complex 2 (1 Viewer)

=)(= · Oct 1, 2022

for c i tried getting it in quad factors and finding coefficient of z^4 but end up with the expression equal to -5/8

sarfaraz_a · Oct 1, 2022

=)(= said:
View attachment 36434
for c i tried getting it in quad factors and finding coefficient of z^4 but end up with the expression equal to -5/8

Substitute z=i into the expression with quadratic factors. That should get you the product.

okkk · Oct 1, 2022

Because it says hence, you have to use part B. Try dividing both sides by w^4 and then using the fact that w^n+w^(-n)=2cos(npi/9) then use sums to products until you get a product of three cos terms, keeping in mind that cos(5pi/9)=-cos(4pi/9).

Drongoski · Oct 2, 2022

Q3(c)

$1+\omega + {\omega}^2 + {\omega}^3 + \cdots {\omega}^8 = 0 \implies \omega + \omega^2 + \omega ^3 + \cdots + \omega ^8 = -1\\ \\ z = cos \theta + i sin\theta \implies \bar z = cos \theta - isin\theta \implies z + \bar z = 2cos \theta \\ \\ $Now $ \omega $ and $ \omega ^8 ,\omega ^2 $ and $ \omega ^7, \omega ^3 $ and $ \omega ^6, \omega ^4 $ and $ \omega ^5 \text { are complex conjugate pairs. }\\ \\ \therefore \omega + \omega^8 = 2 cos \frac {2\pi}{9}\\ \\ \omega^2 + \omega^7 = 2cos \frac {4\pi}{9} \\ \\ \omega^3 + \omega^6 = 2cos \frac {6\pi}{9} = 2cos {2\pi}{3} = -1 \text {( this line not required )} \\ \\ \omega^4 + \omega^5 = 2cos \frac{8\pi}{9} = -2cos(\pi - \frac {8\pi}{9}) = -2cos\frac {\pi}{9} \\ \\ \therefore -2cos\frac{\pi}{9}\times 2cos \frac {2\pi}{9} \times 2cos \frac {4\pi}{9} = (\omega^4 + \omega^5)\times (\omega + \omega^8) \times (\omega^2 + \omega^7) \\ \\ = ( \omega^5 + \omega^{12} +\omega^6 + \omega^{13}) \times (\omega^2 + \omega^7)$

$= \omega ^7 + \omega ^{14} + \omega ^8 +\omega ^{15} + \omega ^{12} + \omega ^{19} + \omega ^{13} + \omega ^{20} \\ \\ = \omega^7 + \omega ^5 + \omega ^8 + \omega ^6 + \omega ^3 + \omega + \omega ^4 + \omega ^2\\ \\ = \omega + \omega^2 + \omega ^3 + \omega ^4 + \omega ^5 + \omega ^6 + \omega ^7 + \omega ^8 \\ \\ = -1\\ \\ \therefore -8 cos \frac{\pi}{9} \times cos \frac {2\pi}{9} \times cos \frac {4\pi}{9} = -1 \\ \\ \therefore cos\frac{\pi}{9} \times cos \frac {2\pi}{9} \times cos \frac {4\pi}{9} = \frac {1}{8}$

QED

=)(= · Oct 2, 2022

Ohh wait i got it subbing in i thx

=)(= · Oct 2, 2022

also i am having a bit of trouble with this

part c

okkk · Oct 2, 2022

Remember z+1/z=2cos2pi/5 and 2cos4pi/5. So by solving the quadratic in (b), you solved a polynomial with roots 2cospi/5 and 2cos4pi/5. Then to get the product you just do sum of roots.

=)(= · Oct 2, 2022

ahh i get it tysm

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