Complex Roots of Unity Q (1 Viewer)

Axio · Mar 21, 2015

Hi,

Can anyone help me with this question? It asks to find 1+2w+3w^2 and 1+3w+2w^2, where 1,w,w^2 are cube roots of unity.

I've found that they equal (-3+-isqrt(3))/2, but I can't figure out which answer (positive or negative) matches which expression.

Thanks

integral95 · Mar 21, 2015

$$since w is the cube root of unity,w^3 = 1 \ \ \ \ \ \ 1+w+w^2 = 0$\ \ and \ \ \ w = cis(\frac{2\pi}{3}) \\ \\ 1+2w+3w^2 = 1+w+w^2 +w+2w^2 = w+2w^2 = w^2+w^2+w+1-1 \\ \\ \\ = w^2-1 = cis(\frac{4\pi}{3}) - 1 = \frac{-1-i\sqrt{3}}{2} - 1 = \frac{-3-i\sqrt{3}}{2}$

and you do the similar working for the other one

FrankXie · Mar 21, 2015

integral95 said:
$\ w = cis(\frac{\pi}{3})$

no

integral95 · Mar 21, 2015

FrankXie said:
no

ahhh crap...my rust is pretty evident

EDITED!!

braintic · Mar 22, 2015

integral95 said:
ahhh crap...my rust is pretty evident

EDITED!!

Nothing a little late night liver lubrication won't fix.

InteGrand · Mar 22, 2015

integral95 said:
ahhh crap...my rust is pretty evident

EDITED!!

We should use $\tau$ instead!

integral95 · Mar 22, 2015

braintic said:
Nothing a little late night liver lubrication won't fix.

InteGrand said:
We should use $\tau$ instead!

Wut

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Complex Roots of Unity Q (1 Viewer)

Axio

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integral95

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FrankXie

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