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!
Best Answer
Hint: $\,0 = \alpha^5 -1 = (\alpha-1)(\alpha^4+\alpha^3+\alpha^2+\alpha+1) \implies \alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0\,$. Then:
$u + v = \alpha + \alpha^2 + \alpha^3 + \alpha^4 = -1$
$uv = \alpha^3+\alpha^4+\alpha^6+\alpha^7 = \alpha^3+\alpha^4+\alpha+\alpha^2 = -1$
Knowing their sum and product, it follows that $\,u,v\,$ are the roots of the quadratic $\,t^2 + t - 1 = 0\,$, so $\,u,v = \frac{-1 \pm \sqrt{5}}{2}\,$, then $\,|u-v| = \sqrt{5}\,$. What's left to prove is that $\,u \gt v\,$, so $\,u-v = \sqrt{5}\,$.