Are these values of the following correct? (Complex cube root of unity)

[Math100]

Homework Statement

Let $x$ be a complex cube root of unity, so that $x^3=1$ but $x\neq 1$. What are the values of the following?
a) $1+x^2+x^4$
b) $1+x^5+x^{10}$
c) $1+x^9+x^{18}$

Relevant Equations

None.

Given that x is a complex cube root of unity, we have $x^{3}=1$ but $x\neq 1$, where the cube roots of unity are $1, \omega, \omega^2$ and $\omega, \omega^2$ are the imaginary roots such that $\omega^3=1, 1+\omega+\omega^2=0$.
Now we will consider three cases of the following.
Case #1: Consider the value of $1+x^2+x^4$.
Let $x=\omega$ or $x=\omega^2$ and $\omega^3=1$,
so $x^4=\omega^4=\omega\cdot \omega^3=\omega$.
This gives $1+x^2+x^4=1+\omega^2+\omega$ for $x=\omega$ and $1+x^2+x^4=1+(\omega^2)^2+(\omega^2)^4=1+\omega+\omega^2$ for $x=\omega^2$.
Thus, $1+x^2+x^4=1+\omega+\omega^2=0$.
Case #2: Consider the value of $1+x^5+x^{10}$.
Let $x=\omega$ or $x=\omega^2$ and $\omega^3=1$,
so $x^5=\omega^5=\omega^2\cdot \omega^3=\omega^2\cdot 1=\omega^2$ and $x^{10}=(\omega^5)^2=(\omega^2)^2=\omega^4=\omega\cdot \omega^3=\omega$.
This gives $1+x^5+x^{10}=1+\omega^2+\omega$ for $x=\omega$ and $1+x^5+x^{10}=1+(\omega^2)^5+(\omega^2)^{10}=1+\omega+\omega^2$.
Thus, $1+x^5+x^{10}=1+\omega+\omega^2=0$.
Case #3: Consider the value of $1+x^9+x^{18}$.
Let $x=\omega$ or $x=\omega^2$ and $\omega^3=1$,
so $x^9=(\omega)^9=(\omega^3)^3=1^3=1$ and $x^{18}=(\omega)^{18}=(\omega^3)^6=1^6=1$.
Thus, $1+x^9+x^{18}=1+1+1=3$.
Therefore, the values of $1+x^2+x^4, 1+x^5+x^{10}, 1+x^9+x^{18}$ are $0, 0, 3$.

 

[Mark44]

Given that x is a complex cube root of unity, we have $x^{3}=1$ but $x\neq 1$, where the cube roots of unity are $1, \omega, \omega^2$ and $\omega, \omega^2$ are the imaginary roots such that $\omega^3=1, 1+\omega+\omega^2=0$.

1. Is the condition that $1 + x + x^2 = 0$ given as part of the problem? If so you should have stated it in the problem description.
2. Strictly speaking, all of the cube roots are 1 are complex (not imaginary) but the third has no imaginary part; i.e., 1.
BTW, the nonreal cube roots of 1 are $-\frac 1 2 + i\frac{\sqrt 3}2$ and $-\frac 1 2 + i\frac{\sqrt 3}2$. Another representation (polar form) for them is $e^{\frac{2\pi}3i}$. You can merely plug these values in the three separate expressions.

 

[Math100]

1. Is the condition that $1 + x + x^2 = 0$ given as part of the problem? If so you should have stated it in the problem description.
2. Strictly speaking, all of the cube roots are 1 are complex (not imaginary) but the third has no imaginary part; i.e., 1.
BTW, the nonreal cube roots of 1 are $-\frac 1 2 + i\frac{\sqrt 3}2$ and $-\frac 1 2 + i\frac{\sqrt 3}2$. Another representation (polar form) for them is $e^{\frac{2\pi}3i}$. You can merely plug these values in the three separate expressions.

1) The condition that $1+x+x^2=0$ is not given as part of the problem, it's something I came up with.
2) Based on these complex cube roots of $1$, I plugged them into the given expressions and got the exact same results as $0, 0, 3$ for the final values. I was wrong thinking that the cube roots of $1$ is imaginary when it's complex.

 

[Math100]

Thank you so much for your help, @Mark44 !

 

[Mark44]

1) The condition that $1+x+x^2=0$ is not given as part of the problem, it's something I came up with.

It happens to be true, but it's something that you should provide some evidence for. It's immediately obvious if you set x to $-\frac 1 2 + i\frac{\sqrt 3}2$, one of the cube roots of 1. $x^2$ turns out to be the complex conjugate of x, so the imaginary parts cancel and$x + x^2$ results in -1.

 

[Steve4Physics]

Given that x is a complex cube root of unity, we have $x^{3}=1$ but $x\neq 1$, where the cube roots of unity are $1, \omega, \omega^2$ and $\omega, \omega^2$ are the imaginary roots such that $\omega^3=1, 1+\omega+\omega^2=0$.
Now we will consider three cases of the following.
Case #1: Consider the value of $1+x^2+x^4$.
Let $x=\omega$ or $x=\omega^2$ and $\omega^3=1$,
so $x^4=\omega^4=\omega\cdot \omega^3=\omega$.
This gives $1+x^2+x^4=1+\omega^2+\omega$ for $x=\omega$ and $1+x^2+x^4=1+(\omega^2)^2+(\omega^2)^4=1+\omega+\omega^2$ for $x=\omega^2$.
Thus, $1+x^2+x^4=1+\omega+\omega^2=0$.
Case #2: Consider the value of $1+x^5+x^{10}$.
Let $x=\omega$ or $x=\omega^2$ and $\omega^3=1$,
so $x^5=\omega^5=\omega^2\cdot \omega^3=\omega^2\cdot 1=\omega^2$ and $x^{10}=(\omega^5)^2=(\omega^2)^2=\omega^4=\omega\cdot \omega^3=\omega$.
This gives $1+x^5+x^{10}=1+\omega^2+\omega$ for $x=\omega$ and $1+x^5+x^{10}=1+(\omega^2)^5+(\omega^2)^{10}=1+\omega+\omega^2$.
Thus, $1+x^5+x^{10}=1+\omega+\omega^2=0$.
Case #3: Consider the value of $1+x^9+x^{18}$.
Let $x=\omega$ or $x=\omega^2$ and $\omega^3=1$,
so $x^9=(\omega)^9=(\omega^3)^3=1^3=1$ and $x^{18}=(\omega)^{18}=(\omega^3)^6=1^6=1$.
Thus, $1+x^9+x^{18}=1+1+1=3$.
Therefore, the values of $1+x^2+x^4, 1+x^5+x^{10}, 1+x^9+x^{18}$ are $0, 0, 3$.

May I suggest simplifying by using the fact that $x^3 =1$.

a) $1+x^2+x^4$ = $1+x^2+x^3x = 1+x^2 + x$

b) $1+x^5+x^{10} = 1+x^3x^2 + (x^3)^3x = 1+x^2+x$

So a) and b) are the same!

c) $1+x^9 + x^{18} = 1+(x^3)^3 + (x^3)^6 = 1+1+1=3$

 

[Mark44]

Aside from, and agreeing with, what @Steve4Physics suggests, I don't see any advantage in bringing another variable, $\omega$ into the mix.

 

[pasmith]

May I suggest simplifying by using the fact that $x^3 =1$.

Together with $$1 + x^n + x^{2n} = \begin{cases} \frac{1 - x^{3n}}{1 - x^n} & x^n \neq 1, \\ 3 & x^n = 1.\end{cases}$$

 

[Vanadium 50]

It happens to be true, but it's something that you should provide some evidence for. It's immediately obvious if you...

Agreed. It's even more obvious with Viela's Theorem, which sadly, is only lightly covered in the Mad Rush to Calculus.

Using it, one can show that for all integers n > 1, the sum of the nth roots of 1 is zero. In two lines, one of which is stating the problem.

 

[Mark44]

Viela's Theorem

Vieta's Theorem?