Range of an analytic function on a unit disc Let $f(z)$ be an analytic function on an open set of the complex plane containing the closed unit disc $D=\{z\in \mathbb{C}:|z|\leq 1\}$. Let $m$ be the minimum of $\{f(z)\ |\ z\in D\}$ and $M$ be the minimum of $\{|f(z)|\ |\ z\in C\}$ where $C=\{z\in \mathbb{C}:|z|=1\}$ of $D$. Assume that $m<M$. Then state whether the following are true or false?
(i) $f(z)$ admits a zero on $D$.
$(ii)$ $f(z)$ attains every complex number $w$ on $D$ such that $|w|<M$.
My attempt: I know that statement $(i)$ is true because of the Minimum modulus principle. But I am not able to find the explanation of statement $(ii)$. Please help.
 A: (i) is true, as you correctly said. If $f$ had no zero in $D$ then the maximum modulus principle applied to $1/f$ would give a contradiction.
(ii) is true as well, this can be seen as follows:
It follows from the open mapping theorem (and the fact that $D$ is bounded) that
$$
 \partial f(D) \subseteq f(\partial D) \, ,
$$
see for example Is it always true that $\partial f(U)=f(\partial U)$ when $f$ is holomorphic?.
It follows that the disk $B_M(0)$ does not contain a boundary point of $f(D)$, and since $0 \in f(D)$, as shown in part (i), the disk $B_M(0)$ is completely contained in $f(D)$.
In other words, $f$ takes every complex value $w$ with $|w| < M$.
A: For $\vert w\vert \lt M$ the line segment $[0,w]$ does not meet the image of $f\circ \gamma$ -- where $\gamma(t) = \exp\big(2\pi i\cdot t\big)$ for $t\in [0,1]$--, so they have the same winding numbers, i.e. $n\big(f\circ \gamma,w\big)=n\big(f\circ \gamma,0\big)\neq 0$ where the right hand side follows by (i) combined with the Argument Principle.  Conclude $w\in f\big(D\big)$ again by the Argument Principle.
