A family of analytic function about open mapping theorem Suppose $\mathbb{F}$ is a family of analytic function $f$ on unit disk D such that $f(0)=0$,$f'(0)=1$ and $|f|<5$.Fist show that there exsits $0<b<1$ and $c>0$ such that $|f|>c$ on $|z|=b$ for all $f\in\mathbb{F}$;Secondly show that for all $|\omega|<c$,$\omega$ is assumed by every $f\in\mathbb{F}$.
I can tell that $f(ann(0;b,1))$ is an open set,but I don't know how to show $0$ is not in such a open set,also I'm not sure how to choose $b$ and what if there are some other zeros of $f$ which is very near the boundary,then how can we find such $b$?
 A: As it is stated now, we can prove it.
Let's make the second step before the first: If $h$ is holomorphic in a neighbourhood of $\{z : \lvert z\rvert \leqslant b\}$, with $f(0) = 0$ and $\lvert f(z)\rvert \geqslant c > 0$ for $\lvert z\rvert = b$, then
$$D_c(0) \subset f(D_b(0)),$$
i.e. $f$ attains every $w$ with $\lvert w\rvert < c$ on the disk around $0$ with radius $b$. That follows since
$$N(w) = \frac{1}{2\pi i} \int_{\lvert z\rvert = b} \frac{f'(z)}{f(z)-w}\,dz$$
is continuous on the complement of $f(\{z : \lvert z\rvert = b\})$, hence locally constant, since it is integer-valued. By assumption $N(0) > 0$, and $D_c(0)$ is contained in the complement of $f(\{z : \lvert z\rvert = b\})$, so $N(w) = N(0) > 0$ for all $w\in D_c(0)$.
So once we've established the existence of $b$ and $c$, it follows that $D_c(0) \subset f(\mathbb{D})$ for all $f\in \mathbb{F}$.
To show the existence of $b$ and $c$, we use the normality of $\mathbb{F}$. Since enlarging $\mathbb{F}$ makes the proposition stronger, we assume that $$\mathbb{F} = \left\{ f\in \mathscr{O}(\mathbb{D}) : f(0) = 0, f'(0) = 1, \lvert f(z)\rvert < 5 \text{ for all } z \in \mathbb{D}\right\}.$$
For $f\in\mathbb{F}$, let 
$$\beta(f) = \sup \left\{ r \in (0,1) : \bigl(0 < \lvert z\rvert < r \implies f(z)\neq 0\bigr)\right\}$$
be the modulus of the zero of $f$ closest to $0$ if any, and $1$ if $0$ is the only zero of $f$. Further, let $b_0 = \inf \{ \beta(f) : f\in\mathbb{F}\}$. We show that $0 < b_0 <1$: $f(z) = z - 2z^2$ shows $b_0 \leqslant \frac{1}{2}$. Let $(f_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{F}$ with $\beta(f_n) \to b_0$. By passing to a subsequence, we can assume that $f_n \to f$ locally uniformly. We have $f(0) = 0$ and $f'(0) = 1$, so there is a $\delta > 0$ such that $f$ has no other zeros than $0$ in $\overline{D_\delta(0)}$. By Hurwitz's theorem, for all large enough $n$, $f_n$ also has exactly one zero (counting mutliplicities) in $\overline{D_\delta(0)}$, so $\beta(f_n) > \delta$, and hence $b_0 \geqslant \delta > 0$.
We choose any $b$ with $0 < b < b_0$. Then
$$\gamma(f) := \inf \left\{ \lvert f(z)\rvert : \lvert z\rvert = b\right\} > 0$$
for all $f\in\mathbb{F}$. We show that
$$c := \inf \{ \gamma(f) : f \in \mathbb{F}\} > 0.$$
Once again, let $(f_n)$ be a sequence in $\mathbb{F}$ with $\gamma(f_n) \to c$, and for each $n$, let $\zeta_n$ a point with $\lvert\zeta_n\rvert = b$ and $\lvert f_n(\zeta_n)\rvert = \gamma(f_n)$. By passing to a subsequence, we may assume that $f_n \to f$ locally uniformly, $\zeta_n \to \zeta$, and $f_n(\zeta_n) \to \eta$. Since $f$ is not constant, it is an open mapping, and therefore $f\in \mathbb{F}$. Further, by locally uniform convergence,
$$f(\zeta) = \lim_{n\to\infty} f_n(\zeta_n) = \eta,$$
so
$$0 < \gamma(f) \leqslant \lvert f(\zeta)\rvert = \lvert\eta\rvert = \lim_{n\to\infty} \lvert f_n(\zeta_n)\rvert = \lim_{n\to\infty}\gamma(f_n) = c.$$
