Show that the punctured unit disk \(\{z:0<|z|<1\}\) and the annulus \(\{z:1<|z|<2\}\) cannot be conformally equivalent.
Solution
Write \(D=\{z:|z|<1\}\) and \(A=\{w:1<|w|<2\}\). Suppose there were a biholomorphism \(f:D\setminus\{0\}\to A\) with holomorphic inverse \(g:A\to D\setminus\{0\}\).
1. The map \(f\) extends holomorphically to \(D\), with its value at zero in \(A\).
The bound \(|f|<2\) makes the singularity at zero removable [@SS03]. Denote the extension by \(F\). Continuity gives \(1\leq|F(0)|\leq2\). The extension is nonconstant since its restriction \(f\) is bijective onto \(A\).
If \(|F(0)|=2\), the holomorphic function \(F\) attains its maximum modulus at an interior point, so it is constant by the maximum modulus principle, a contradiction. If \(|F(0)|=1\), then \(F\) is nonzero on \(D\), and \(1/F\) is holomorphic with modulus at most one and with modulus one at zero. The same principle makes \(1/F\), and hence \(F\), constant, again a contradiction [@SS03]. Thus \(1<|F(0)|<2\), so \(F(0)\in A\).
2. The inverse map gives a contradiction at the added point.
For every \(z\in D\setminus\{0\}\), one has \(g(F(z))=g(f(z))=z\). Since \(F(0)\in A\), the map \(g\) is continuous at \(F(0)\). Letting \(z\to0\) therefore gives \[ g(F(0))=0. \] But \(g\) takes all its values in \(D\setminus\{0\}\). This contradiction rules out the biholomorphism.