- The family of mappings introduced here plays an important role in complex analysis. These mappings, sometimes called Blaschke factors, will reappear in various applications in later chapters.
- Let \(z , w\) be two complex numbers such that \(\overline { { z } } w \ne 1\) . Prove that
\[ \left| \frac {w - z}{1 - \overline {{w}} z} \right| < 1 \quad \text { if } | z | < 1 \text { and } | w | < 1, \]
and also that
\[ \left| \frac {w - z}{1 - \overline {{w}} z} \right| = 1 \quad \text { if } | z | = 1 \text { or } | w | = 1. \]
[Hint: Why can one assume that z is real? It then sufices to prove that
\[ (r - w) (r - \overline {{w}}) \leq (1 - r w) (1 - r \overline {{w}}) \]
with equality for appropriate r and \(\abs w\).]
- Prove that for a fixed w in the unit disc D, the mapping
\[ F: z \mapsto \frac {w - z}{1 - \overline {{w}} z} \]
satisfies the following conditions:
F maps the unit disc to itself (that is, \(F : \mathbb { D } \to \mathbb { D } )\) , and is holomorphic.
F interchanges 0 and \(w ,\) namely \(F ( 0 ) = w\) and \(F ( w ) = 0\)
\(| F ( z ) | = 1 { \mathrm { ~ i f ~ } } | z | = 1 .\)
\(F : \mathbb { D } \to \mathbb { D }\) is bijective. [Hint: Calculate \(F \circ F . ]\)
Solution
1. (a) If \(\abs z<1\) and \(\abs w<1\), then \(\abs{\frac{w - z}{1 - \bar w z}} < 1\); if \(\abs z=1\) or \(\abs w=1\), then \(\abs{\frac{w - z}{1 - \bar w z}} = 1\).
1.1. It suffices to treat \(z = r\ge0\) real.
Replacing \((z,w)\) by \((e^{i\theta}z,e^{i\theta}w)\) multiplies \(w-z\) by \(e^{i\theta}\) and leaves \(\bar wz\) unchanged, so it preserves \(\abs{\frac{w - z}{1 - \bar w z}}\) and the hypotheses. Choose \(\theta\) with \(e^{i\theta}z=\abs z\).
1.2. For \(r\ge0\), \(\abs{1-\bar w r}^2-\abs{w-r}^2=(1 - r^2)(1 - |w|^2)\).
Expanding, \(\abs{w-r}^2=(r - w)(r - \bar w)=r^2 - r(w + \bar w) + |w|^2\) and \(\abs{1-\bar wr}^2=(1 - rw)(1 - r\bar w)=1 - r(w + \bar w) + r^2|w|^2\); subtract.
1.3. Q.E.D.
By step 1 take \(z=r=\abs z\). If \(r<1\) and \(\abs w<1\), the right side of step 2 is positive, so \(\abs{w-r}<\abs{1-\bar wr}\). If \(r=1\) or \(\abs w=1\), it is zero, so \(\abs{w-r}=\abs{1-\bar wr}\), and this common value is nonzero because \(\bar wz\ne1\).
2. (b) For fixed \(w\in\mathbb D\), the map \(F(z) = \frac{w - z}{1 - \bar w z}\) satisfies (i)–(iv).
2.1. (i) \(F\) is holomorphic on \(\mathbb D\) and \(F(\mathbb D)\subseteq\mathbb D\).
For \(\abs z<1\), \(\abs{\bar wz}<1\), so the denominator does not vanish and \(F\) is a rational function without poles in \(\mathbb D\). Step 1 gives \(\abs{F(z)}<1\).
2.2. (ii) \(F(0) = w\) and \(F(w) = 0\); (iii) \(\abs{F(z)} = 1\) if \(\abs z = 1\).
Substitution gives (ii), and step 1 gives (iii).
2.3. (iv) \(F\circ F = \operatorname{id}_{\mathbb D}\), so \(F\colon\mathbb D\to\mathbb D\) is bijective.
For \(z\in\mathbb D\), \[F(F(z)) = \frac{w - \frac{w - z}{1 - \bar w z}}{1 - \bar w \frac{w - z}{1 - \bar w z}} = \frac{w(1 - \bar w z) - (w - z)}{(1 - \bar w z) - \bar w(w - z)} = \frac{z(1 - |w|^2)}{1 - |w|^2} = z.\] By step 1, \(F\) maps \(\mathbb D\) into itself, so \(F\) is its own inverse on \(\mathbb D\).