Conditions implying that image of $f$ contains the unit disc I'm stuck with this problem from Stein-Shakarchi:
Let $f$ be non-constant and holomorphic in an open set containing the closed unit disc.
a) Show that if $|f(z)| = 1$ whenever $|z| = 1$, then the image of $f$ contains the unit disc.
b) If $|f(z)| \geq 1$ whenever $|z| = 1$ and there exists $z_{0} \in D(0,1) $ such that $|f(z_{0})| < 1$, then the image of $f$ contains the unit disc.
Any idea ? 
 A: This is philosophically the same as Daniel Fischer's answer but it follows the hint given in the textbook.
(a). We need to show $f(z) = w_0$ has a root for every $w_o\in\Bbb D$. Since $|w_0|<1$ and $|f(z)| =1$ on $z\in\partial\Bbb D$ by assumption, by Rouche's theorem, the number of roots of $f$ and $f-w_0$ are the same. Hence, it suffices to show that $f(z)$ has a root on $\Bbb D$. If $f$ has no root on $\Bbb D$, then ${1\over f}$ is holomorphic on $\Bbb D$ with $\left|{1\over f}\right|\leq 1$ by the maximum modulus principle. Hence, $1\leq |f(z)|$ on $\Bbb D$. But this contradicts the open mapping theorem as if $z\in\partial\Bbb D$ then $|f(z)| =1$. If $U$ is a small neighborhood of $z$ contained in the domain, then $f(U)$ is an open neighborhood of $f(z)$ which contradicts $1\leq |f(z)|$ on $\Bbb D$. Hence, $f$ has a root on $\Bbb D$.
(b). Same as before, Rouche's theorem implies the number of roots of $f$ and $f-w$ is the same for $|w|<1$. Since the number of roots of $f$ is constant, the number of roots of $f-w$ is constant for all $w\in\Bbb D$. Since $f - w_0$ has at least one root, we conclude that $f-w$ has at least one root.
A: a) is a special case of b), since the maximum modulus principle implies $\lvert f(z)\rvert < 1$ for all $z$ in the open unit disk when $f$ is non-constant.
Consider $h \colon \mathbb{D}\to \mathbb{C}$,
$$h(w) = \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f'(z)}{f(z)-w}\,dz.$$
By the residue theorem (in the form of the argument principle), $h(w)$ is the number of times the value $w$ is attained by $f$ in the open unit disk (counting multiplicities). In particular, $h$ is integer-valued. On the other hand, since the integrand depends continuously on $w$, $h$ is continuous on $\mathbb{D}$. It follows that $h$ is constant, and by assumption $h(f(z_0)) \geqslant 1$, so $h(w) \geqslant 1$ for all $w\in \mathbb{D}$, which means $w\in f(\mathbb{D})$ for all $w\in \mathbb{D}$.
