How to prove that $\frac{1}{2} \left( 1 - \sqrt{3} \pm i \sqrt{2} \sqrt[4]{3} \right)$ is not a root of unity? I have a proof for the complex roots $\sqrt{2} - 1 \pm i \sqrt{2\sqrt{2}-2}$ which involve a contradiction using the minimal polynomial of the real part of the root, but the proof does not work on the requested number, as the minimal polynomial of $\frac{1 - \sqrt{3}}{2}$ is $t^2 - t - \frac{1}{2}$. (not integral, so a lemma I used does not work).
The lemma:
If $\alpha$ is a root of unity whose real part is an algebraic integer, then $\alpha^4 = 1$.
Perhaps such a proof is impossible?
If this unfortunately turns out to be the case, the most elementary proof possible will be accepted.
 A: Let $\alpha:=\frac12(1-\sqrt{3}+i\sqrt{2}\sqrt[4]3)$, then $|\alpha|=1$, whence $\overline{\alpha}=\frac1\alpha$. Therefore,
$$1-\sqrt{3}=2\operatorname{Re}(\alpha)=\alpha+\overline{\alpha}=\alpha+\frac1\alpha.$$
We conclude that
$$3\alpha^2=\left(\alpha^2-\alpha+1\right)^2=\alpha^4+\alpha^2+1+2(-\alpha^3+\alpha^2-\alpha),$$
which may be simplified to
$$\alpha^4-2\alpha^3-2\alpha+1=0.$$
Now, we use @JyrkiLahtonen's remark that the above polynomial is irreducible because $\alpha+\overline{\alpha}\not\in\mathbb{Q}$, and it has two real zeroes, at approximately $0.435$ and $2.297$. Those are clearly not roots of unity, and if $\alpha$ were a root of unity, all its conjugates would be too.
Alternatively, once you know the polynomial is (monic) irreducible, it can only have a root of unity as one of its roots if it is cyclotomic. Take a quick look at the first $20$ or so cyclotomic polynomials, and you'll see that's not the case.
A: I am going to "piggyback" on Mastrem's answer.  The object is to show explicitly that the minimal polynomial derived there has roots not on the unit circle.
Quote the minimal polynomial equation:
$\alpha^4-2\alpha^3-2\alpha+1=0$
Now we have a little fun with this.  Suppose $\alpha$ is on the unit circle.  If we render this as the ratio $(1+z)/(1-z)$ then $|1+z|=|1-z|$ and thus $z$ is pure imaginary.  Let us make this substitution and determine whether all roots for $z$ meet this requirement.
$(1+z)^4-2(1+z)^3(1-z)-2(1+z)(1-z)^3+(1-z)^4=0$
$2(3z^4+6z^2-1)=0$
We have good news and bad news.  The good news is that both roots for $z^2$ are real.  The bad news is that they are opposite in sign.  The negative root for $z^2$, properly corresponding to pure imaginary $z$, comes from both complex conjugates for $\alpha$ that lie on the unit circle.  But... the positive root for $z^2$ represents nonzero real roots for $z$, and these correspond to real reciprocal roots for $\alpha$ that do not lie on the unit circle.

Upon further review, there actually is an approach to make the proof with the OP's quoted lemma. Simply square the given value of $\alpha$. Using standard complex-number multiplication gives
$\alpha^2=\color{blue}{(1-\sqrt3)}+i[(1-\sqrt3)\sqrt2\sqrt[4]3/2]$
where now, the real part is an algebraic integer but the fourth power is not $1$ (the real part does not match with any fourth root of unity). So $\alpha^2$ is not a root of unity and $\alpha$ can't be one either.
