$K/\mathbb{Q}$ contains a complex number - is complex conjugation in $\text{Aut}(K/\mathbb{Q})$? Suppose $K$ is a finite extension of $\mathbb{Q}$.
Suppose there is some complex number in $K$.
Is it necessary that $\tau$, complex conjugation, is a member of $\text{Aut}(K/\mathbb{Q})$?
I feel like this should be true, but consider $\mathbb{Q}(\sqrt{2} + i\sqrt{3})$. I'm not even sure if $\sqrt{2} - i\sqrt{3}$ is in this field.
 A: Not necessarily.  Consider the extension of $\mathbb{Q}$ by $\alpha = \sqrt[3]{2}\cdot \zeta $, where $\zeta$ is a primitive cube root of unity. Then $\alpha$ is a non-real cube root of $2$.
Let us show that $\mathbb{Q}(\alpha)$ is not stable under complex conjugation.  Indeed, we have that
$$ \overline{\sqrt[3]{2}\cdot \zeta} = \sqrt[3]{2}\cdot \zeta^2, $$
so if $\mathbb{Q}(\alpha)$ were stable under conjugation, we would have that all three cube roots of $2$
$$\sqrt[3]{2}\cdot \zeta, \sqrt[3]{2}\cdot \zeta^2, \sqrt[3]{2} = -\sqrt[3]{2}\cdot \zeta-\sqrt[3]{2}\cdot \zeta^2$$
would be in $\mathbb{Q}(\alpha)$, and $\mathbb{Q}(\alpha)$ would be a Galois extension of $\mathbb{Q}$.  However, we know that it isn't.  
Hence $\overline{\alpha}$ is not in $\mathbb{Q}(\alpha)$.

Added: Note, as Qiaochu Yuan has remarked, that the claim is true if $K$ is a Galois extension of $\mathbb{Q}$.  One way to see this is to view $K$ as the splitting field of a polynomial $F$ with coefficients in $\mathbb{Q}$.  Since the conjugate of a root of $F$ is still a root of $F$, we get that $K$ is stable under conjugation, and since $K$ contains (by hypothesis) a non-real element, we get that conjugation is a non-trivial automorphism of $K$.
