Conjugate of $\sqrt{a+\sqrt{b}}$ How can we compute the conjugate of $\frac{1}{\sqrt{a+\sqrt{b}}}$ ?
I know that the conjugate of $(a+\sqrt{b})$ is $(a-\sqrt{b})$  but what about the global square ?
For example how do we go from : $ \frac{\sqrt{2}}{2 \sqrt{2+\sqrt{2}}}$ to $ \frac{\sqrt{2-\sqrt{2}}}{2} $?
 A: Upon multiplying $\sqrt{a+\sqrt{b}}$ by itself you get the term that you know the conjugate of (i.e. the term $a+\sqrt{b}$. Now multiply again by $(a-\sqrt{b})$. Two multiplications were finally involved:
$$(\sqrt{a+\sqrt{b}})(a-\sqrt{b})$$
A: The conjugates of $\sqrt{a+\sqrt b}$ are calculated as follows, for generic $a,b\in\mathbb Q$. The minimal polynomial of $\sqrt{a+\sqrt b}$ is $(x^2-a)^2-b=0$, by inspection. The other roots of this equation are $\pm\sqrt{a\pm\sqrt b}$.
Note that this is for generic $a,b\in\mathbb Q$. For instance, when $a=3$, $b=8$, we have $\sqrt{a+\sqrt b}=1+\sqrt2$, which only has a single conjugate $1-\sqrt2$.
P.S. The calculation above also shows that the Galois group is the order $8$ group $\mathrm{Gal}\big(\mathbb Q(\sqrt{a\pm\sqrt b})/\mathbb Q\big)\cong D_8$. This is because $\mathbb Q(\sqrt{a\pm\sqrt b})/\mathbb Q(\sqrt b)$ is a Kummer extension, with Galois group $C_2\times C_2$, and $\mathbb Q(\sqrt b)/\mathbb Q$ has Galois group $C_2$. Thus the Galois group $G$ splits as follows: $1\to C_2\to G\to C_2\times C_2\to 1$.
It is generated by $\sigma\colon \sqrt{a+\sqrt b}\mapsto\sqrt{a+\sqrt b},\sqrt{a-\sqrt b}\mapsto-\sqrt{a-\sqrt b}$ of order $2$ and $\tau\colon\sqrt{a+\sqrt b}\mapsto-\sqrt{a-\sqrt b},\sqrt{a-\sqrt b}\mapsto\sqrt{a+\sqrt b}$, which has order $4$.
