# $f^2$ and $f^3$ are holomorphic implies $f$ is holomorphic.

Suppose $f$ is a continuous complex valued function on a domain $\Omega$. Suppose $f^2$ and $f^3$ are holomorphic in $\Omega$. Show that $f$ is also holomorphic in $\Omega$.

Assume $f=u+iv$. I see that if $u,v$ are in $C^1$ then $f^2$ is holomorphic can imply $f$ is holomorphic considering Cauchy-Riemann equations. But I don't know how to get $u,v$ are in $C^1$ by the adding condition $f^3$ is holomorphic. (I can't get $u,v$ from some combinations of real and imaginary part of $f^2$ and $f^3$). Do someone know how to do this?

• Hint: Consider $f^3(z)/f^2(z)$ for all $z$ such that $f^2(z) \neq 0$. What can you say about the resulting function? Can you extend it to points where $f^2(z) = 0$? – Alex G. May 6 '14 at 12:55
• @AlexG. Thank you! I know how to do now. – Koma May 6 '14 at 13:06

The zeros of $f^2$ are a closed discrete set by the identity theorem. On their complement, $f^3/f^2$ is defined, holomorphic and coincides with $f$.
Now use the continuity of $f$ to conclude that all the singularities of $f^3/f^2$ are in fact removable, proving $f = f^3/f^2$ on all $ℂ$.
• To extend $f$ to the zeros of $f^{2}$, could we use that if $f^{2}$ has a zero of order $m$ then $f^{3}$ must also have a zero of order $m$? – user135520 Aug 16 '18 at 16:54