How to show $ \Bbb Q（ζ_n,i）＝ \Bbb Q（ζ_N）$, where $N＝LCM（n,4）$?
And what is the intuitive understanding of this equal ? I Guess $i$ has period 4 has important meaning.
I tried a problem ' Is $\Bbb Q（sin2π/n）/\Bbb Q $ Galois ? And I reached this question .Thank you for helping me!
In general, $\mathbb Q(\zeta_{n},\zeta_m) = \mathbb Q(\zeta_{lcm(n,m)})$. There are many ways to see this depending on what you know. One inclusion is clear since $\zeta_{lcm(n,m)}^{lcm(n,m)/n} = \zeta_n$ and similarly for $m$.
For the other direction, note that $\zeta_n\zeta_m = e^{2\pi i(1/n+1/m)}$ is a root of unity of order $lcm(n,m)$.
As for the original problem you started with (and you probably know this!), you can observe that since $\mathbb Q(\zeta_n,i)/\mathbb Q$ is Galois, any subextension has to be Galois (by Galois theory, this corresponds to the fact that all subgroups of abelian groups are normal!).