Let \(f\) be holomorphic on a neighborhood of the closed unit disk and suppose \[ \abs{f(z)}<1 \qquad(\abs z=1). \] Show that \[ f(z)=z^3 \] has exactly three solutions, counted with multiplicity, in \(\abs z<1\).
Solution
Set \[ g(z)\coloneqq z^3-f(z). \]
1. The functions \(g(z)\) and \(z^3\) have the same number of zeros in \(\abs z<1\), counted with multiplicity.
On the unit circle, \[ \abs{g(z)-z^3} =\abs{f(z)} <1 =\abs{z^3}. \] Both functions are holomorphic on a neighborhood of the closed unit disk. Rouché’s theorem therefore implies that \(g\) and \(z^3\) have the same number of zeros in the open unit disk, counted with multiplicity.
2. The equation \(f(z)=z^3\) has exactly \(\boxed{3}\) solutions in \(\abs z<1\), counted with multiplicity.
The function \(z^3\) has one zero at \(0\) of multiplicity \(3\). By step 1, \(g(z)=z^3-f(z)\) therefore has exactly three zeros in the unit disk, counted with multiplicity. These are precisely the solutions of \(f(z)=z^3\).
3. Q.E.D.
Step 2 is the required count.