Proving relation with zeros of holomorphic functions The question is the following:
Let f be holomorphic on $\Omega$. Assume that $a \in \Omega $ is a zero of order $m$ of $f$.
a) Prove that $f$ can be written as $f = g^m$ for some function $g$ holomorphic in a neighborhood of $a$ satisfying $g'(a) \neq 0$
b)Can $f$ be written as $f = g^m$ on $\Omega$?
I have managed to solve part a) by writing $f$ as $f = (z - a)^mh(z)$ where $h(z)$ is holomorphic on $D(a, r)$. Then letting $g(z) = (z - a)h(z)^{1/m}$ I obtain the relation needed and gives $g'(a) \neq 0$.
Does that sound okay? and how do I go about solving part b?? 
 A: For part (b), you need to be a little careful.
Consider the example where $\Omega = \mathbb C$, $a = 0$ and
$$ f(z) = z^2 (z-1)$$
Then $f$ vanishes with multiplicity two at $a = 0$. You might be tempted to write
$$ f(z) = g(z)^2$$
where
$$ g(z) = z \sqrt{z - 1}.$$
Unfortunately, this doesn't quite work because $z \sqrt{z-1}$ has a branch cut: there is no way to pick signs for the square root such that $z \sqrt{z-1}$ is continuous, let alone holomorphic, on the whole of $\mathbb C$.

However, for part (a), it is generally possible to write $f(z) = g(z)^m$ on a small open neighbourhood of $a$.
To prove this, first note that there exists a convex open neighbourhood $U$ of $a$ where $h(z)$ does not vanish; this follows by continuity arguments.
[As defined in your question, $h(z)$ is the holomorphic function, non-vanishing at $a$, such that $f(z) = (z-a)^m h(z)$.] 
Now $h'(z)/h(z)$ is well-defined and holomorphic on $U$. Since $U$ is convex, a corollary of Cauchy's theorem tells you that $h'(z)/h(z)$ has an antiderivative $H(z)$ such that $$H'(z) = \frac{h'(z)}{h(z)}. $$
This implies that $$\frac{d}{dz}(h(z)e^{-H(z)}) = 0,$$ which implies that $$h(z) = e^{H(z)},$$
possibly after absorbing a constant of integration into the definition of $H(z)$.
You can then define the function
$$g(z) = (z - a) e^{H(z)/m},$$
which holomorphic on the convex open neighbourhood $U$ of $a$. Indeed, this function satisfies
$$ f(z) = g(z)^m$$
for $z \in U$. But of course, $g(z)$ is only defined on $U$; it is not defined on the whole of $\Omega$.
The upshot is: $f(z) = g(z)^m$ locally but not globally.
