Is $\mathbb Q(\sqrt{4+i\sqrt{20}},\sqrt{4-i\sqrt{20}})=\mathbb Q (\sqrt{4+i\sqrt{20}})$? I don't know if this is true but it is trivial that the right is contained in the left. With the other inclusion, I think the only non trivial thing to check is if $A=\sqrt{4-i\sqrt{20}} \in \mathbb Q (\sqrt{4+i\sqrt{20}})$
Can we just take the square or A and then do $A^2 - 2 A^2 = -4 -i\sqrt{20}$ and then because we are in fields, we can add $8$ to both sides to get $4-i\sqrt{20}$ and then can we just take the root of this can prove the claim?
 A: If you multiply the two elements you get
$$\sqrt{4+i\sqrt{20}}\cdot\sqrt{4-i\sqrt{20}} = \sqrt{16-20i^2}=\sqrt{16+20}=\pm 6. \label{A} \tag{A} $$
So you have that
$$\sqrt{4-i\sqrt{20}}=\frac{\pm 6}{\sqrt{4+i\sqrt{20}}}\in \mathbb{Q}\left(\sqrt{4-i\sqrt{20}}\right).$$
Note that the first equality in my equation \eqref{A} is not really valid since we are working with complex numbers; the square roots you mention are not well-defined, as you can't really know what root of $X^2-(4+\sqrt{20})$ you have chosen as $\sqrt{4+i\sqrt{20}}$. But since the two possible choice are opposite for both square roots, the product is well-defined up to a sign, thus the argument is still valid.
A: Let $a = \sqrt{4 - \mathrm{i} \sqrt{20}}$, $b = \sqrt{4 + \mathrm{i} \sqrt{20}}$, and $F = \Bbb{Q}(b)$.
You have shown that some polynomial in $a$ is in $F$.  This is not the same as showing $a \in F$.  Too see this, consider that $\mathrm{i}^2 \in \Bbb{Q}$ but $\mathrm{i} \not \in \Bbb{Q}$.
There are two generic ways to attack this.  The first is to find an expression in $F$ that produces $a$.  This is what Arnaud D. did in his answer.
The more straightforward/tedious, but less insight-requiring/-providing method is: Observe that the minimal polynomial of $b$ over $\Bbb{Q}$ is $x^4 - 8x^2 + 36$, so, $\Bbb{Q}(b)$ is a four-dimensional vector space over $\Bbb{Q}$.  (We don't actually have to find the minimal degree.  We could look at $b$ and say, "the outer root gives degree $2$, the $\mathrm{i}$ doubles that, and the inner root doubles that, giving degree $8$."  This is an overestimate, but an overestimate will work; you'll just spend more time discovering that some of the powers of $b$ are redundant in the elimination steps below.  Alternatively, $b$ is simple enough that writing down a degree $4$ polynomial is pretty easy: $b^2-4 = \mathrm{i}\sqrt{20}$, so $(b^2-4)^2 = -20$.  This might still be an overestimate, but that won't matter.)  
That is, $\{1, b, b^2, b^3\}$ spans $\Bbb{Q}(b)$  If you can find $q_0, q_1, q_2, q_3 \in \Bbb{Q}$ such that $$  a = q_0 + q_1 b + q_2 b^2 + q_3 b^3  \text{,}  $$ then $a \in F$.  Looking at the real parts (and grouping by $\sqrt{5}$, since that number is not rational), this says $$ q_0 + 4q_2 + \sqrt{5}(q_1 + 2q_3) = \sqrt{5}  \text{,} $$ from which we find $q_0 = - 4 q_2$ and $q_1 = 1 - 2q_3$.  Plugging those in to our general form and looking at imaginary parts, we find $$1 + 2\sqrt{5}q_2 + 12 q_3 = -1  \text{,}  $$ so $q_2 = 0$ and $q_3 = -1/6$.  Backsubstituting, we find $q_0 = 0$ and $q_1 = 4/3$.  Hence, $a = 4b/3 - b^3/6 \in F$.
