1. ## Complex Numbers.

$\alpha$ is the complex root (with the smallest positive argument) of the equation $z^5 - 1 = 0$

show that $\alpha^4 = \alpha^*$. hence or otherwise obtain $z^5 - 1 = 0$ as a product of real linear and quadratic facto giving the coefficients in terms of integers and cosines.
I am good with all of this expect the last part about the quadratic factors. A pointer would be good.

Many Thanks, Bobak.

2. Originally Posted by bobak
I am good with all of this expect the last part about the quadratic factors. A pointer would be good.

Many Thanks, Bobak.
since $1$ is a zero of the equation and we know that there are five roots that are evenly spaced around the unit cirlce by

$\frac{2 \pi}{5}$

so the five roots are

$1$ , $e^{\frac{2 \pi}{5}i}$, $e^{\frac{4 \pi}{5}i}$ , $e^{\frac{6 \pi}{5}i}$, and $e^{\frac{8 \pi}{5}i}$.

so $\alpha =e^{\frac{2 \pi}{5}i}$ and $\alpha^* =e^{\frac{-2 \pi}{5}i}=e^{\frac{8 \pi}{5}i}$

so one of the quadraric factors is
$(z-e^{\frac{2 \pi}{5}i})(z-e^{\frac{8 \pi}{5}i})=z^2-(e^{\frac{8 \pi}{5}i}+e^{\frac{2 \pi}{5}i})z+1$

Now finally use Eulers formula to write the z coeffient is terms of sine and cosine.

I hope this helps

Brett

3. The imaginary parts cancel so all coefficients are real, for the other root I used $(z-\alpha^2)(z-\alpha^3)$ use the conjugate relation and simplified. (and the linear root is just obvious)

Thanks for the help Brett

Bobak