Prove $\mathbb{Q}(\sqrt2,\sqrt2i)=\mathbb{Q}(\sqrt2+\sqrt2i)$

Question

I want to know why the following two are equivalent:

$$\mathbb{Q}(\sqrt2,\sqrt2i)=\mathbb{Q}(\sqrt2+\sqrt2i)$$, where $\mathbb{Q}$ is the rational number field, and $\mathbb{Q}(\sqrt2,\sqrt2i)=a+b\sqrt2+c\sqrt2i+di,a,b,c,d\in\mathbb{Q}$.

Answer 1

An inclusion is obvious: $$\mathbb{Q}(\sqrt{2}(1+i))\subseteq \mathbb{Q}(\sqrt{2},i\sqrt{2})$$ and the reverse inclusion follows from: $$\left(\sqrt{2}(1+i)\right)^2 = 4i,\qquad \left(\sqrt{2}(1+i)\right)-i\left(\sqrt{2}(1+i)\right)=2\sqrt{2}.$$