Prove the open mapping theorem for holomorphic functions: If \(f\) is a non-constant holomorphic function on an open set \(U\) in \(\mathbb{C}\), then \(f(U)\) is also an open set.
Solution
Idea:
Let \(f: U\to \CC\).Using the argument principle. Pick \(w_0\in W\) with \(f(z_0) = w_0\) for some \(z_0\in U\); we want to show that \(w_0\) is an interior point of \(f(U)\), so we’re looking for a disc containing \(w_0\) and contained in \(f(U)\).
Write \[ g_0(z) \da f(z) - w_0 ,\] so \(g_0\) is holomorphic and has a zero at \(z_0\). Since zeros of holomorphic functions are isolated, there is some \(U' \da \DD_r(z_0)\) where \(g_0\) is nonvanishing. The claim is that if we choose \(\eps\) small enough, we can arrange so that \(W_\eps \da \DD_\eps(w_0) \subseteq f(U)\). This will follow if for every \(w\in W_\eps\), the equation \(f(z) = w\) has a solution in \(U\), i.e. Define a function that counts the number of zeros: \[ F(w) &\da {1\over 2\pi i}\int_{\bd U' } {f(z) \over f(z) - w_1 }\dz\\ &= {1\over 2\pi i}\int_{\bd U' } {\dd{}{z}\qty{f(z) - w} \over f(z) - w }\dz\\ &= \size Z(f(z) - w, U' ) ,\] which is the number of zeros of \(f(z) - w\) in \(U'\) by the argument principle. Now \(F\) is a \(\ZZ\dash\)valued function, and the only obstruction to continuity is if \(f(z) - w = 0\) in the integrand for some \(z\). The claim is that \(\eps\) can be chosen such \[ z\in \bd U' \implies \abs{f(z) - w} > 0 \qquad \forall w\in W_\eps .\] The theorem then follows: \(F(w): U' \to W_\eps\) is a continuous and \(\ZZ\dash\)valued function on the connected set \(W_\eps\) (a disc), and a continuous integer-valued function on a connected set is constant (its image is a connected subset of the discrete space \(\ZZ\), hence a single point). Then noting that \(F(w_0) = 1\) since \(z_0\in U'\) and \(w_0\in W_\eps\), we have \(F\equiv 1 > 0\) for all \(w\).
(of claim)
Choose \[ \eps \da \min_{z\in \bd U'}\abs{f(z) - w_0} .\] Now if \(\abs{w-w_0} < \eps\) and \(\abs{z-z_0} = r\), we have \(\abs{f(z) - w} > \eps > 0\).