Let \(f\) be complex-valued on the open unit disk \(D\). Suppose \[ f^2 \quad\text{and}\quad f^3 \] are both analytic on \(D\). Prove that \(f\) is analytic on \(D\).
Solution
Set \[ g\coloneqq f^2, \qquad h\coloneqq f^3. \]
1. If \(g\equiv0\), then \(f\) is analytic on \(D\).
For every \(z\in D\), \[ f(z)^2=g(z)=0, \] so \(f\equiv0\), which is analytic.
2. Assume \(g\not\equiv0\). On\[ U\coloneqq\{z\in D:g(z)\ne0\}, \]one has\[ f=\frac{h}{g}, \]so \(f\) is analytic on \(U\).
If \(z\in U\), then \(g(z)=f(z)^2\ne0\), hence \(f(z)\ne0\). Therefore \[ \frac{h(z)}{g(z)} = \frac{f(z)^3}{f(z)^2} = f(z). \] Since \(g\) and \(h\) are analytic and \(g\) does not vanish on \(U\), the quotient \(h/g\) is analytic on \(U\).
3. Let \(a\in D\) satisfy \(g(a)=0\). Then \(f\) is analytic in a neighborhoodof \(a\).
Because \(g\) is analytic and not identically zero, its zeros are isolated. Choose \(r>0\) such that \[ \{z:0<\abs{z-a}<r\}\subseteq U. \] On this punctured disk, step 2 gives \[ \frac{h(z)}{g(z)}=f(z). \] Moreover, \[ \abs{f(z)}^2 = \abs{f(z)^2} = \abs{g(z)}. \] Since \(g\) is continuous and \(g(a)=0\), \[ \abs{f(z)} = \sqrt{\abs{g(z)}} \longrightarrow0 \] as \(z\to a\). Hence \(h/g\) is bounded near the isolated singularity \(a\) and in fact tends to \(0\). By the removable-singularity theorem, \(h/g\) extends analytically across \(a\) with value \(0\).
Finally, \[ f(a)^2=g(a)=0, \] so \(f(a)=0\). Thus the analytic extension agrees with the original function \(f\) at \(a\), proving that \(f\) is analytic near \(a\).
4. The function \(f\) is analytic on all of \(D\).
By step 2, \(f\) is analytic at every point where \(g\) is nonzero. By step 3, it is analytic at every zero of \(g\). These two sets cover \(D\).
5. Q.E.D.
Step 1 handles the case \(g\equiv0\), and step 4 handles the case \(g\not\equiv0\).