Proving $\mathbb{Q}(\eta,i)=\mathbb{Q}(\xi)$ where $\xi=e^{\pi i/10}$ and $\eta=e^{2\pi i/5}$ Proving $\mathbb{Q}(\eta,i)=\mathbb{Q}(\xi)$ where $\xi=e^{\pi i/10}$ and $\eta=e^{2\pi i/5}$.
All I have got to prove is $[\mathbb{Q}(\eta):\mathbb{Q}]=4$, $[\mathbb{Q}(\xi):\mathbb{Q}]=8$ and $\mathbb{Q}(\eta)\subset\mathbb{Q}(\xi)$. We could prove $\mathbb{Q}(\eta,i)=\mathbb{Q}(\xi)$ by showing $i\in\mathbb{Q}(\xi)$ and $i\notin\mathbb{Q}(\eta)$ or showing $\xi\in\mathbb{Q}(\eta,i)$, but I do not know how to do it.
 A: There is no need to use the degree of the field extensions, simply prove the two inclusions $\mathbb{Q}(\eta,i) \subset \mathbb{Q}(\xi)$ and $\mathbb{Q}(\xi) \subset \mathbb{Q}(\eta,i)$. For this, it will be useful to know that $e^{\pi i/2}  = i$.
A: Alternatively, it’s just a matter of observing that in the circle group, i.e. the group of complex numbers of absolute value $1$, the subgroup of order $4$ and the subgroup of order $5$ generate the subgroup of order $20$.
A: Similar to how @rghthndsd approached this you could note that
$$
\mathbb{Q}(\eta, i) = \mathbb{Q}(\eta \cdot i)
$$
and then prove the inclusions for this extension. Note that
$$
e^{9\pi i/10} = -\xi^{-1}
$$
A: The other answers are all usefull, but I thought I'd solve it your way using cyclotomic polynomials. Note that $\eta$ is a $5$th root of unity, so its minimum polynomial over $\mathbb{Q}$ is the $5$th cyclotomic polynomial $X^4 + \ldots + X + 1$, so $[\mathbb{Q}(\eta):\mathbb{Q}] = 4$. 
Next, $\xi$ is a $20$th unit root, so its minimum polynomial is the $20$th cyclotomic polynomial, for which it is not hard to show that it equals $\Phi_4(X^5)/\Phi_4(X)$, which has degree $(5\cdot 2) - 2 = 8$. 
You still have to show that $i\notin\mathbb{Q}(\eta)$ and that either $\mathbb{Q}(\eta,i)\subseteq\mathbb{Q}(\xi)$ or the other way around, which the other answers give hints to.
