How to obtain $\sqrt{5}$ (without a calculator) using complex numbers & roots of unity? For context, here is the entire question:

15. (a) Find the fifth roots of unity in exponential form.$\hspace{14.5pt}$(b) Let $\alpha$ be the complex fifth root of unity with the smallest positive argument, and suppose that $u = \alpha + \alpha^4$ and $v = \alpha^2 + \alpha^3$,$\hspace{30pt}$(i) Find the values of $u + v$ and $u - v$.

Particularly, I am struggling with the second part of (b)(i) where the answer is $\sqrt{5}$ for $u-v$. However, I cannot fathom how this answer is obtained without using a calculator.
What I have done so far: $u-v = \alpha + \alpha^4 - \alpha^2 - \alpha^3$, then through substitution of $\alpha = e^{i(2\pi/5)}$, I eventually obtained $u - v = 2\cos(2\pi/5) - 2\cos(4\pi/5)$.
How do I go further from here to calculate √5, or is it a distinct approach to the question entirely?
Any help greatly appreciated!

The cosine (or sine) of a rational multiple of $\pi$ is always algebraic. Often a simple way to express such a number as a polynomial root is using the Chebyshev polynomials.
Here the second and third Chebyshev polynomials of the first kind give the identities
$$ \cos(2x) = 2 \cos^2 x - 1 $$
$$ \cos(3x) = 4 \cos^3 x - 3 \cos x $$
If $x = \frac{2\pi}{5}$, then we know $\cos(2x) = \cos \frac{4\pi}{5} = \cos \frac{6\pi}{5} = \cos(3x)$, so $t = \cos \frac{2\pi}{5}$ is a solution to
$$ 2 t^2 - 1 = 4 t^3 - 3 t $$
$$ 4 t^3 - 2 t^2 - 3t + 1 = 0 $$
$t=1$ is one obvious root, so the cubic is factored:
$$ (t-1)(4t^2 + 2t - 1) = 0 $$
$$ 4(t-1)\left(t-\frac{-1+\sqrt{5}}{4}\right)\left(t-\frac{-1-\sqrt{5}}{4}\right) = 0 $$
Since $0 < \cos \frac{2\pi}{5} < 1$, we must have
$$ \cos \frac{2\pi}{5} = \frac{-1+\sqrt{5}}{4} $$
Then the rest is just plugging in and computations.
$$ \cos \frac{4\pi}{5} = 2 \cos^2 \frac{2\pi}{5} - 1 = \frac{-1-\sqrt{5}}{4} $$
(This makes sense since $x=\frac{4\pi}{5}$ is another value where $\cos(2x)=\cos(3x)$.)
$$ 2\cos \frac{2\pi}{5} - 2\cos \frac{4\pi}{5} = \sqrt{5} $$
