Let \[ S=\{z=x+iy:|x|<1,\ y>0\}. \] Give an explicit conformal map from \(S\) to the upper half-plane \(\mathbb H\).
Solution
Take \[ \boxed{F(z)=\sin\left(\frac{\pi z}{2}\right)}. \]
1. The map \(F\) sends \(S\) into \(\mathbb H\).
Write \(z=x+iy\) with \(|x|<1\) and \(y>0\). Using \[ \sin(a+ib)=\sin a\cosh b+i\cos a\sinh b, \] we obtain \[ \operatorname{Im}F(z) =\cos\left(\frac{\pi x}{2}\right) \sinh\left(\frac{\pi y}{2}\right). \] Because \(|x|<1\), \[ \cos\left(\frac{\pi x}{2}\right)>0, \] and because \(y>0\) the hyperbolic sine factor is also positive. Hence \(\operatorname{Im}F(z)>0\).
2. The derivative of \(F\) never vanishes on \(S\).
We have \[ F'(z)=\frac{\pi}{2}\cos\left(\frac{\pi z}{2}\right). \] The zeros of \(\cos(\pi z/2)\) are the odd integers. They lie on the real axis, and none lies in \(S\). Thus \(F'(z)\neq0\) throughout \(S\).
3. The map \(F\) is injective on \(S\).
Suppose \(F(z_1)=F(z_2)\) and put \[ u_j=\frac{\pi z_j}{2}. \] Then \[ \sin u_1=\sin u_2. \] Hence either \[ u_1-u_2=2\pi k \] or \[ u_1+u_2=(2k+1)\pi \] for some \(k\in\mathbb Z\). Since \[ -\frac\pi2<\operatorname{Re}u_j<\frac\pi2, \] the first equation forces \(k=0\), hence \(u_1=u_2\). The second equation is impossible because \(\operatorname{Im}u_1,\operatorname{Im}u_2>0\), whereas its right-hand side is real. Therefore \(z_1=z_2\).
4. The map \(F:S\to\mathbb H\) is proper.
For \(z=x+iy\), \[ |F(z)|^2 =\sin^2\left(\frac{\pi x}{2}\right) +\sinh^2\left(\frac{\pi y}{2}\right). \] Thus \(|F(z)|\to\infty\) as \(y\to\infty\).
Now let \(K\subset\mathbb H\) be compact. Then there are constants \(M<\infty\) and \(\delta>0\) such that \[ |w|\le M, \qquad \operatorname{Im}w\ge\delta \] for all \(w\in K\). The preceding formula bounds \(y\) on \(F^{-1}(K)\). Also, if a sequence in \(F^{-1}(K)\) approached either the bottom edge \(y=0\) or one of the vertical edges \(x=\pm1\), then the formula in 1 would give \[ \operatorname{Im}F(z)\longrightarrow0, \] contradicting the lower bound \(\delta\). Hence \(F^{-1}(K)\) stays a positive distance from every boundary component of \(S\) and is bounded. Being closed, it is compact.
5. The map \(F\) is a conformal homeomorphism from \(S\) onto\(\mathbb H\).
By 4, \(F\) is a nonconstant proper holomorphic map between connected Riemann surfaces, so it is surjective. By 3 it is injective, and by 2 its derivative never vanishes. Therefore \(F\) is an analytic homeomorphism with analytic inverse, hence a conformal map from \(S\) to \(\mathbb H\).