Distinct solutions of $f(z) - w_0 = 0$ for $f$ holomorphic, $w_0$ in a punctured neighborhood of $f(z_0) = 0$ Problem: Let $f: \Omega \rightarrow \mathbb{C}$ be a holomorphic function ($\Omega$ is open), and $f$ has a zero of order k at $z_0 \in \Omega$. Show that there is a neighborhood $U$ of $z_0$ and a neighborhood $V$ of $f(z_0)$ such that if $w_0 \in V - \{f(z_0)\} $, then $f(z) - w_0 = 0$ has $k$ distinct roots in $U - \{z_0 \}$.
Attempt at solution: I was thinking about showing that there are $z_1,...,z_k$ such that $f(z_i) = w_0$ and $f'(z_i) = (f(z_i) - w_0)' \neq 0$ thus showing that there are $k$ distinct solutions to the equation. Maybe I can make use of the Cauchy integral formula and that $f(z_0) = 0$ and $w_0$ being near $f(z_0)$ to show that the derivatives are indeed nonzero, however I am having trouble making this intuition precise and actually finding the $k$ solutions. Any help is appreciated!
 A: If $z_0 \in \Omega$ is a zero of order $k$, then $f^{(j)}(z_0) = 0$ for each $j = 0, 1 , \cdots, k-1$ and $f^{(k)}(z_0) \neq 0$. 
Since, the zeros of holomorphic functions are discrete and isolated, for each $0 \le j \le k-1$, there is a deleted neighborhood $D(z_0,r_j)\setminus\{z_0\}$ in $\Omega$ where, $f^{(j)}$ is non-vanishing. 
Choose, $\displaystyle r < \min_{0 \le j \le k-1} r_j$, then, $f$ and all it's first $(k-1)$ derivatives are non vanishing in a $\overline{D(z_0,r)}\setminus\{z_0\}$.
For, $\displaystyle 0 < |w| < \inf_{|z-z_0| = r} |f(z)| = m$ (note, $m > 0$ since, $|f|$ does not vanish on boundary of the disc) using Rouche's theorem we compare $f$ and $f-w$ in $\overline{D(z_0,r)}$,
$$|f(z) - (f(z) - w)| = |w| < |f(z)| \quad \forall z \in \partial D(z_0,r)$$
i.e., $f$ and $f-w$ have the same number of zeros (counting multiplicities) in $D(z_0,r)$ (note, both $f$ and $f-w$ have no zeros on the boundary of the disc $\partial D(z_0,r)$). 
I.e, $f-w$ has exactly $k$ zeros in $D(z_0,r)$ and each of them are distinct (otherwise, $(f-w)' = f'$ vanishes at a multi-zero, which was ruled out by our choice of the neighborhood).
Stated differently, $f(z) - w$ has $k$ distinct zeros in $D(z_0,r)$, for any $w \in D(0,m)\setminus \{0\}$ for choice of $r,m>0$ as above.
