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If $f(z)$ is defined in the unit disk such that both $f^2(z)$ and $f^3(z)$ are analytic in the disk. is $f(z)$ analytic in the disk?

Attempt: If $f(z) \ne 0$ in the disk, then $f(z) = \dfrac{f^3(z)}{f^2(z)}$ must be analytic.

Hence, any counterexample involving $f$ must be such that $f(z) = 0$ at least one point in the disk.

Any hints on how to to move forward from here?

MathMan
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  • Can you show it is still continuous at the points where $f(z)=0?$ – Thomas Andrews Sep 26 '21 at 20:58
  • @ThomasAndrews Yes, for if $f$ was not continuous, $f^3$ won't be continuous as well – MathMan Sep 26 '21 at 21:03
  • Hint $f$ is meromorphic in the disc as the ratio of analytic function but its possible poles are only the zeroes of $f$ etc – Conrad Sep 26 '21 at 21:07
  • That’s a bit vague. It’s not true that $f^3(z)$ is not continuous when $f$ is not continuous. It has something to do with $0$ specifically. – Thomas Andrews Sep 26 '21 at 21:09
  • This has been asked and answered several times: https://math.stackexchange.com/questions/linked/298951/ – duplicate question found with a Google search for “f^2 and f^3 analytic” – Martin R Sep 27 '21 at 05:24

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If $f$ is the zero function then the statement is obviously true. Otherwise, $f^2$ is a nonzero holomorphic function, and so its zeros are isolated. Equivalently, the zeros of $f$ are isolated. The function $f$ is clearly analytic at all the other points of the disk, and its zeros are isolated singularities. So in order to show that $f$ is analytic at the whole disk, you just have to show that these singularities are removable. And this is indeed the case. If $f$ had a pole at such a point then so would $f^2$. If $f$ had an essential singularity then using Casorati-Weierstrass (or Picard's theorem if you know it) it is easy to show that $f^2$ would also have an essential singularity there. Both contradict the assumption that $f^2$ is holomorphic.

Mark
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  • This is an over-kill. No need for anything as deep as Picard's theorem for this elementary result. – Kavi Rama Murthy Sep 26 '21 at 23:22
  • @KaviRamaMurthy I didn't write to use Picard's theorem. I wrote in brackets that IF somehow OP knows it then he might use it. Casorati-Weierstrass is more than enough, and this is taught in pretty much any complex analysis course. This should be absolutely clear. Leave alone the fact that my answer is $100$% correct. – Mark Sep 27 '21 at 00:21
  • This has been asked and answered before. – Martin R Sep 27 '21 at 05:27