# Does an imaginary root,$\omega$ ,of the equation $x^n-1=0$ satisfy both $\omega=1$ and $\omega \ne 1$?

I am studying the $n$th roots of unity $1$ and I have a question about this exercise which is showed in my book:

If $\omega$ is one of the imaginary roots of the equation $x^3=1$ ,then find the product $(1-\omega +\omega^2 )\cdot(1+ \omega-\omega^2)$.

Solution: Writing the given equation as $x^3-1=0$,we factor and find $(x-1)(x^2+x+1)=0$.

Since $\omega$ is imaginary,it is a root of $x^2+x+1$ ,so $\omega^2 +\omega +1 =0$.

Hence we have $1+\omega=-\omega^2$ and $1+\omega^2=-\omega$ and $$(1-\omega+\omega^2)(1+\omega -\omega^2)=(-\omega -\omega)(-\omega^2-\omega^2)=4\omega^3=4,$$ since $\omega^3 =1$

Now,my question is about the last step where we have $\omega^3=1$ .If we solve this we have $\omega=1$ ,but previously in my book it's showed that for any root $\omega \ne 1$ we have $\omega^{n-1}+\omega^{n-2}+...+1=0$ as a direct consequence of the fact that $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+...+1)=0$$

So why do we have now that $\omega=1$ works ?

This leads me to the conclusion that $\omega$ ,being an imaginary root, is such that $\omega=1$ and $\omega \ne 1$ since we have that $\omega^{n-1}+\omega^{n-2}+...+1=0$ if and only if $\omega \ne 1$ and also in the previous exercise we have $\omega^3=1$.

I know I must be wrong but I can't see where's the fall in the logic (I guess I can't draw the conclusion $\omega^3=1 \implies \omega=1$ but why ?).

I am perplexed and confused at the same time...

• "...we have $\omega^3=1$ .If we solve this we have $\omega=1$..." This is where the flaw is. Once you are working with complex numbers, you can't just take cube roots like you did, similarly to how you couldn't deduce $x=1$ from $x^2=1$. Commented Dec 20, 2015 at 11:32
• The complex numbers have $n$ $n$th roots. Your deduction $\omega^3 = 1 \implies \omega = 1$ is not correct. Commented Dec 20, 2015 at 11:32

You can't do the $\omega^3=1 \implies \omega=1$ step for the same reason you can't do $\omega^2=1 \implies \omega=1$ where $-1$ also works. There are $3$ solutions, only 1 of which is $\omega=1$
• Got it.So we're taking the one which isn't equal to $1$.One last question: the property $\omega^{n-1}+\omega^{n-2}+\cdots+1=0$ is then satisfied only by imaginary numbers ? I ask this because my book only states for any root $\ne 1$,so I guess they want me to figure out that they are referring to imaginary roots. Commented Dec 20, 2015 at 11:36
• Not imaginary, complex. They don't have to be multiples of $i$. They are arranged in a circle. Commented Dec 20, 2015 at 11:40