[January 2008, Problem \(\# 5 \mathrm { b } ]\) Suppose \(f : \mathbb { D } \to \mathbb { D }\) is analytic, has a zero of order \(k\) at the origin, has no other zeros, and satisfies \(\lim_{\abs{z} \to 1} \abs{f(z)} = 1\). Give, with proof, a formula for \(f(z)\).
Solution
Define \[ g(z) = \begin{cases} \dfrac{f(z)}{z^k},&z\neq0,\\ \dfrac{f^{(k)}(0)}{k!},&z=0. \end{cases} \]
1. The function \(g\) is analytic and zero-free on \(\DD\).
Because \(f\) has a zero of order exactly \(k\) at the origin, there is an analytic function \(g\) near \(0\) with \[ f(z)=z^kg(z) \] and \(g(0)\neq0\). This is exactly the extension displayed above. Since \(f\) has no zeros away from the origin, \(g\) has no zeros anywhere in \(\DD\).
2. One has\[ \lim_{\abs{z}\to1}\abs{g(z)}=1. \]
For \(z\neq0\), \[ \abs{g(z)} = \frac{\abs{f(z)}}{\abs{z}^k}. \] As \(\abs{z}\to1\), the numerator tends to \(1\) by hypothesis and the denominator tends to \(1\). Hence the quotient tends to \(1\).
3. For every \(z_0\in\DD\),\[ \abs{g(z_0)}\leq1. \]
Fix \(z_0\in\DD\) and choose \(r\) with \[ \abs{z_0}<r<1. \] By the maximum modulus principle applied to the disk \(\abs{z}\leq r\), \[ \abs{g(z_0)} \leq \max_{\abs{z}=r}\abs{g(z)}. \] Step 2 implies \[ \max_{\abs{z}=r}\abs{g(z)} \longrightarrow1 \] as \(r\uparrow1\). Letting \(r\uparrow1\) gives the claimed inequality.
4. For every \(z_0\in\DD\),\[ \abs{g(z_0)}\geq1. \]
By step 1, the function \(1/g\) is analytic on \(\DD\). Step 2 gives \[ \lim_{\abs{z}\to1}\abs{1/g(z)}=1. \] Applying step 3 to \(1/g\) yields \[ \abs{1/g(z_0)}\leq1, \] equivalently \(\abs{g(z_0)}\geq1\).
5. The function \(g\) is constant with unimodular value.
Steps 3 and 4 give \[ \abs{g(z)}=1 \] for every \(z\in\DD\). By the open mapping theorem, a nonconstant analytic function has open image, whereas the unit circle is not open. Hence \(g\) is constant. Write \[ g\equiv\lambda, \qquad \abs{\lambda}=1. \]
6. The required formula is\[ \boxed{ f(z)=\lambda z^k, \qquad \abs{\lambda}=1. } \]
By the definition of \(g\), \[ f(z)=z^kg(z). \] Substitute the constant value from step 5.
7. Q.E.D.
Step 6 is the requested formula.