Let \(f\) be holomorphic on \(A=\{1<|z|<2\}\) and continuous on its closure. Assume that \(f\) is nonconstant and \(|f|=5\) on both boundary circles. Show that \(f\) has at least two zeros in \(A\), counted with multiplicity.
Solution
Set \(F=f/5\).
1. The normalized map sends the annulus properly into the unit disk.
The function \(F\) is holomorphic on \(A\), continuous on \(\overline A\), and has \(|F|=1\) on both boundary circles. By the maximum modulus principle, \(|F|\le1\) on \(A\). Since \(F\) is nonconstant, it cannot attain modulus one at an interior point, so \[ |F(z)|<1\qquad(z\in A). \] Thus \(F:A\to\Delta\).
Continuity and the boundary condition imply that for every \(0<\rho<1\) there is a collar of each boundary circle on which \(|F|>\rho\). Hence the inverse image under \(F\) of the compact disk \(\{|w|\le\rho\}\) lies in a compact subannulus. It is closed there, so it is compact. Therefore \(F:A\to\Delta\) is proper. In particular its zeros form a finite set; let \(N\) be their total multiplicity.
2. Every value \(w\in\Delta\) has exactly \(N\) preimages counted with multiplicity.
Fix \(w\in\Delta\). Choose \(\rho\) with \(|w|<\rho<1\). By the boundary continuity from step 1, choose radii \(1<r_1<r_2<2\) so that every zero of \(F\) lies in \(r_1<|z|<r_2\) and \[ |F(z)|>\rho>|w| \] on both circles \(|z|=r_1,r_2\).
For \(0\le t\le1\), the function \(F-tw\) has no zero on these two circles. The argument-principle count of its zeros in the subannulus is therefore constant in \(t\): the boundary integral \[ \frac1{2\pi i}\int_{\partial A_{r_1,r_2}} \frac{F'(z)}{F(z)-tw}\,dz \] depends continuously on \(t\) and takes integer values. At \(t=0\) it equals \(N\); at \(t=1\) it counts the solutions of \(F(z)=w\), with multiplicity. No solution lies in the omitted boundary collars because there \(|F|>\rho>|w|\). Hence every \(w\in\Delta\) has exactly \(N\) preimages counted with multiplicity.
3. The degree \(N\) cannot be zero or one.
Choose any \(z_0\in A\) and put \(w_0=F(z_0)\in\Delta\). Step 2 shows that \(w_0\) has exactly \(N\) preimages counted with multiplicity, and it has at least the preimage \(z_0\). Thus \(N\ge1\).
Suppose \(N=1\). Then step 2 says every \(w\in\Delta\) has exactly one preimage, and that preimage has multiplicity one. Consequently \(F\) is bijective and has nonzero derivative everywhere. The holomorphic inverse function theorem gives a holomorphic local inverse at every point, so the set-theoretic inverse \(F^{-1}:\Delta\to A\) is holomorphic. Thus \(F\) would be a biholomorphism.
But a biholomorphism is a homeomorphism and therefore preserves simple connectedness. The unit disk is simply connected, whereas the annulus is not: the loop \(t\mapsto(3/2)e^{2\pi it}\) has winding number one about zero and cannot be contracted inside \(A\). This contradiction rules out \(N=1\). Therefore \(N\ge2\), so \(f\) has at least two zeros in the annulus, counted with multiplicity.