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The answer is here: $f^2$ and $f^3$ are holomorphic implies $f$ is holomorphic. However, they assume continuity of $f$.

I just wanted to make sure that that is not necessary. Since $f^2$ is holomorphic, it is bounded around its zeroes. Thus $f$ is bounded around its zero. Hence now we can conclude that that all of the singularities are removable as $f^3/f^2$ is bounded around $f$'s zeroes. Thus it follows $f$ is holomorphic

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    assuming $f \ne 0$ (hence $f^2 \ne 0$) identically, $f^3/f^2$ is meromorphic so $f$ is meromorphic and then $f^2$ has no poles, so $f$ doesn't either – Conrad Jul 13 '20 at 01:21

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The continuity of $f$ is indeed not needed. As you correctly said, $f$ is bounded around each zero, and therefore has a removable singularity at each zero. One has to argue carefully that the holomorphic continuation coincides with $f$.

I would proceed as follows: Assume that $f^2$ and $f^3$ are holomorphic in $D \subset \Bbb C$, and that $f$ is not identically zero.

  • The set $A = \{ z \in D \mid f(z) = 0 \}$ has no accumulation point in $D$ because $f^2$ is holomorphic.
  • $f = f^3/f^2$ is holomorphic in $D \setminus A$.
  • At every point $z_0 \in A$ we have $$ \lim_{z \to z_0}{f^2(z)} = 0 \implies \lim_{z \to z_0}{f(z)} = 0 = f(z_0) \, . $$ so that we can conclude that $f$ is continuous at $z=z_0$. Riemann's theorem on removable singularities then shows that $f$ is holomorphic at $z=z_0$.

It follows that $f$ is holomorphic in $D$.

Martin R
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  • out here: "The set $A = { z \in D \mid f(z) = 0 }$ has no accumulation point in $D$ because $f^2$ is holomorphic." did you mean: "The set $A = { z \in D \mid f^2(z) = 0 }$ has no accumulation point in $D$ because $f^2$ is holomorphic."? – thefool Dec 30 '22 at 23:13
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    @mathman12: $ { z \in D \mid f(z) = 0 }$ and $ { z \in D \mid f(z)^2 = 0 }$ are the same sets. – Martin R Dec 31 '22 at 01:43