Find a conformal map from the strip \(\theset{z\in \CC \suchthat 0 < \Im(z) < 1}\) to \(\HH\).
Solution
Goal: Find a conformal (biholomorphic) map from the strip \(S = \theset{z \suchthat 0 < \Im z < 1}\) onto the upper half-plane \(\HH\).
1. \(e^{\pi z}\) maps \(S\) onto \(\HH\).
For \(z = x + iy \in S\), \(e^{\pi z} = e^{\pi x} e^{i\pi y}\); since \(y \in (0,1)\), the argument \(\pi y\) lies in \((0, \pi)\) and the modulus \(e^{\pi x}\) is any positive number, so \(e^{\pi z}\) ranges over all of \(\HH\) (indeed over all nonzero complex numbers with argument in \((0,\pi)\)).
2. \(e^{\pi z}\) is injective on \(S\).
If \(e^{\pi z_1} = e^{\pi z_2}\), then \(\pi(z_1 - z_2) \in 2\pi i \ZZ\), i.e. \(z_1 - z_2 \in 2i\ZZ\); but for \(z_1, z_2 \in S\), \(\Im(z_1 - z_2) \in (-1, 1)\), so the only integer multiple of \(2i\) in that range is \(0\), hence \(z_1 = z_2\).
3. \(e^{\pi z}\) is holomorphic with nowhere-vanishing derivative on \(S\).
The exponential is entire and \(\ddd{z} e^{\pi z} = \pi e^{\pi z} \neq 0\) for all \(z\).
4. \(T(z) = e^{\pi z}\) is a conformal map from \(S\) onto \(\HH\).
By 1 it is onto, by 2 it is injective, and by 3 it is holomorphic with nonzero derivative; a bijective holomorphic map with nonzero derivative is a biholomorphism, i.e. conformal.
5. Q.E.D.
4 gives the conformal map \(z \mapsto e^{\pi z}\) from the strip to the upper half-plane.