I want to prove that $E'=\mathbb{Q}(\sqrt2, \sqrt{1+i})$ is a normal closure of the degree-4 extension $\mathbb{Q}(\sqrt{1+i})/\mathbb{Q}$.
I'm aware that $[E':\mathbb{Q}]=8 $ but I can't go further! I would really appreciate it if someone helped me for solving this question.
First let $p(x)=x^4-2x^2+2$. Then we have that $p((1+i)^{1/2})=0$. Moreover this is the minimal polynomial. One can compute that $p(x)=\prod (x-\pm (1\pm i)^{1/2})$, so we have (since the normal closure of $F[a]$ is the extension generated by the roots of the minimal polynomial of $a$) that is $E'=\mathbb{Q}(\pm (1\pm i)^{1/2})=\mathbb{Q}((1\pm i)^{1/2})$. Now note that $(1-i)^{1/2}(1+i)^{1/2}=2^{1/2}$, so that $\mathbb{Q}((1\pm i)^{1/2})=\mathbb{Q}((1+ i)^{1/2}, 2^{1/2})$