Since the four roots of $X^4-3X^2+4$ are equal to $\pm(\frac{\sqrt7\pm i}{2})$, we find $\mathbb Q(\alpha,i)=\mathbb Q(\sqrt7,i)$ and the degree of the extension is $4$. As the Galois group has two elements of order $2$ (one sending $i$ to $-i$ and another sending $\sqrt7$ to $-\sqrt7$), it must be $\mathbb Z/(2)\times\mathbb Z/(2)$.
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