 Solution

A primitive eighth root satisfies ${\zeta }_8=(1+i)/\sqrt{2}$, while $i={\zeta }_8^2$ and $\sqrt{2}={\zeta }_8+{\zeta }_8^{-1}$. Thus $K=\mathbb{Q}(i,\sqrt{2})$. The real elements of $\mathbb{Q}(i)$ are rational, so it does not contain $\sqrt{2}$. The tower law gives