Suppose analytic functions \(f\) and \(g\) on a region \(U\) satisfy \[ |f(z)|=|g(z)| \qquad(z\in U). \] Prove that \(f(z)=a g(z)\) for some constant \(a\) with \(|a|=1\).
Solution
If \(g\equiv0\), then \(|f|=|g|\) gives \(f\equiv0\), and the conclusion holds for any constant \(a\) with \(|a|=1\). Assume henceforth that \(g\) is not identically zero.
1. The zeros of \(f\) and \(g\) coincide.
For every \(z\in U\), \[ |f(z)|=|g(z)|. \] Thus \(f(z)=0\) if and only if \(g(z)=0\).
2. At every common zero, \(f\) and \(g\) have the same order of vanishing.
Let \(z_0\) be a common zero. Since neither function is identically zero on the region \(U\), write locally \[ f(z)=(z-z_0)^m F(z), \qquad g(z)=(z-z_0)^n G(z), \] where \(F(z_0)\neq0\) and \(G(z_0)\neq0\). On a punctured neighborhood of \(z_0\), the equality of moduli gives \[ |z-z_0|^{m-n}\left|\frac{F(z)}{G(z)}\right|=1. \] The second factor tends to the finite nonzero number \(|F(z_0)/G(z_0)|\). If \(m>n\), the left side tends to \(0\); if \(m<n\), it tends to \(\infty\). Both are impossible. Hence \(m=n\).
3. The quotient \(h=f/g\), initially defined away from the zeros of \(g\),extends to an analytic function on all of \(U\).
Away from the zeros of \(g\), the quotient is analytic. At a common zero \(z_0\), 2 gives local factorizations with the same power: \[ f(z)=(z-z_0)^mF(z), \qquad g(z)=(z-z_0)^mG(z). \] Thus on the punctured neighborhood \[ h(z)=\frac{F(z)}{G(z)}, \] and the right side is analytic at \(z_0\) because \(G(z_0)\neq0\). It therefore provides the removable extension of \(h\).
4. The extended function \(h\) satisfies \(|h(z)|=1\) for every \(z\in U\).
Away from the common zeros, \[ |h(z)|=\frac{|f(z)|}{|g(z)|}=1. \] The extension is continuous, so the same equality holds at the removable points as well.
5. The function \(h\) is constant.
If \(h\) were nonconstant, the open mapping theorem would imply that \(h(U)\) is open in \(\mathbb C\). But 4 gives \[ h(U)\subset S^1, \] and the unit circle contains no nonempty open subset of \(\mathbb C\). Therefore \(h\) is constant, say \(h\equiv a\). Since \(|h|=1\), one has \(|a|=1\).
6. Hence \(f=ag\) on \(U\) for a constant \(a\) with \(|a|=1\).
By definition of the extended quotient, \(f=hg\) everywhere on \(U\). Apply 5.