Let $f:\Omega\to\Bbb C$ 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 $z_1,\ldots,z_k$ such that $f(z_i)=w_0$ and $f'(z_i) = (f(z_i) - w_0)'\ne 0,$ thus, showing that there are $k$ distinct solutions to the equation. Maybe I can make use of the Cauchy integral formula and $f(z_0) = 0$ 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!

  • $\begingroup$ I'd suggest rouche's theorem $\endgroup$
    – r9m
    Commented Dec 7, 2017 at 22:32
  • $\begingroup$ How would you propose to use it then? I'm not sure how this theorem helps in finding the distinct solutions... $\endgroup$
    – Longti
    Commented Dec 7, 2017 at 22:41
  • $\begingroup$ Choose a neighborhood of $z_0$ where none of the first $k-1$ derivatives have a zero (since, the zeros are isolated). That ensures the zeros that rouche's theorem counts are distinct (no multiple roots). $\endgroup$
    – r9m
    Commented Dec 7, 2017 at 22:48
  • $\begingroup$ @r9m but besides distinctness, what functions should we compare (by Rouche's theorem) to show that $f(z) - w_0$ actually has k roots inside that neighborhood? $\endgroup$
    – Longti
    Commented Dec 8, 2017 at 1:49
  • $\begingroup$ compare $f(z)$ and $f(z) - w$, note that $|f(z) - (f(z) - w)| = |w| < |f(z)|$ on the boundary of a small disc $D(z_0,r)$ as long as we choose $\displaystyle |w| < \inf_{z \in \partial D(z_0,r)} |f(z)|$ (which is non zero since we are looking at boundary of disc $D(z_0,r)$ inside a neighborhood where $z_0$ is the only isolated zero). $\endgroup$
    – r9m
    Commented Dec 8, 2017 at 2:35

1 Answer 1


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.


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