Question: Let $U\subset \mathbb{C}$ be an open set $p \in U$ and $f$ be a holomorphic function defined on $U$ so that $f(p) = f'(p) = 0$. Use the argument principle to prove the following statement: there exists $\delta>0$ so that if $0 < |Q|< \delta$ then $f^{-1}(Q)$ contains at least two points.
My attempt: I know $f$ has a zero of order at least $2$ at $p$, however I am unsure how to obtain the $\delta$ described in the question. Any hints are appreciated.
Let $\gamma$ denote $\{|z-p| = r\}$ where $r> 0$ so that $\overline{D(p,r)} \subset U$ and $f$ is non-vanishing on the deleted closed ball $\overline{D(p,r)} – \{p\}$. By the argument principle, we have $n(f \circ \gamma, f(p)) \geq 2$.
Best Answer
Assume that $f$ is non constant, hence the zeros are isolated. Choose $r>0$ such that both $f$ & $f'$ have no zeros in $\overline{B}(p,r)$ other than $p$ (the constraint on $f'$ is to ensure that any zero has multiplicity one). Let $\gamma(t) = p+r e^{2 \pi i t}$, and let $\Gamma$ be the range of $\gamma$. Note that ${1 \over 2 \pi i}\int_\gamma {f' \over f} \ge 2$.
Let $\mu = \min_{z \in \Gamma} |f(z)|$ and choose $Q$ such that $0<|Q|<\mu$. Let $f_Q = f-Q$. Note that for $z \in \Gamma$, $|f_Q(z)| \ge |f(z)| -|Q| \ge \mu -|Q| > 0$.
Note that $\int_\gamma {f' \over f} = \int_{f \circ \gamma} {dz \over z} $. Define the homotopy $H(t,s) = f_{sQ}(\gamma(t)) = f(\gamma(t)) - sQ$ and note that from the previous paragraph, the range of $H$ does not include zero, hence $\int_{f \circ \gamma} {dz \over z} = \int_{f_Q \circ \gamma} {dz \over z} \ge 2$ and so $f_Q$ contains at least two zeros (counting multiplicities).
Finally, note that if $z_Q$ is a zero of $f_Q$ in ${B}(p,r)$, then $f_Q'(z_Q) \neq 0$, and hence the zeros are simple.