Subtracting roots of unity. Specifically $\omega^3 - \omega^2$

Question

This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headway into how to subtract these? The answer they give is $2cos(\dfrac{3\pi}{5})$

Please, some help will be greatly appreciated. I cant seem to find help over the internet either.

Answer 1

Hint: $\cos(3\pi/5)$ and $\cos(2\pi/5)$ are closely related, as are $\sin(3\pi/5)$ and $\sin(2\pi/5)$.

Remark If you are familiar with the complex exponential, note that $\omega^2\omega^3=e^{i\pi}=-1$. So we are looking at $e^{3\pi i/5}+e^{-3\pi i/5}$, which is $2\cos (3\pi/5)$.

Answer 2

$$\omega=\cos(\pi/5)+i\sin(\pi/5)$$ $$\omega^3-\omega^2=\cos(3\pi/5)+i\sin(3\pi/5)-(\cos(2\pi/5)+i\sin(2\pi/5))$$ $$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$ $$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$ $$\omega^3-\omega^2=-2\sin(\pi/2)\cdot\sin(\pi/10)-2i\cos(\pi/2)\cdot\sin(\pi/10)$$ $$\omega^3-\omega^2=-2\sin(\pi/10)$$ $$\omega^3-\omega^2=-2\cdot\dfrac{\sqrt{5}-1}{4}$$ $$\omega^3-\omega^2=\dfrac{1-\sqrt{5}}{2}$$

use some trig. formula: $\cos C-\cos D=2\cdot \sin\dfrac{C+D}{2}\cdot \sin \dfrac{D-C}{2}$

and $\sin C-\sin D=2\cdot \cos\dfrac{C+D}{2}\cdot \sin \dfrac{C-D}{2}$

and $\sin (\pi/10)=\dfrac{\sqrt5-1}{4}$

Answer 3

$$\overline{\omega}=\omega^9 \Rightarrow \overline{\omega^3}=\omega^{27}=\omega^7=-\omega^2 \,.$$

Thus

$$\omega^3-\omega^2= \omega^3+\overline{\omega^3}=2 Re(\omega^3) \,.$$