1. ## Complex nunber question

$\alpha=cos\frac{2\pi}{5}+isin\frac{2\pi}{5}$.
Find the quadratic equation with roots $\alpha+\alpha^4$ and $\alpha^2+\alpha^3$.
Hence show that:
$cos\frac{2\pi}{5}=\frac{\sqrt5-1}{4}$.

So far all I can come up with for the quadratic is:
$(x-\alpha-\alpha^4)(x-\alpha^2-\alpha^3)$, but I cannot think of how that will help with the second part. Help would be much appreciated!

2. Yes I know de Moivre's formula its what all of this sheet has been about. I just cannot see how this helps to find $cos\frac{2\pi}{5}$

3. I will use a for $alpha$ and c for $cos(2*pi/5)$

If you use DeMoivre's and some trig, then you can show that
root1 = a + a^4 = 2cos(2*pi/5) = 2c
and root2 = a^2 + a^3 = 2cos(4*pi/5) = 2(2cos^2(2*pi/5) - 1) = 2(2c^2 - 1)
All the imaginary parts disappear.

So your quadratic is (x - root1)(x - root2) = (x - a - a^4)(x - a^2 - a^3)
Now, the constant parts of the quadratic eq must be equal, so
root1 * root2 = a^2 + a^4 + a^6 + a^7

We can show that a^2 + a^4 + a^6 + a^7 = -1

This means that
root1 * root2 = 2c * 2(2c^2-1) = -1

Now, solve for c. This will give you 3 solutions, but only one should make sense for cos(2*pi/5), and the problem already tells you the answer

4. Awesome, thank you so much!