Let $K$ be a field, $F$ be complex conjugation, and let $K_{\mathbb{C}} = \prod_{\tau} \mathbb{C}$, where $\tau$ is taken over the set of embedding $K \hookrightarrow \mathbb{C}$. Then for $z = (z_{\tau}) \in K_{\mathbb{C}}$, $Fz_{\tau} = \overline{z_{\overline{\tau}}}$. My question is, why do both $z$ and $\tau$ get conjugated. It seems like only one or the other should be conjugated.
Edit: I've come to the realization that conjugating both $z$ and $\tau$ is just the way that the action of $F$ is defined, but I'm still a curious as to why.
Somewhat more canonically, $K_{\mathbb C} = \mathbb C\otimes_{\mathbb Q} K$. Complex conjugation then acts just by its action on the first factor in the tensor product.
This tensor product is a product of copies of $\mathbb C$, labelled by the embeddings $\tau$; the projection onto the $\tau$th factor is then just the embedding $\tau$. The isomorphism is given by $$ z\otimes x \mapsto ( z \tau(x) )_{\tau}$$ (where $z \in \mathbb C$ and $x \in K$).
Concretely, then, conjugation take a tuple $(z_{\tau})_{\tau}$ to the tuple $(\overline{z}_{\overline{\tau}})_{\tau}$, just because $$z\otimes x \mapsto \overline{z} \otimes x \quad \text{(this is complex conjugation on the tensor product) }$$ $$\mapsto \bigl( \overline{z} \tau(x) \bigr)_{\tau} = \bigl(\overline{z\, \overline{\tau}(x)} \bigr)_{\tau}$$ (using the formula $\overline{\tau(x)} = \overline{\tau}(x),$ which can be rewritten as $\tau(x) = \overline{\overline{\tau}(x)}$).
Summary: the complex conjugation is the obvious, natural, one if you think of $K_{\mathbb C}$ as $\mathbb C\otimes_{\mathbb Q} K.$
As Alex points out in his answer, the fixed part of complex conjugation is then $\mathbb R\otimes_{\mathbb Q} K$, which can be described (as in his answer) as the product of $r_1$ copies of $\mathbb R$ and $r_2$ copies of $\mathbb C$.