Let $K=\mathbb{Q}(\sqrt[8]{2},i)$ and let $F=\mathbb{Q}(i)$. I need to show that $\operatorname{Gal}(K/F)\cong Z_8$ where $Z_8$ is the cyclic group of order 8.
I have already shown that $[K:F]=8$, i.e. the dimension of $K$ over $F$ is 8. I have also already shown that $K/F$ is Galois so that $|\operatorname{Gal}(K/F)|=[K:F]=8$.
I am not sure how to go about picking the automorphisms I want. I think I want to pick all the automorphisms that fix $\mathbb{Q}(i)$. Thus I want to choose $\sigma$ so that $$\sigma(\sqrt[8]{2})=\zeta_8\sqrt[8]{2},$$ where $\zeta_8$ is a primitive $8^{\text{th}}$ root of unity. Thus $\langle \sigma\rangle=\operatorname{Gal}(K/F)\,\,$ is cyclic of order 8 and we conclude that $\operatorname{Gal}(K/F)\cong Z_8$.
Am I on the right track?