Let \(T\subset\mathbb C\) be a square. Find a formula, which may involve an integral, for a holomorphic covering map \[ f:\mathbb H\longrightarrow\mathbb C\setminus T. \]
Solution
Let \[ E=\{w:|w|>1\}. \] Choose on \(E\) the branch \[ q(w)=\sqrt{1-w^{-4}} \] with \(q(w)\to1\) as \(w\to\infty\), and define \[ \Phi(w) =w+\int_\infty^w\bigl(q(\zeta)-1\bigr)\,d\zeta. \] Then \(\Phi\) maps \(E\) conformally onto the complement of a square \(S\). If \[ A(z)=az+b, \qquad a\ne0, \] is an orientation-preserving similarity carrying \(S\) onto the prescribed square \(T\), then \[ \boxed{ f(z)=A\!\left(\Phi(e^{-iz})\right) } \] is the required covering map \(\mathbb H\to\mathbb C\setminus T\).
1. The formula for \(\Phi\) is single-valued and satisfies\[ \Phi'(w)=\sqrt{1-w^{-4}}, \qquad \Phi(w)=w+O(w^{-3}) \]at infinity.
The chosen square root is holomorphic on \(E\): in the coordinate \(u=1/w\), it is the branch of \[ \sqrt{1-u^4} \] equal to \(1\) at \(u=0\). Moreover \[ q(w)-1=O(w^{-4}). \] Its Laurent expansion contains only powers \(w^{-4k}\), so in particular its residue is zero. Hence its integral around the generator of \(\pi_1(E)\) vanishes, and the integral defining \(\Phi\) is path-independent. Differentiation gives the displayed formula for \(\Phi'\), and integration of \(O(w^{-4})\) gives the stated asymptotic expansion.
2. The unit circle is mapped by \(\Phi\) to four straight segments meeting at right angles.
On the first quadrant of the unit circle, write \[ w=e^{i\theta}, \qquad 0<\theta<\frac\pi2. \] Then \[ 1-w^{-4} =1-e^{-4i\theta} =2i e^{-2i\theta}\sin(2\theta). \] For the branch fixed in 1, \[ q(e^{i\theta}) =\sqrt{2\sin(2\theta)}\,e^{i(\pi/4-\theta)}. \] Since \(dw=i e^{i\theta}\,d\theta\), \[ d\Phi =q(w)\,dw =\sqrt{2\sin(2\theta)}\,e^{3\pi i/4}\,d\theta. \] Thus the image of this quarter-circle is a straight segment.
Also \[ q(iw)=q(w). \] The functions \(\Phi(iw)\) and \(i\Phi(w)\) have the same derivative with respect to \(w\), and their asymptotics at infinity agree; hence \[ \Phi(iw)=i\Phi(w). \] Therefore the four quarter-arcs map to four congruent straight segments obtained successively by rotation through \(\pi/2\). They form a square, denoted \(S\).
3. The map\[ \Phi:E\longrightarrow\mathbb C\setminus S \]is a conformal homeomorphism.
1. The Schwarz–Christoffel theorem therefore identifies \(\Phi\) as the conformal map from the exterior disk onto the exterior of the square found in
2.
4. The map\[ p(z)=e^{-iz} \]is a holomorphic covering map from \(\mathbb H\) onto \(E\).
If \(z=x+iy\in\mathbb H\), then \[ |e^{-iz}|=e^y>1, \] so \(p(\mathbb H)\subset E\). Conversely, if \(w\in E\), choose a logarithm \[ \log w=\log|w|+i\arg w. \] Then \[ z=i\log w \] has imaginary part \(\log|w|>0\) and satisfies \(e^{-iz}=w\). The derivative \[ p'(z)=-ie^{-iz} \] never vanishes, and the fibers differ by integer multiples of \(2\pi\). Thus \(p\) is the standard universal covering of the exterior disk.
5. The displayed formula is a covering of the required square complement.
By 3, \(\Phi\) is a conformal homeomorphism from \(E\) onto \(\mathbb C\setminus S\), and the similarity \(A\) is a conformal homeomorphism \[ A:\mathbb C\setminus S\longrightarrow\mathbb C\setminus T. \] Composing these homeomorphisms with the covering \(p\) from 4 gives \[ f=A\circ\Phi\circ p:\mathbb H\longrightarrow\mathbb C\setminus T, \] which is again a holomorphic covering map.