Let \(f\) and \(g\) be non-zero analytic functions on a region \(\Omega\). Assume \(|f(z)| = |g(z)|\) for all \(z\) in \(\Omega\). Show that \(f(z) = e^{i \theta} g(z)\) in \(\Omega\) for some \(0 \leq \theta < 2 \pi\).
Solution
Goal: Prove that if \(f, g\) are nonzero analytic functions on a region \(\Omega\) with \(\abs{f(z)} = \abs{g(z)}\) for all \(z \in \Omega\), then \(f(z) = e^{i\theta} g(z)\) for some \(0 \leq \theta < 2\pi\).
1. Define \(h := f / g\); then \(h\) is analytic and never vanishes on \(\Omega\).
\(g \neq 0\) on \(\Omega\) by hypothesis, and quotients of analytic functions are analytic.
2. \(\abs{h(z)} = 1\) for all \(z \in \Omega\).
\(\abs h = \abs f / \abs g = 1\) by hypothesis.
3. \(h\) is constant.
3.1. \(h(\Omega)\) is contained in the unit circle \(S^1\).
2.
2.2. \(h(\Omega)\) is open in \(\CC\) if \(h\) is nonconstant.
Open mapping theorem: a nonconstant holomorphic map on a region is open.
2.3. \(S^1\) has empty interior in \(\CC\).
The unit circle contains no open disk.
2.4. Hence \(h\) is constant.
<2>1–<2>3: a nonconstant \(h\) would map \(\Omega\) onto an open set inside \(S^1\), impossible.
4. \(h \equiv e^{i\theta}\) for some \(0 \leq \theta < 2\pi\).
3 and \(\abs h \equiv 1\) (<1>2): the constant value lies on the unit circle.
5. Q.E.D.
1 and 4 give \(f = hg = e^{i\theta} g\) on \(\Omega\).
Solution
Define \(F(z) \da {f(z) \over g(z)}\).
\(F\) is holomorphic on \(\Omega\).
(of claim)
Note that \(g(a) = 0\) iff \(f(a) = 0\), so \(F\) has no poles. If \(F\) has a singularity at \(z_0\), noting that \(\abs{F(z_0)} = 1\), \(F\) is bounded in a neighborhood of \(z_0\) and thus the singularity must be removable. By Riemann’s removable singularity theorem, \(F\) extends to a holomorphic function.
Given this, note that \(\abs{F(z)} = 1\) for all \(z\), so \(F(\Omega) \subseteq S^1\), which is codimension 1 in \(\CC\) and not open. By the open mapping theorem, \(F\) must be constant, so \(F(z) = \lambda\), and in particular since \(\abs{F(z)} = 1\), \(\lambda = e^{it}\in S^1\) for some \(t\). Then \(f(z) = \lambda g(z)\).