In my Galois theory course we are currently finding fixed fields. In an exercise we defined the following isomorphisms $$\sigma(\alpha) = \alpha i,\quad \sigma (i) = i, \quad o(\alpha)=4$$ $$\tau(\alpha) = \alpha,\quad \tau (i) = -i, \quad o(\tau)=2$$
We have that the fixed field of $\langle \sigma^2 \rangle$ is $\mathbb{Q}(\alpha^2, i)$ and that the fixed field of $\langle \tau \rangle$ is $\mathbb{Q}(\alpha)$ and I understand how we found them.
Then our teacher said that the fixed field of $\langle \sigma^2, \tau \rangle$ is $\mathbb{Q}(\alpha^2, i) \cap \mathbb{Q}(\alpha) = \mathbb{\alpha^2}$. How do we find $\mathbb{Q}(\alpha^2, i) \cap \mathbb{Q}(\alpha)$?
It's hard to write this up precisely because you don't say what $\alpha$ is. Since $\sigma(\alpha) = i \alpha$ is an automorphism, it's reasonable to guess that $\alpha = \sqrt[4]{2}$ or similar. I'll be working under this assumption, but it's not hard to modify this approach for whatever your $\alpha$ happens to be.
As a completely brute-force approach, recall
$$ \mathbb{Q}(\alpha) = \{ a + b \alpha + c \alpha^2 + d \alpha^3 \} $$
$$ \mathbb{Q}(\alpha^2, i) = \{ e + b i + c \alpha^2 + d \alpha^2 i \} $$
where all of the coefficients come from $\mathbb{Q}$.
But then what does it mean for an element to be contained in both of these fields? Well it must be of the form
$$ \{ a + c \alpha^2 \} $$
So we get $\mathbb{Q}(\alpha^2)$ as the intersection. Which agrees with what your teacher said.
I hope this helps ^_^