Finding the conjugates, why can they argue this way?(exercise) In one exercise I am supposed to find the conjugate of $\sqrt{2}+i$ over $\mathbb{Q}$. 
I found the answer by finding irr$( \sqrt{2}+i,\mathbb{Q})$, and then solving the polynomial finding all the roots(this is the definition of conjugates), this gave the correct answer, but in the solutions manual, they use a trick I do not understand.
They say that:
The conjugate are $\sqrt{2}+i,\sqrt{2}-i,-\sqrt{2}+i,-\sqrt{2}-i$. This is clear beacuse $\mathbb{Q}(\sqrt{2}+i)=(\mathbb{Q}(\sqrt{2}))(i)$.
Can someone say how this is clear and why it follows from the fact that: $\mathbb{Q}(\sqrt{2}+i)=(\mathbb{Q}(\sqrt{2}))(i)$?
 A: By what you mention in your comments, the conjugates of $(\sqrt{2}+i)$ can be computed if we knew the image of $(\sqrt{2}+i)$ under every homomorphism
$$
\varphi : \mathbb{Q}(\sqrt{2}+i) \to \overline{\mathbb{Q}}
$$
Since
$$
\mathbb{Q}(\sqrt{2}+i) = \mathbb{Q}(\sqrt{2})(i)
$$
we need to determine $\varphi(i)$ and $\varphi(\sqrt{2})$. However, since $\varphi$ is a field homomorphism
$$
\varphi(\sqrt{2})^2 = \varphi(2) = 2
$$
since any field homomorphism must fix $\mathbb{Q}$ (this takes a short proof). However, $2$ has exactly 2 square roots in $\overline{\mathbb{Q}} \subset \mathbb{C}$, so
$$
\varphi(\sqrt{2}) = \pm \sqrt{2}
$$
Similarly,
$$
\varphi(i) = \pm i
$$
and so the list of possible conjugates is
$$
\{\pm\sqrt{2}\pm i\}
$$
To prove that each of these is, in fact, a conjugate of $(\sqrt{2}+i)$, you need the fact that the number of such homomorphisms is precisely equal to the degree
$$
[\mathbb{Q}(\sqrt{2}+i):\mathbb{Q}]
$$
My answer is long enough already, so let me know if this helps and if you have any questions.
A: Let $F=\mathbb Q(\sqrt2+i)$.  If you square $\sqrt{2}+i$ you get $1+2\sqrt2i$.  Thus $\sqrt{2}i\in F$.  Multiply this by $\sqrt2+i$ and you get $2i-\sqrt2\in F$.  Add $\sqrt2+i$ and you get $3i\in F$ thus $i\in F$.  Thus subtracting $i$ from $\sqrt{2}+i$ you get $\sqrt2\in F$.
Thus both $i$ and $\sqrt2$ are in $\mathbb Q(\sqrt2+i)$
