Let \(| a _ { k } | < 1 ( k = 1 , 2 , \ldots , n ) , | b | < 1\) and
\[ f ( z ) = { \frac { z - a _ { 1 } } { 1 - { \overline { { a } } } _ { 1 } z } } { \frac { z - a _ { 2 } } { 1 - { \overline { { a } } } _ { 2 } z } } \cdots { \frac { z - a _ { n } } { 1 - { \overline { { a } } } _ { n } z } } . \]
Show that \(f(z) = b\) has \(n\) solutions in \(\abs{z} < 1\).
Solution
For \(1\leq k\leq n\), set \[ \phi_k(z) = \frac{z-a_k}{1-\overline{a_k}z}, \] so that \[ f(z)=\prod_{k=1}^{n}\phi_k(z). \]
1. Every denominator\[ 1-\overline{a_k}z \]is nonzero on the closed unit disk.
If \(\abs{z}\leq1\), then \[ \abs{\overline{a_k}z} \leq \abs{a_k} < 1. \] Hence \(\overline{a_k}z\neq1\), so the denominator cannot vanish.
2. If \(\abs{z}=1\), then\[ \abs{\phi_k(z)}=1 \]for every \(k\).
For \(\abs{z}=1\), \[ z-a_k = z(1-a_k\overline{z}). \] Taking moduli and using \(\abs{z}=1\) gives \[ \abs{z-a_k} = \abs{1-a_k\overline{z}}. \] But \[ \abs{1-a_k\overline{z}} = \abs{\overline{1-\overline{a_k}z}} = \abs{1-\overline{a_k}z}. \] Therefore the numerator and denominator of \(\phi_k\) have equal modulus.
3. On the unit circle,\[ \abs{f(z)}=1. \]
By step 2, \[ \abs{f(z)} = \prod_{k=1}^{n}\abs{\phi_k(z)} = 1. \]
4. The function \(f\) has exactly \(n\) zeros in the open unit disk,counting multiplicity.
By step 1, none of the denominators vanishes in the closed unit disk. Thus the zeros of \(f\) there come exactly from the numerator factors \[ z-a_1,\ldots,z-a_n. \] Each \(a_k\) lies in the open unit disk by hypothesis. Counting repeated values among the \(a_k\) with their multiplicities gives exactly \(n\) zeros.
5. On the unit circle,\[ \abs{b}<\abs{f(z)}. \]
The hypothesis gives \(\abs{b}<1\), while step 3 gives \(\abs{f(z)}=1\).
6. The equation\[ f(z)=b \]has exactly \(n\) solutions in the open unit disk, counting multiplicity.
1. Step 5 is the strict Rouché inequality\[ \abs{(f-b)-f} = \abs{b} < \abs{f} \]on the unit circle. Hence \(f-b\) and \(f\) have the same number of zeros inthe unit disk, counting multiplicity. Step 4 says that this number is\(n\).
7. Q.E.D.
Step 6 is the required conclusion.