Let $G$ be the Galois group of $X^4-2\in\mathbb Q[X]$. Then the splitting field is given by $\mathbb Q(\alpha, i)$ where $\alpha=\sqrt[4]{2}$ and thus $G\cong D_4$. Now, by applying the extension theorem to first $\mathbb Q\rightsquigarrow \mathbb Q(i)$ and then $\mathbb Q(i)\rightsquigarrow\mathbb Q(i,\alpha)$ I can get all $8$ elements of $G$ and I am interested in their fixed fields. I can find all but two properly. Let's say $$\tau=\begin{cases}i\mapsto -i\\\alpha\mapsto\alpha\end{cases}$$is the reflection and $$\sigma=\begin{cases}i\mapsto i\\\alpha\mapsto i\alpha\end{cases}$$is the rotation. Then fixed fields $\mathcal{F}(\langle\sigma\tau\rangle)$ and $\mathcal{F}(\langle \sigma^3\tau\rangle)$ are what give me a little trouble.
For this, we first see that $$\sigma\tau=\begin{cases}i\mapsto -i\\\alpha\mapsto i\alpha\end{cases},\ \sigma^3\tau=\begin{cases}i\mapsto -i\\\alpha\mapsto -i\alpha\end{cases}$$ and both generate a subgroup of index $4$ so we need a primitive element of order $4$. Now, simply by fooling around I can find $\mathcal{F}(\langle\sigma\tau\rangle)=\mathbb Q((1+i)/\alpha)$ and $\mathcal{F}(\langle\sigma^3\tau\rangle)=\mathbb Q((1-i)/\alpha)$ but I would like to have some sort of consistent method to obtain such elements. I know about the "trace method" where I choose a root $\beta$ and compute $\text{Tr}_H(\beta)=\sum_{h\in H}h(\beta)$ for a subgroup $H\subset G$. This, for example, yields $\text{Tr}_{\langle \sigma^2\tau\rangle}(i\alpha)=2i\alpha$ which immediately gets me $\mathcal{F}(\langle\sigma^2\tau\rangle)=\mathbb Q(i\alpha)$ but obviously no root does the trick for the groups mentioned above. So - is there a consistend method?