conformally equivalent flat tori

Solution 1:

In general the answer is no, even in a slightly more general case. The tori you describe can be represented by $\mathbb{C} / \Gamma_\tau$, where $\Gamma_\tau$ is the lattice generated by $1$ and $\tau$ with $\Im \tau > 0$. (Rectangles correspond to $\tau$ being purely imaginary.) If you have a conformal map between two such tori, you can lift it to their universal covers, and so you get an affine map $T(z)=az+b$ in the plane. Composing with a translation (which projects to a conformal self-map of any such torus), you can assume that $T(0)=0$, so $T(z) = az$. This map has to map the lattice $\Gamma_{\tau_1}$ to $\Gamma_{\tau_2}$, so the question is which of these lattices are equivalent under multiplication with some complex scalar $a$. Obviously $\tau \Gamma_{-1/\tau} = \Gamma_\tau$, and $\Gamma_\tau = \Gamma_{\tau+1}$. The maps $\tau \mapsto -1/\tau$ and $\tau\mapsto \tau+1$ generate $PSL(2,\mathbb{Z}) = \left\{ \tau \mapsto \frac{a\tau+b}{c\tau+d}: a,b,c,d \in \mathbb{Z}, \, ad-bc=1 \right\}$, and it can be shown that $\tau_1$ and $\tau_2$ generate equivalent lattices iff they are in the same orbit of $PSL(2,\mathbb{Z})$ acting on the upper halfplane $\Im \tau > 0$. For the special case of rectangles, we get that $\tau_1 = i t_1$ and $\tau_2 = i t_2$ generate equivalent lattices iff $t_1=t_2$ or $t_1 = 1/t_2$, i.e., iff the rectangles are similar.