Calculating the roots of a complex polynomial

Question

the problem is the following:

Given a complex polynomial P = z⁴ - $\alpha$, with $\alpha \in \mathbb {C} $. We know that $ e^{i\frac{2\pi}{3}} $ is a complex root of 2P.
Select all the correct options:

(A) 1 is a complex root of P

(B) $ e^{i\frac{\pi}{6}} $ is a complex root of P

(C) - $\sqrt{3}$ - $i$ is a complex root of 2P

(D) $\sqrt{2}$ + $ i\sqrt{2}$ is a complex root of 2P

(E) None of the above

I think that subtracting $ \alpha $ to z⁴ and then multiply by 2 will not change the number of complex roots (which is 4). So I divided 2$\pi$ by 4 to get the angle and calculated the other roots using the $ e^{i\frac{2\pi}{3}} $. After this I had to exclude (C) and (D) because the modulus didn't match: | - $\sqrt{3}$ - $i$ | = |$\sqrt{2}$ + $ i\sqrt{2}$| = 2 $\neq$ 1
Then I calculated the complex number 2P wich is equal to $ (e^{i\frac{2\pi}{3}})^4 $ and from that I got P and its complex roots. I finally excluded (A) and (B) because 1 is a positive real number and therefore the only option is $ e^{i0 + 2k\pi} $ (which isn't a root of P) and $ e^{i\frac{\pi}{6}} $ is not a root of P either.
After this, there's only one option left. Option (E)

Can you tell me if my logic is correct? And if not, what am I missing here?

Thank you

Answer 1

$\exp\left(i\frac{2\pi}3\right)$ is a root of the polynomial $2P$

$\implies$ $\exp\left(i\frac{2\pi}3\right)$ is a root of the polynomial $P$

$\implies$ $\left(\exp\left(i\frac{2\pi}3\right)\right)^4-\alpha=0$

$\implies$ $\alpha=\exp\left(i\frac{8\pi}3\right)=\exp\left(i\frac{2\pi}3\right)$

$\implies$ $P=z^4-\exp\left(i\frac{2\pi}3\right)$

$\implies$ Option (B) is true.