x^2+ x+ 1 which you know is 0.
 
Asla said:
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.
 
  • #10
Nice I got it.Thanks
 

Similar threads

Back
Top