Is $i$ contained in this field extension? As part of a larger problem I need to show that > $i$ is not contained in the field extension $\mathbb Q(\sqrt[3]{2},\zeta)$, where $\zeta$ is the third root of unity. I understand that the third root of unity is equal to $${-1}/{2} + i\sqrt{3}/2.$$ I'm unsure how to procure a contradiction now though. Do I have to consider the degree? How would I do that? Thanks

Both $K=\mathbb{Q}(\sqrt[3]{2},\zeta)$ and $L=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$ have degree $6$ over $\mathbb{Q}$. But the latter is obviously real, so $K\
eq L$, and in particular $\sqrt{3}\
otin K$.

Now, if $i\in K$, can we show that $\sqrt{3}\in K$ for a contradiction?