Let $f: \Omega \to \mathbb{C}$ be analytic, and assume that $f'$ has a zero of order $k \geq 1$ at $z_0 \in \Omega$.
I want to prove that there is an analytic $g: \Omega_1 \to \mathbb{C}$, $\Omega_1 \subset \Omega$, with the property that $g$ has a zero of order $1$ at $z_0$, and that $f(z) - f(z_0) = g(z)^{k+1}$.
I've tried writing $f'(z) = (z-z_0)^k h(z)$ where $h(z)$ is non-zero on some open set around $z_0$, and played with ideas of finding an analytic logarithm for $h$, but it didn't lead me anywhere. Hints please?
$f'$ has a zero of order $k$ so in the taylor representation of $f$ around $z_0$ we have the $f \ ^ j (z_0) = 0$ for $j=1,..,k$ so we can write $f(z) = f(z_0) + \sum_{n =k+1}^{\infty} \dfrac{f \ ^ n(z_0)}{n!} (z-z_0) \ ^ n$.
Now, we can take $(z-z_0) \ ^ {k+1} $ out from the sum to get $f(z) = f(z_0) + (z-z_0) \ ^ {k+1}h(z)$ where $h(z)$ is analytic around $z_0$ and $h(z_0) \ne 0$.
So we can find a logarithm for $h$ , denote it by $g$. so $h(z) = e\ ^ {g(z)}$.
Now write $l(z) = (z-z_0)e \ ^ {g(z)/(k+1)}$.
So $l$ has a simple zero at $z_0$ and we have $l (z) \ ^ {k+1} = (z-z_0) \ ^ {k+1} h(z)$ so we get the desired result.
$$f(z) = f(z_0) + [l(z)] \ ^ {k+1}$$ ,$l$ analytic function with simple zero.