3.2 Schwarz Reflection Principle

Before talking about the principle, it helps to recall (from Calc 3) what the total derivative matrix of a function is. Suppose \(h(x,y) = h_1(x,y)\mathbf{i} + h_2(x,y)\mathbf{j}\) if a function \(h:U\to \mathbb {R}^2\) where \(U\) is an open subset of \(\mathbb {R}^2\) containing a point \((x_0,y_0)\). Then the total derivative matrix at \((x_0,y_0)\) is the matrix

\[ Dh(x_0,y_0) = \begin{pmatrix} h_{1x}(x_0, y_0) & h_{1y}(x_0,y_0) \\ h_{2x}(x_0, y_0) & h_{2y}(x_0,y_0) \end{pmatrix}. \]

We can view \(U\) as an open subset of \(\mathbb {C}\) and \(h\) as a function from \(U\) to \(\mathbb {C}\) if we set \(1 = \mathbf{i}\) and \(i = \mathbf{j}\). Then the Cauchy-Riemann equations say that \(h_{1x} = h_{2y}\) and \(h_{1y} = - h_{2x}\).

Proposition 3.12

Suppose \(U\) is an open subset of \(\mathbb {C}\). Write

\[ \bar U = \{ z\in \mathbb {C}: \bar z \in U\} . \]

Suppose \(f\in A(U)\), and define \(g(z) = \overline{f(\bar z)}\) for \(z\in \bar U\). Then \(g\in A(\bar U)\).

Proof ▼

Write \(C:\mathbb {C}\to \mathbb {C}\) for the map \(z\mapsto \bar z\). Then \(g = C\circ f\circ C\). Write \(f(z) = u(x,y) + iv(x,y)\) with \(u\) and \(v\) real-valued functions on \(U\).

We have

\[ DC = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \]

Then, by the Chain Rule, for \(z_0\in U\),

\begin{align*} Dg(\bar z_0) & = D(C\circ f \circ C) (\bar z_0)\\ & = DC(f(z_0))\circ Df(z_0)\circ DC(\bar z_0)\\ & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} u_x(x_0,y_0) & u_y(x_0,y_0) \\ v_x(x_0,y_0) & v_y(x_0,y_0) \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\\ & = \begin{pmatrix} u_x(x_0,y_0) & -u_y(x_0,y_0) \\ -v_x(x_0,y_0) & v_y(x_0,y_0) \end{pmatrix}. \end{align*}

So, since the Cauchy-Riemann equations hold for \(f\) on \(U\), they hold for \(g\) on \(\bar U\).

Lemma 3.13

Suppose \(D\) is a domain which is symmetric about the real axis in the sense that \(\bar D = D\). Then \(D\cap \mathbb {R}\) contains an open interval \((a,b)\). In particular, \(D\cap \mathbb {R}\) has a limit point in \(D\).

Proof ▼

First let’s show that \(D\cap \mathbb {R}\neq \emptyset \). Since \(D\) is (by definition) nonempty, we can find \(z_0\in D\). Since \(D\) is connected, we can find a piecewise linear contour \(\gamma :[0,1]\to D\) with \(\gamma (0) = z_0\) and \(\gamma (1) = \bar z_0\). Define \(h:[0,1]\to \mathbb {R}\) by \(h(t) = \operatorname{\mathrm{Re}}(\gamma (t))\). If \(z_0\not\in \mathbb {R}\), then \(h(0)\) and \(h(1)\) have opposite signs. So, by the Intermediate Value Theorem, there exists \(t\in [0,1]\) such that \(h(t) = 0\). Thus \(x = h(t) \in \mathbb {R}\).

Now, suppose \(w\) is any point in \(D\cap \mathbb {R}\). Since \(D\) is open, there exists \(r {\gt} 0\) such that \(|z-w|{\lt} r \Rightarrow z\in D\). Therefore, \((w-r, w+r)\subseteq D\cap \mathbb {R}\). Thus \(r\) is a limit point of \(D\cap \mathbb {R}\) in \(D\).

Theorem 3.14 Schwarz Reflection Principle

Suppose \(D\) is a domain which is symmetric about the real axis in the sense that \(\bar D = D\), and suppose \(f\in A(D)\) is a function such that \(f(x)\) is real for all \(x\in D\cap \mathbb {R}\). Then \(\overline{f(\bar z)} = f(z)\) for all \(z\in D\).

Proof ▼

Set \(h(z) = f(z) - \overline{f(\bar z)}\). By Proposition 3.12, \(h\in A(D)\), and, by Lemma 3.13, the zero set of \(h\) has a limit point in \(D\). So \(h\equiv 0\) on \(D\).