Element of a splitting field that can be construct using ruler and compas Looking at the splitting field over $\Bbb{Q}$ of $X^5-2$ which is $D:=\Bbb{Q}(e^{2ik\pi/5},2^{1/5}).$
What element of $D$ can be construct using ruler and compas ? For the reals I know it means that the degree of the slipping field is a power of $2$, but what about complexes elements? 
 A: Note $\zeta_5 = e^{2ik\pi/5}$. You have
$$[\mathbb Q(\zeta_5):\mathbb Q]=4, \ [\mathbb Q(\sqrt[5]{2}):\mathbb Q]=5.$$
As $4, 5$ are coprime, $[\mathbb Q(\zeta_5,\sqrt[5]{2}):\mathbb Q]=20$. Therefore the Galois group $\text{Gal}(\mathbb Q(\zeta_5,\sqrt[5]{2})/\mathbb Q)$ is of order $20$. 
Now, a complex $z$ number is constructible over $\mathbb Q$ (like for a real) if and only if $[\mathbb Q(z):\mathbb Q]$ is a power of $2$. This was one of your question.
So we're left to find the subfields $K \subseteq \mathbb Q(\zeta_5,\sqrt[5]{2})$ such that $[K:\mathbb Q] \in \{2,4\}$.
Suppose that that $[K:\mathbb Q] =4$
Then according to Fundamental theorem of Galois theory
$$(\text{Gal}(\mathbb Q(\zeta_5,\sqrt[5]{2})/\mathbb Q):G)=4$$ where $G=\text{Aut}(\mathbb Q(\zeta_5,\sqrt[5]{2})/K)$. That means that $G$ is a $5$-Sylow of $\text{Gal}(\mathbb Q(\zeta_5,\sqrt[5]{2}))$. According to Sylow theorems, there is only one $5$-Sylow and therefore one subfield $K$ having the required properties which is in fact $\mathbb Q(\zeta_5)$. The associated constructible number provides a construction of the pentagon.
Suppose that that $[K:\mathbb Q] =2$
To be done.
