Online Course Discussion Forum
Math Challenge III 2.20 Algebra
For 2.20, it says let ω be the imaginary root of degree 3. Calculate $(1-ω)(1-ω^2)(1-ω^4)(1-ω^8)$. I'm not sure what an imaginary root of degree 3 means.
The wording does need to be more clear. It means that $\omega$ is a root of the following equation: $z^3=1$, and $\omega$ is an imaginary number. A better way to describe $\omega$ is "imaginary cube root of unity".
Ok thank you
In terms of doing the problem, I get that $\omega= 1, \mathrm{cis} \frac{2 \pi}{3}, \mathrm{cis} \frac{4 \pi}{3}$. Would it be a good idea to convert this to polar\rectangular form and directly calculate the product? I guess because $\omega$ is a cube root of unity, I can simplify the expression to $(1-\omega)^2(1-\omega^2)^2$.
$1$ doesn't count because $\omega$ is imaginary. The other two give you the same result. Working with roots of unity, you don't need to convert to rectangular form. Instead, use the fact that $\omega^3=1$, and $\omega^2+\omega+1=0$, you will find the algebra pretty easy.
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