How to find all field homomorphisms $\mathbb Q(\sqrt(5+ \sqrt 5 ) \rightarrow \mathbb C$

After computing the minimal polynomial of $\sqrt{(5+\sqrt5}$ in $\mathbb C$, ie. $X^4 -10X^2 +20$. I know that $\mathbb Q \left(\sqrt{(5+\sqrt5} \right)$ = $ \left\{ a+b\alpha + c\alpha^2 + d\alpha^3 | a,b,c,d \in \mathbb Q, \alpha = \sqrt{(5+\sqrt5} \right\}$

Lets say $\phi : \mathbb Q \left(\sqrt{(5+\sqrt5} \right) \rightarrow \mathbb C$ is a homomorphism. Then it satisfies that for $e$ and $\forall a,b \in Q(\sqrt{(5+\sqrt5})$ $$\phi(e) = e,$$ $$\phi(a+b) = \phi(a)+\phi(b),$$ $$\phi(ab) = \phi(a)\phi(b). $$

How do I find all $\phi?$ I know that $\phi (a+b\alpha + c\alpha^2 + d\alpha^3) = \phi (a) + \phi(b\alpha) + \phi(c\alpha^2) + \phi (d\alpha^3)$. How should I proceed finding them?

Taking your reasoning one step further: \begin{eqnarray*}\phi (a+b\alpha + c\alpha^2 + d\alpha^3) &=& \phi (a) + \phi(b\alpha) + \phi(c\alpha^2) + \phi (d\alpha^3)\\&=&a + b\phi(\alpha) + c\phi(\alpha)^2 + d\phi (\alpha)^3.\end{eqnarray*} Here we have used the fact that any field homomorphism $\mathbb{Q}\to\mathbb{C}$ must be inclusion, as $\phi(1)=1$, so $\phi(n)=n$ for $n\in\mathbb{Z}$ and hence $\phi(\frac1n)n=\phi(\frac1n)\phi(n)=1$ for $n\in \mathbb{Z}$.

From this we see that the field homomorphisms are determined completely by $\phi(\alpha)$. We know $\phi(\alpha)$ satisfies the minimal polynomial of $\alpha$: \begin{eqnarray*} \alpha^4-10\alpha^2+20&=&0 \\ \implies \phi(\alpha)^4-10\phi(\alpha)^2+20&=&0. \end{eqnarray*} Here we apply $\phi$ to both sides of the first line, to get the second line.

Thus there are four possibilities for $\phi(\alpha)$: $\pm\sqrt{5\pm \sqrt{5}}$, hence four possibile field homomorphisms.