To find the degree of $\mathbb{Q}(\sqrt{5},i)/ \mathbb{Q}$ I have to use Tower law to find the degree of  $\mathbb{Q}(\sqrt{5},i)/ \mathbb{Q}$
So far, this is what I get:
First to find the degree of $[\mathbb{Q}(\sqrt{5}): \mathbb{Q}]$, the sequence $(1,\sqrt{5})$ clearly span $\mathbb{Q}(\sqrt{5})$. Assume that $(1,\sqrt{5})$ is L.D, i.e, $u+v\sqrt{5}=0$, with $u, v$ not both zero. Hence, $\sqrt{5}=-u/v$ which is a contraction. Hence, $(1,\sqrt{5})$ is L.I. and it follows that $(1,\sqrt{5})$ is a basis of $\mathbb{Q}(\sqrt{5})$, hence $[\mathbb{Q}(\sqrt{5}): \mathbb{Q}]=2$
Next, to find the degree of $[\mathbb{Q}(\sqrt{5},i): \mathbb{Q}(\sqrt{5})]$, same as the previous part, let $(1,i)$ be a basis of $\mathbb{Q}(\sqrt{5},i)$ over $\mathbb{Q}$. Hence $[\mathbb{Q}(\sqrt{5},i): \mathbb{Q}(\sqrt{5})]=2$.
so, by the tower law, the answer is $4$. I am not sure about the second part of my solution. If there is wrong, can someone point it out and help me with it please?
Thank you so much
 A: In the second part, you need to determine the degree of $\mathbb{Q}(\sqrt{5},i)$ over $\mathbb{Q}(\sqrt{5})$, not over $\mathbb{Q}$. 
Your overall approach is sound, but you need to be careful with explaining what it is exactly you're proving, and prove exactly this. The coefficients of linear combinations in the second step come from $\mathbb{Q}(\sqrt{5})$, not merely from $\mathbb{Q}$. 
A: Both of these are simplified by the fact that you're dealing with extensions of degree two. Let $K$ be a field of characteristic 0. If $r$ satisfies $r^2 + br +c =0$, with $b,c \in K$, then $r$ is either in $K$ or of degree two over $K$. This is due to the fact that the above polynomial is irreducible over $K$ iff it does not have a root in $K$. 
For the first step, $K = \mathbb{Q}$, and since $\sqrt{5}$ satisfies $x^2 -5 =0$ and is not rational, the first extension is degree two. That's basically what your first part shows. 
For the second part, you know that $i$ satisfies $x^2 +1  =0$, so you just need that $i \not \in \mathbb{Q}(\sqrt{5})$. But, you kind this since $\mathbb{Q}(\sqrt{5}) \subseteq \mathbb{R}$  
