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.