Author Topic: Complex roots question (Read 1378 times) Tweet Share

sarahtorrington · « **on:** July 26, 2016, 10:32:00 am »

We got b)i), but are struggling with ii)

We've converted w4 and w5 to w and w2 respectively... But are now stuck!

RuiAce · « **Reply #1 on:** July 26, 2016, 11:11:26 am »

Quote from: sarahtorrington on July 26, 2016, 10:32:00 am

We got b)i), but are struggling with ii)

We've converted w4 and w5 to w and w2 respectively... But are now stuck!

(Image removed from quote.)

$\text{Since }\omega\text{ is the cube root of unity, }1+\omega+\omega^2=0$
$\begin{align*} \therefore (1+\omega)^6 &= (-\omega)^6\\ &= \omega^6 \\ &= \left( \omega^3\right)^2\\ &= 1^2\\ &=1\end{align*}$
$\text{Note, since }z=\omega\text{ satisfies }z^3=1, \quad \omega^3=1$
$\begin{align*}(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)&= (1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2)\\ &= \left[ (1-\omega)(1-\omega^2)\right]^2\\ &= (1-\omega-\omega^2+\omega^3)^2\\ &= (1-(-1)+1)^2\\ &= 9\end{align*}$
I haven't figured out a hence method yet but I did this question halfway through a uni lecture so just tell me if something makes no sense

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Author Topic: Complex roots question (Read 1378 times) Tweet Share

sarahtorrington

Complex roots question

RuiAce

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