consider the function $\displaystyle f:\mathbb{D}\rightarrow \mathbb{C}$

where $\displaystyle f$ is an analytic function on the space $\displaystyle \mathbb{D}$ (open unit disk in $\displaystyle \mathbb{C}$)

can we show that $\displaystyle f$ is in fact surjective?

In the bigger picture I would like to thow that $\displaystyle \mathcal{A}$, the Banach algebra whose elements are analytic functions on $\displaystyle \mathbb{D}$ with pointwise defined operations is contained in the space $\displaystyle H^{\infty}(\mathbb{D})$ of all bounded analytic function on $\displaystyle \mathbb{D}$

The plan is as follows:

Since any $\displaystyle f\in\mathcal{A}$ is analytic it will also be continuous. $\displaystyle \mathbb{D}$ is compact, so if I can show that theses functions $\displaystyle f$ are in fact surjective then I can use preservation of compactness which implies the image $\displaystyle f(\mathbb{D})$ will be compact, and hence closed and bounded.