Let $\theta:=\frac{\sqrt{2}+\sqrt{3}+i\sqrt{3}}{\sqrt{5}+\sqrt{7}+i\sqrt{7}}\in \mathbb{C}$. Show that $(\mathbb{Q}(\theta)/\mathbb{Q})$ is a divisor of 32.
Studying for my upcoming algebra exam, I stumbled upon this exercise from an old algebra exam.
My idea:
$\mathbb{Q}(\theta)$ from $\mathbb{Q}$ can be constructed out of a series of quadratic field extensions. But I'm not sure how to do that properly so maybe someone could give me a hint? Thanks in advance!
Edit:
SoI think the the quadratic field extensions should look something like this
$\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{5}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{5}, \sqrt{7},i)$
Is this better, since the $i$ only appears once?
Multiply numerator and denominator by $(\sqrt{5}+\sqrt{7}-i\sqrt{7})$ to get a real denominator. Then multiply numerator and denominator by a suitable number (which one ?) to get a rational denominator.
Then, simplifying numerator if necessary, you should get that you only need to add five square roots: $\sqrt{2},\sqrt{14},\sqrt{15},\sqrt{21}$ and $i$. Since $2^5=32\cdots$