I am struggling with the following problem: Let $f \in F[x]$ be an irreducible quintic polynomial with splitting field $K$, where $\mathbb{Q} \subseteq F$. Supposing that $f$ has three real roots and two complex roots, prove that $Aut(K/F)$ contains a $2$-cycle.
My attempt: I suspect the $2$-cycle will be an automorphism which fixes all real roots and sends the complex roots to their conjugates. Let the real roots of $f$ be $a_1, a_2, a_3$. Construct a tower of fields $F \subset E \subset K$ where $E = F[a_1, a_2, a_3]$. There must be a minimal polynomial $g \in E[x]$ with the two complex roots of $f$ as its only roots. Call these roots $\omega$ and $\omega^*$.
Since $K$ is a Galois extension of $F$, then so too is it a Galois extension of $E$. Hence, $|Aut(K/E)| = [K:E]$. Since the extension $K$ over $E$ is nontrivial, then there must be some nontrivial $E$-automorphism that permutes the roots $\omega, \omega^*$. The only nontrivial permutation of the roots would send $\omega \mapsto \omega^*$.
Thus, we conclude two things. First, $[K:E] = 2$. Second, $\phi(\omega) = \omega^*$ is a legitimate $F$-automorphism of $K$. Therefore, $\phi$ is our $2$-cycle. Is this correct?
In general, say that an irreducible polynomial $f \in F[x]$ has $n$ pairs of complex roots and that its splitting field is $K$. Is complex conjugation of every root a legitimate $F$-automorphism of $K$?