How to prove $\mathbb Q(i)=\mathbb Q(\frac{2i+1}{i-1})$? I know the elements in these two groups, but how to prove that $\mathbb Q(i)=\mathbb Q(\frac{2i+1}{i-1})$ ?
Can the second thing be replaced by other complex numbers?
 A: Notice that $\frac{2i+1}{i-1} = \frac12 - \frac32 i$.
Hence $$\frac{2i+1}{i-1} = \underbrace{\frac12}_{\in\mathbb{Q}} - \underbrace{\frac32 i}_{\in\mathbb{Q}(i)} \in \mathbb{Q}(i)$$
so $\mathbb{Q}\left(\frac{2i+1}{i-1}\right) \subseteq \mathbb{Q}(i)$.
Conversely, we have $$i = \underbrace{\frac13}_{\in\mathbb{Q}} - \underbrace{\frac23\frac{2i+1}{i-1}}_{\in\mathbb{Q}\left(\frac{2i+1}{i-1}\right)} \in \mathbb{Q}\left(\frac{2i+1}{i-1}\right)$$
so $\mathbb{Q}(i) \subseteq \mathbb{Q}\left(\frac{2i+1}{i-1}\right)$.
A: Note that $\mathbb{Q}(\frac{2i+1}{i-1})=\mathbb{Q}(2+\frac{3}{i-1})=\mathbb{Q}(\frac{i+2}{-2})=\mathbb{Q}(-\frac{i}{2})=\mathbb{Q}(\frac i 2)$
Can you continue from here?
A: I'll suppose your first question is already answered. Now take a complex number $x+iy, x,y \neq 0$ (if $x=0, y\neq 0$ it's trivial and if $y =0$ it's impossible) and suppose $\Bbb Q(i) = \Bbb Q(x+iy)$. So $(a+bx) + i\cdot by = i$ for some choice of $a, b$. Then $b = -a/x$ and then $-ay/x = 1$ for some choice of $a$. Then $a = -x/y \in \Bbb Q$ is our choice and with $\Bbb Q(i) \supseteq Q(x+iy)$ trivially, we do also get by closure of sum that $\Bbb Q(x+iy) \supseteq Q(i)$ so they're equal. 
So this occurs iff there's a way to write $y = 1/b$ and $x = a/b$, i.e., $(x,y) \in \Bbb Q \times \Bbb Q_*$. 
