We will prove two theorems using the argument principle. We begin by showing a corollary of it, as well as a useful proposition. The corollary says that the argument principle can be generalized to count the number of points that \(f\) maps to a specific complex number \(q\). In the original version, \(q\) was zero.
The corollary is immediately obtained by applying the argument principle on \(f(z)-q\), and observing that \[W((f-q) \circ \ell, 0)=W(f \circ \ell-q, 0)= W(f \circ \ell, q)\]
From now on, whenever we say "argument principle", we mean its general version, given in Corollary 1.
Let \(z(s):= (1-s)z_0+sz_1\), for \(s\in[0,1]\). This is a parametrization of \([z_0,z_1]\). From the definition of winding number, we have
Consider an open set \( B \subseteq A \). We will show that for any \( z_0 \in B \), the point \( f(z_0) \) lies in the interior of \( f(B) \). This immediately implies that \( f(B) \) is open.
Fix a \( z_0 \in B \). Without loss of generality, we assume that \( f(z_0) = 0 \). As in the proof of the argument principle, since \( f \) is non-constant, the point \( z_0 \) is an isolated root. Take \( \varepsilon > 0 \), so that the closed disc \( \overline{D}_{z_0,\varepsilon} \) , centered at \( z_0 \) and of radius \( \varepsilon \), lies in \( B \) and does not contain any other root of \( f \). Its positively oriented boundary is the circle \( C_{z_0, \varepsilon}(t) = \varepsilon e^{i2\pi t} + z_0 \), \( \ t \in [0,1] \). Arguing exactly as in the proof of the local version of the argument principle, we get that on the image space there exists a small open disc \( D_{0,\delta} \) that does not intersect the loop \( f \circ C_{z_0,\varepsilon} \). Thus, Proposition 1 implies that for all \( q \in D_{0,\delta} \ \),
By the argument principle, for any \( q \in D_{0,\delta} \), there is a \( z \in D_{z_0,\varepsilon} \) such that \( f(z) = q \). Thus, \( D_{0,\delta} \subseteq f(B) \). \(\ \square\)
We remind the reader the definition of a biholomorphism. Let \(A, B\subseteq \mathbb{C}\) open. If a holomorphism \(f:A\rightarrow B\) is a bijection, then we say that \(f\) is a biholomorphism.
Suppose there is a \( z_0 \in A \) such that \( f'(z_0) = 0 \). Since \( f \) is injective, the roots of \( f' \) are isolated. Let \( \varepsilon > 0 \), such that the closed disc \( \overline{D}_{z_0,\varepsilon} \) is inside \( A \), and also does not contain any other root of \( f' \). Again because of injectivity, the loop \( f \circ C_{z_0,\varepsilon} \) does not hit \( f(z_0) \). Combining this with the fact that \( f'(z_0) = 0 \), by the argument principle:
\[ W\left( f \circ C_{z_0,\varepsilon} \ , f(z_0) \right) \ge 2 \]
Now, arguing exactly as in the proof of the local version of the argument principle, we get that if we choose a \( z_1 \in D_{z_0,\varepsilon} \) that is different from \( z_0 \) but close enough to it, then the segment \([f(z_0), f(z_1)]\) does not intersect the loop \( f \circ C_{z_0,\varepsilon} \). Thus, from Proposition 1:
\[ W\left( f \circ C_{z_0,\varepsilon} \ , f(z_1) \right) \ge 2 \]
Since \( f'(z_1) \neq 0 \), by the argument principle there is another \( z_2 \in D_{z_0,\varepsilon} \) such that \( f(z_2) = f(z_1) \), contradiction.
Now we show that \( f^{-1} \) is holomorphic. Take two distinct points \( w, w_0 \in B \), and let \( z := f^{-1}(w) \) and \( z_0 := f^{-1}(w_0) \). By Theorem 1, \( f^{-1} \) is continuous. Thus, as \( w \to w_0 \) we have \( z \to z_0 \ \), and so
\[ \lim_{w \to w_0} \frac{f^{-1}(w) - f^{-1}(w_0)}{w - w_0} = \lim_{z \to z_0} \frac{z - z_0}{f(z) - f(z_0)} = \frac{1}{f'(z_0)} \]
Thus, \( f^{-1} \) is holomorphic with \( (f^{-1})'(w) = 1/f'(f^{-1}(w)) \). \(\ \square\)