Let \(A=\{z\in\mathbb C:1<|z|<2\}\) and let \(f\) be holomorphic on a neighborhood of \(\overline A\), with \(f|_A\) nonconstant and \(|f|=1\) on \(\partial A\). Prove that \(f\) has at least two zeros in \(A\), counted with multiplicity.
Solution
1. The function maps \(A\) into the unit disk and has a positive finite zero count.
The maximum modulus principle on the compact closed annulus gives \(|f|\leq1\), with strict inequality in \(A\) because \(f|_A\) is nonconstant [@SS03]. If \(f\) had no zero in \(A\), it would have none on \(\overline A\), and the same principle applied to \(1/f\) would give \(|f|\geq1\) in \(A\), a contradiction. Its number \(N\) of zeros in \(A\), counted with multiplicity, is finite and positive: zeros cannot accumulate in the compact subset \(\overline A\) of the holomorphy domain unless \(f\) vanishes identically on its component, by the identity theorem [@SS03].
2. Every point of the unit disk has exactly \(N\) preimages in \(A\), with multiplicity.
Give \(\partial A\) the positive orientation, outer circle counterclockwise and inner circle clockwise. For a fixed \(w\) with \(|w|<1\) and \(0\leq t\leq1\), one has \(|f-tw|\geq1-|w|>0\) on this boundary. The argument principle therefore gives \[ N_w(t)=\frac1{2\pi i}\int_{\partial A} \frac{f'(z)}{f(z)-tw}\,dz \] as the finite integer counting the zeros of \(f-tw\) in \(A\) [@SS03]. The positive denominator bound makes this integral continuous in \(t\), hence constant. Thus \(N_w(1)=N_w(0)=N\).
3. A zero count of one is impossible.
Suppose \(N=1\). Step 2 makes \(f:A\to D\) bijective, where \(D\) is the unit disk, and every fiber point is simple. Hence \(f'\) never vanishes in \(A\), and its inverse \(g:D\to A\) is holomorphic by the inverse function theorem [@SS03]. As \(g\) has no zero on the simply connected disk, it has a holomorphic logarithm \(L\) there. Explicitly, a primitive of \(g'/g\), with its constant chosen at zero, has exponential \(g\).
The holomorphic function \(K=L\circ f\) on \(A\) then satisfies \(e^{K(z)}=g(f(z))=z\). Differentiating gives \(K'(z)=1/z\). But the integral of a derivative around the circle \(|z|=3/2\) is zero, whereas direct parametrization gives \[ \int_{|z|=3/2}\frac{dz}{z}=2\pi i. \] This contradiction excludes \(N=1\). Together with \(N\geq1\), it proves \(N\geq2\).