I am trying to show that the Galois group is isomorphic to $D_4$. I was wondering if there is a better way than brute forcing it.
What I did: $Q(i,\sqrt[^4]{11})$ Is the splitting field. It has degree $8$ so we know that we can send each root to all other roots of minimal polynomial. I found $s$ sends $i \to -i$ and $\sqrt[^4]{11} \to -\sqrt[^4]{11}$ while $r$ fixes $i$ and $\sqrt[^4]{11} \to i\sqrt[^4]{11}$. I have shown that $sr=r^{-1}s$ and that all $1,r,r^2,r^3,s,sr..$ are distinct ending the proof by appealing to representation. That took a while, is there a more efficient away?
Let $k=\mathbf Q(i), K=\mathbf Q(\sqrt [4] {11}),L=k.K$ your splitting field of degree $8$. To show that $G:=Gal(L/\mathbf Q)$ is $D_8$ (I prefer this notation for the dihedral group of order $8$), no need to explicitly compute $G$, only recall the description of $D_8$ as an extension of groups. Viewing $L$ as $L/K/\mathbf Q$ (resp. $L/k/\mathbf Q)$, you see that $G$ admits a normal subgroup $H=Gal(L/K)$ (resp. a quotient group $Q=Gal(k/\mathbf Q))$ which are both of order $2$, and moreover $K\cap k =\mathbf Q$ because $K$ is totally real and $k$ is totally imaginary. It follows (just draw the Galois diagram) that $H$ is a lift of $Q$ to $G$, more precisely, the exact sequence of groups $1\to H\to G\to Q\to 1$ splits, i.e. $G$ is a semi-direct product of $H$ and $Q$. It cannot be a direct product because it is not abelian (since $K/\mathbf Q$ is not normal), hence $G\cong D_8$.