I'm stuck with this subquestion: $F(a) = a^6 - \sqrt{3}a^3 + 1$ express as product of linear factors where $a^3 = \displaystyle \frac{\sqrt{3} - i}{2}$.
I've worked out that $a = re^{π/18}, re^{13π/18}, re^{25π/18}, re^{-π/18}, re^{11π/18}, re^{23π/18}$.
I've tried to search how to do product of linear factors online but there's nothing similar to this at all. Any help would be appreciated!
To decompose $F(a)$, let $t=a^3$, you have a quadratic equation in $t$: $$t^2-\sqrt{3}t+1=0$$ The solutions are: $$t=\frac{\sqrt{3}\pm\sqrt{3-4}}{2}=\frac{\sqrt{3}\pm i}{2}$$ Now, you sustitute $a^3=t$ and you have: $$a=\frac{\sqrt{3}\pm i}{2}$$ The factorization of $F(a)$ is so: $$F(a)=\left(a^3-\frac{\sqrt{3}+ i}{2}\right)\left(a^3-\frac{\sqrt{3}-i}{2}\right)$$ This factorization is not complete, in fact in the complex plane we can solve $a^3=\frac{\sqrt{3}+ i}{2}$ and $a^3=a^3=\frac{\sqrt{3}-i}{2}$.
For the first, let $x=a$. The modulus of $z=\frac{\sqrt{3}+ i}{2}$ is $$|z|=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=1$$ By DeMoivre's rule, we have: $$z=\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)=r^3(\cos(3\theta)+i\sin(3\theta))$$ The radius $r$ is $r=1$, while for $\theta$, we have: $$3\theta=\frac{\pi}{6}+2k\pi, k=0,1,2\leftrightarrow \: \theta_0=\frac{\pi}{18} \vee \theta_1=\frac{13\pi}{18} \vee \theta_2=\frac{25\pi}{18}$$
The three complex numbers are: $$p_0=\cos\left(\frac{\pi}{18}\right)+i\sin\left(\frac{\pi}{18}\right) \vee p_1=\cos\left(\frac{13\pi}{18}\right)+i\sin\left(\frac{13\pi}{18}\right) \vee p_2=\cos\left(\frac{25\pi}{18}\right)+i\sin\left(\frac{25\pi}{18}\right)$$
For the second let $x=a$. The modulus of $z=\frac{\sqrt{3}-i}{2}$ is again $1$, but in this case: $$z=\cos\left(\frac{5\pi}{6}\right)+i\sin\left(\frac{5\pi}{6}\right)=r^3(\cos(3\theta)+i\sin(3\theta))$$ The solutions are: $$p_3=\cos\left(\frac{5\pi}{18}\right)+i\sin\left(\frac{5\pi}{18}\right) \vee p_4=\cos\left(\frac{11\pi}{18}\right)+i\sin\left(\frac{11\pi}{18}\right) \vee p_5=\cos\left(\frac{23\pi}{18}\right)+i\sin\left(\frac{23\pi}{18}\right)$$ Finally, we arrive at: $$F(a)=\left(a-\cos\left(\frac{\pi}{18}\right)-i\sin\left(\frac{\pi}{18}\right)\right)\cdot\left(a-\cos\left(\frac{13\pi}{18}\right)-i\sin\left(\frac{13\pi}{18}\right)\right)\cdot\left(a-\cos\left(\frac{\pi}{18}\right)-i\sin\left(\frac{25\pi}{18}\right)\right)\cdot\left(a-\cos\left(\frac{5\pi}{18}\right)-i\sin\left(\frac{5\pi}{18}\right)\right)\cdot\left(a-\cos\left(\frac{11\pi}{18}\right)-i\sin\left(\frac{11\pi}{18}\right)\right)\cdot\left(a-\cos\left(\frac{23\pi}{18}\right)-i\sin\left(\frac{23\pi}{18}\right)\right)$$
Or, using exponential form: $$F(a)=(a-e^{\pi/18})\cdot(a-e^{13\pi/18})\cdot(e^{25\pi/18})\cdot(a-re^{5\pi/18})\cdot(a-e^{11\pi/18})\cdot(a-e^{23\pi/18})$$