While Studying Elliptic function I found this theorem which I post below
There exists a basis$ (\omega _1 , \omega _2)$ such that the ratio $\tau=\frac{\omega _2}{\omega _1}$ satisfies the following conditions :
- $ Im(\tau )>0$
- $\frac{1}{2} < Re (\tau) \leq \frac{1}{2}$
- $|\tau|\geq1 $
- $Re( \tau) \geq 0 $ if $|\tau|=1$
the ratio $\tau$ can be uniquely determined by these conditions, and there is a choice of two , four or six corresponding bases
And the proof goes like this,
If we select $ (\omega _1 , \omega _2)$ as $\tau=\frac{\omega _2}{\omega _1}$ , Then
$|\omega_1 ,||\omega_2 | \leq |\omega_1 + \omega_2 |$ and $|\omega_2 | \leq |\omega_1 -\omega_2 |$
In terms of $\tau $ this conditions are equivalent to $|\tau| \geq 1$ and $ Re (\tau ) \leq \frac{1}{2} $
I didn't get how comes the last statement.Any Help leading to answer will be Appreciable ...!
$$ |\omega _1|\le |\omega _2|,\quad |\omega _2|\le |\omega _1+\omega _2|, \quad|\omega _2|\le |\omega _1-\omega _2| \iff |\tau|\ge 1,\quad |\operatorname{Re }\tau|\le \frac{1}{2}$$ Proof. First $$ |\omega _1|\le |\omega _2|\iff |\tau|\ge 1$$ because $\tau=\omega_2 /\omega _1.$
Next \begin{align} &|\omega _2|\le |\omega _1+\omega _2|, \quad|\omega _2|\le |\omega _1-\omega _2|\\ \iff& |\omega_2 /\omega _1|\le |1+\omega_2 /\omega _1|,\quad |\omega_2 /\omega _1|\le |1-\omega_2 /\omega _1| \\ \iff &|\tau|\le |1+\tau|,\quad|\tau|\le |1-\tau|\\ \iff&|\tau|^2\le|1+\tau|^2,\quad|\tau|^2\le |1-\tau|^2\\ \iff&\tau\bar{\tau}\le (1+\tau)(1+\bar{\tau}),\quad \tau\bar{\tau}\le (1-\tau)(1-\bar{\tau})\\ \iff & 0\le 1+2\operatorname{Re }\tau,\quad 0\le 1-2\operatorname{Re }\tau\\ \iff& -1/2\le\operatorname{Re }\tau\le 1/2. \end{align}