Does there exist an analytic function mapping the annulus \[ A=\{z:1\le |z|\le4\} \] onto \[ B=\{z:1\le |z|\le2\} \] and taking the boundary circles \(C_1\) to \(C_1\) and \(C_4\) to \(C_2\), where \(C_r=\{z:|z|=r\}\)?
Solution
Let \[ \Omega\coloneqq\{z\in\CC:1<\abs{z}<4\}. \]
1. If such a function \(f\) existed, then on \(\Omega\) one would have\[ \log\abs{f(z)} = \frac12\log\abs{z}. \]
Because \(f\) maps into the annulus \(1\le\abs{w}\le2\), it has no zeros. Thus \[ u(z)\coloneqq\log\abs{f(z)} \] is harmonic on \(\Omega\). The boundary-circle assumptions give \[ u=0 \quad\text{on }C_1, \qquad u=\log2 \quad\text{on }C_4. \] The function \[ h(z) \coloneqq u(z)-\frac12\log\abs{z} \] is harmonic on \(\Omega\) and continuous on its closure. Since \(\frac12\log4=\log2\), the function \(h\) vanishes on both boundary components. The maximum principle applied to \(h\) and \(-h\) therefore gives \(h\equiv0\).
2. There is a constant \(c\in\CC\) with \(\abs{c}=1\) such that\[ f(z)^2=cz \]for every \(z\in\Omega\).
Define \[ g(z)\coloneqq\frac{f(z)^2}{z}. \] By step 1, \[ \abs{g(z)} = \frac{\abs{f(z)}^2}{\abs{z}} = 1. \] Thus the holomorphic function \(g\) has image contained in the unit circle. By the open mapping theorem, \(g\) must be constant. Writing this constant as \(c\) gives the claim, and its modulus is \(1\).
3. The identity in step 2 is impossible for an analytic \(f\) on\(\Omega\).
Since \(f\) has no zeros, differentiating \(f(z)^2=cz\) and dividing by \(f(z)^2=cz\) gives \[ 2\frac{f'(z)}{f(z)}=\frac1z. \] Let \(\gamma(t)=2e^{it}\) for \(0\le t\le2\pi\). Integrating around this circle gives \[ \frac{1}{2\pi i} \int_\gamma\frac{f'(z)}{f(z)}\,dz = \frac12 \frac{1}{2\pi i} \int_\gamma\frac{dz}{z} = \frac12. \] But the left side is the winding number of the closed curve \(f\circ\gamma\) about \(0\), hence is an integer. This contradiction proves the claim.
4. Therefore\[ \boxed{\text{no such analytic function exists}.} \]
Steps 1 and 2 are consequences of the assumed existence, while step 3 shows those consequences are impossible.
5. Q.E.D.
Step 4 answers the existence question.