Problem: Let $\alpha=\sqrt{\frac{3+i\sqrt{7}}{2}}$ and $K=\mathbb{Q}(\alpha)$. Find the fixed field $M=\{x\in \mathbb{Q}(\sqrt{\frac{3+i\sqrt{7}}{2}})|\sigma(x)=x\}$, where $\sigma$ is the automorphism on $\mathbb{Q}(\sqrt{\frac{3+i\sqrt{7}}{2}})$ such that $\sigma(\alpha)=\frac{2}{\alpha}$.
My Attempt: Note $\alpha$ and $\frac{\alpha}{2}$ are roots of $f(t)=t^4-3t^2+4$ and $K$ is the splitting field of $f(t)$ over $\mathbb{Q}$. By the Fundamental Theorem of Galois Theory, it is easy to see that such a $\sigma$ in the Galois group of $K: \mathbb{Q}$ indeed exists.
Now, every $x\in K$ has the form $x=p+q\alpha+r\alpha^2+s\alpha^3$, where $p, q, r, s\in\mathbb{Q}$, because $K: \mathbb{Q}$ is a simple algebraic extension. Thus $\sigma(x)=x\implies p+q\alpha+r\alpha^2+s\alpha^3=\sigma(p)+q\sigma(\alpha)+r\sigma^2(\alpha)+s\sigma^3(\alpha)=p+\frac{2q}{\alpha}+\frac{4r}{\alpha^2}+\frac{8s}{\alpha^3}$.
My Question: I failed to transform $p+\frac{2q}{\alpha}+\frac{4r}{\alpha^2}+\frac{8s}{\alpha^3}$ into the form of $p+q\alpha+r\alpha^2+s\alpha^3$. Any pointer would be greatly appreciated.
Multiply by $\;a^3\;$ in the identity you got and use $\;a^4=3a^2-4\;$:
$$p+qa+ra^2+sa^3=p+\frac{2q}a+\frac{4r}{a^2}+\frac{8s}{a^3}\stackrel{\cdot a^3}\implies$$
$$qa^4+ra^5+sa^6=2qa^2+4ra+8s\implies3qa^2-4q+3ra^3-4ra+s(3a^4-4a^2)=2qa^2+4ra+8s$$
$$\implies3qa^2-4q+3ra^3-4ra+s(5a^2-12)=2qa^2+4ra+8s\implies$$
$$\implies3ra^3+(q+5s)a^2-8ra-4(q+5s)=0$$
The last equality is possible iff $\;r=0\,,\,\,q=-5s\;$ , so your element is of the form
$$x=p-5sa+sa^3$$
Try now to take it from here...