Let \(f:D(0,1)\to\mathbb C\) be a function such that \(f^2\) and \(f^3\) are holomorphic. Prove that \(f\) is holomorphic.
Solution
Put \(A=f^2\) and \(B=f^3\), both holomorphic by hypothesis. If \(A\) is identically zero, then \(f(z)^2=0\) at every point, so \(f=0\) and the conclusion holds. Suppose \(A\not\equiv0\).
1. The function \(f\) is holomorphic off the discrete zero set of \(A\).
The zeros of the nonzero holomorphic function \(A\) on the connected disk are isolated [@SS03]. At any point where \(A\ne0\), the identity \(f=B/A\) follows from the definitions. The quotient is holomorphic on the open set where its denominator is nonzero. Thus \(f\) is holomorphic there.
2. At every zero of \(A\), the quotient extends with exactly the given value of \(f\).
Let \(a\) be a zero of \(A\) and choose a disk about \(a\) containing no other zero. On its punctured interior, step 1 applies and \[ |f(z)|^2=|A(z)|\longrightarrow |A(a)|=0 \qquad(z\to a). \] Hence \(f(z)\to0\) and in particular \(f\) is bounded there. The removable-singularity theorem extends it holomorphically to \(a\) with value zero [@SS03]. But \(f(a)^2=A(a)=0\) already forces \(f(a)=0\). The extension therefore agrees with the original function even at \(a\). Applying this at every isolated zero proves that the given \(f\) itself is holomorphic throughout the disk.