Exhibit a conformal map from the strip \(\{ z \in \mathbb { C } : 0 < \operatorname { I m } ( z ) < 1 \}\) onto the open unit disk.
Solution
Let \[ S=\{z\in\CC:0<\operatorname{Im}z<1\}, \qquad \mathcal H=\{w\in\CC:\operatorname{Im}w>0\}. \]
1. The map\[ E:S\longrightarrow\mathcal H, \qquad E(z)=e^{\pi z}, \]is a conformal bijection.
Write \[ z=x+iy, \qquad 0<y<1. \] Then \[ E(z) = e^{\pi x}e^{i\pi y}. \] Its argument lies strictly between \(0\) and \(\pi\), so \[ E(z)\in\mathcal H. \]
Conversely, every \(w\in\mathcal H\) has a unique polar representation \[ w=re^{i\theta}, \qquad r>0, \quad 0<\theta<\pi. \] Then \[ z=\frac{\log r}{\pi}+i\frac{\theta}{\pi} \] belongs to \(S\) and satisfies \(E(z)=w\). Thus \(E\) is onto.
If \(E(z_1)=E(z_2)\), then \[ \pi(z_1-z_2)\in2\pi i\ZZ, \] so \[ z_1-z_2\in2i\ZZ. \] But two points of \(S\) have imaginary parts differing by a number strictly between \(-1\) and \(1\), so the only possible multiple of \(2i\) is \(0\). Hence \(z_1=z_2\), and \(E\) is injective.
Finally, \[ E'(z)=\pi e^{\pi z}\neq0 \] on \(S\), so \(E\) is conformal.
2. The map\[ C:\mathcal H\longrightarrow\DD, \qquad C(w)=\frac{w-i}{w+i}, \]is a conformal bijection.
For \(w=u+iv\) with \(v>0\), \[ \abs{w+i}^2-\abs{w-i}^2 = 4v > 0, \] so \[ \abs{C(w)}<1. \]
Solving \[ \zeta=\frac{w-i}{w+i} \] for \(w\) gives \[ w=i\frac{1+\zeta}{1-\zeta}, \] which maps \(\DD\) into \(\mathcal H\). Thus \(C\) is bijective. Also \[ C'(w)=\frac{2i}{(w+i)^2}\neq0 \] on \(\mathcal H\), so \(C\) is conformal.
3. A conformal bijection from the strip \(S\) onto the unit disk is\[ \boxed{ T(z) = \frac{e^{\pi z}-i}{e^{\pi z}+i} }. \]
By steps <1>1–<1>2, \[ T=C\circ E \] is a composition of conformal bijections \[ S\xrightarrow{E}\mathcal H\xrightarrow{C}\DD. \] Therefore \(T\) is a conformal bijection from \(S\) onto \(\DD\).
4. Q.E.D.
Step 3 gives the requested map.