What is the conformal map of a rectangle into a band? What is the conformal map from the rectangle to the "band" (space between two parallel lines) in the complex plane? By the Riemann mapping theorem, such a map exists.
We can define a rectangle as $\{a+bi: a \in (-1,1),b\in(-w,w)\}$ for some $w \in \mathbb R_+$, and the "band" as $\{a+bi: a \in (-1,1), b \in (-\infty,\infty)\}$.
Note in particular that the function should map the rectangles vertical axis to the bands vertical axis (i.e., it should map $\{0\} \times (-w,w)$ to $\{0\} \times (-\infty,\infty)$).
(The reason I want such a map is that I want to transform the band model of Circle Limit III into a desktop wallpaper.)
EDIT: I found this article claiming (on page 4) that you can map from a rectangle to a disk of the form
$$w = \frac{1 + i\sqrt k sn(\frac 1 \alpha (z + ib))}{1 + \sqrt k sn(\frac 1 \alpha (z + ib))}$$
where $z$ is a point the rectangle $[-a,a] \times [0,2b]$ and $w$ is a point in the unit circle ($\alpha = \frac K a$, $K$ is a quarter of the real period of $sn$, and $k$ is parameter drawn from $[0,1]$). It is trivial then to map the disk into the band.
The problem is that I looked up the $sn$ function, and it has two arguments, but in the equation above, only one argument is being given.
EDIT: $sn$ does not preserve symmetry.
 A: Here's just a sketch of an argument. A suitable elliptic function will map the interior of the rectangle conformally and bijectively to the upper half plane $H$. The principal logarithm will map $H$ conformally and bijectively to a strip $\{x+yi:0<y<\pi\}$.
OK, I am skimping on details of the elliptic function. Roughly speaking
you take a fundamental region built of four copies of the given rectangle.
Apply the Weierstrass $\wp$-function. Avoiding half-period points, and using symmetry should give you what you want. I don't have the time or
inclination to check this fully right now $\ddot\frown$.
A: Perhaps $x \mapsto x$ and $y \mapsto \tan \left( \frac{\pi y}{2w} \right) $. This should send $y$ to $\pm \infty$ as $y$ tends to $\pm w$. Is this conformal?
Here $w=f(z)$ with $f = u+iv$ has $u = x$ and $v = \tan \left( \frac{\pi y}{2w} \right)$ so $u_x=1$, $u_y=0$ and $v_x=0$ whereas $v_y = \frac{\pi}{2w}\sec^2\left( \frac{\pi y}{2w} \right)$ so apparently this map is not conformal. That said, it might still be interesting for your application.
A: I used the same formula 
w=(1+ik√sn(1α(z+ib)))/(1+k√sn(1α(z+ib)))
sn function has two arguments, the other argument is k, the elliptic modulus. It was written vaguely in that paper that you have to choose the value of k, such that K'/K has to be 2*b/a. 
K' is the elliptic integral of the first kind with parameter 1-m, 
and K is the elliptic integral of the first kind with parameter m.
m is the elliptic parameter and related to k by the equation:
m = k^2
I have used Matlab function "ellipticK" to get the value of K and K' and then found a suitable value of k.
Hope, that helps.
Thank you.
