Primitive element for $F=\Bbb Q(i,\sqrt{11},\sqrt{3})$ over $\Bbb Q$

Question

The element $k$ is called a primitive element of $K$ over $F$.

Find $k$ $\in$ $F$ such that $F=\mathbb{Q}(k)$.

$F=\mathbb{Q}(I,\sqrt{11},\sqrt{3})$

Will $k=8$?

Answer 1

Hint: Look at $\mathbb{Q}(\sqrt{3}+\sqrt{11})$