Galois conjugate $\sigma(a) = \Re(a)$ implies $\sigma(a) = a$? Let $a$ be a complex number which is algebraic over $\mathbb{Q}$. Let $\sigma(a)$ be a Galois conjugate of $a$, and let $\Re(a)$ be the real part of $a$.
Question: Is it true that $\sigma(a) = \Re(a)$ implies $\sigma(a) = a$?
 A: Let $f(x)$ be an irreducible polynomial over $\mathbf{Q}$. I claim that $f(x)$ cannot have three distinct roots $\alpha, \beta, \gamma$ such that
$$\alpha + \beta = 2 \gamma.$$
Assume otherwise. Let $L$ denote the splitting field. Since the Galois group of $L$  acts transitively on the roots,
there exists an embedding of $L$ into $\mathbf{C}$ such that $|\gamma|$ has the largest absolute value. But then $|\gamma| \ge |\alpha|, |\beta|$, and from the triangle inequality we see that the inequality above cannot hold.
Now consider your question, namely, we have an algebraic number $\alpha$ which is conjugate to $\gamma = \mathrm{Re}(\alpha)$, and thus $\gamma$ also a root of the minimal polynomial $f(x)$ of $\alpha$. Moreover, the complex conjugate $\beta$ of $\alpha$ is also a root. But
$$\alpha + \beta = \alpha + \overline{\alpha} = 2 \mathrm{Re}(\alpha) = 2 \gamma.$$
From the above analysis this cannot happen if these are distinct, which forces $\alpha$ to be real, and hence $\alpha = \gamma$ as desired.
