$f^2$ &$ f^3$ are holomorphic then so is $f$. My problem is :
If $f$ is a complex valued function(Notice that there is any other restriction of $f$) on the open unit disc $D$. If $g=f^2$ and $h=f^3$ are both holomorphic in $D$, then prove that $f$ is holomorphic in $D$.
All my attemps are stuck in the same step which is very clear under an assumption that $f$ is continuous at it's zeros.
But I cannot prove the continuity of $f$ at it's zeros. Anyone can help with this problem? 
Thank you.
 A: the zeros of $f$ are the same as the zeros of $g$ and $h$.
Because $g$ is holomorphic, both $g$ and $|g|$ are continuous at it's zeros, and therefore $|f|=\sqrt{|g|}$ is continuous at it's zeros. It can be shown that $f$ is continuous at it's zeros iff $|f|$ is continuous.
A: As before, zeros of $f$, $g$, and $h$ are exactly the same points.  Therefore the zeros are discrete.  Outside the zeros of $f = h/g$.  I assume you've gotten this far.
So WLOG suppose $0$ is such a zero.  Write
$$
f^2(z) = g(z) = z^k \tilde{g}(z)
$$
where $\tilde{g}(0) \not= 0$ and same for $h$:
$$
f^3(z) = h(z) = z^\ell \tilde{h}(z)
$$
where $\tilde{h}(0) \not= 0$.
From $|g(z)|^{3/2} = |h(z)|$ you will find that $\ell > k$ (use polar coordinates and take limit as $r \to 0$).  Then write what $f$ is and notice it has a removable singularity at the origin.
You would be surprised how many local properties of analytic functions follow from factoring out the order of a zero.  In particular, every function is really just $z^k$ locally.  So always figure out what you'd do if the function was in fact $z^k$, and the point in question was the origin, and then try to generalize from there.  In vast majority of these questions, knowing the proof for $z^k$ is sufficient.
