Thread: cube roots of unity again

hi folks,

many thanks for the quick answer to my original post on this subject.

I tried a different method and get stuck for different reasons. I think my mistake is fundamental, like the last time, so it may be instructive to others.

The cube roots of unity are 1 and $\dfrac{-1}{2} \pm \dfrac{\sqrt{3}}{2}i $

$z^3 = 1$

let $z = (\cos t + i \sin t)$

$(\cos t + i \sin t)^3 = 1$

there are two ways (at least) of expanding this out. If you do, you will get

$\cos^3 t - 3 \cos t \sin^2 t = 1 $ .......... Real part

$3\cos^2 t \sin t - \sin^3 t = 0 $ ........... Imaginary part

so from the imaginary part

$\sin t (3 \cos^2 t - \sin^2 t ) = 0 $

gives

$\sin t = 0$ or $ 3\cos^2 t = \sin ^2 t $

$3 \cos^2 t = 1 - \cos^2 t$

$\cos^2 t = \dfrac{1}{4}$

$\cos t = \pm \dfrac{1}{2}$

Now $\cos t = - \dfrac{1}{2}$ is great. This gives us the -1/2 real bit of the complex root. But we also have a perfectly legitimate + 1/2.

I have been thinking of how to get rid of it but I cannot find a good reason. But we must get rid of it! But how?

$3 \cos^2 t = \sin^2 t $ gives the correct imaginary part.

You only found solutions that satisfy one equation. Try it in the other equation. You now have possible solutions: $\cos t = \pm \dfrac{1}{2}$ and $\sin t = \pm \dfrac{\sqrt{3}}{2}$.

You also have:
$\cos^3 t - 3\cos t \sin^2 t = 1$
Plug in $\cos t = \dfrac{1}{2}$ and $\sin^2 t = \dfrac{3}{4}$ (which is true for either value of $\sin t$ you choose) and you get:

$\dfrac{1}{8} - \dfrac{3}{2}\dfrac{3}{4} = -1 \neq 1$ So, $\dfrac{1}{2}$ is not valid.

Excellent! And so fast! Really appreciated. Stuff like this keeps me up at night.