Can you help me with the following please:
If $f:B_2 (0) \rightarrow \mathbb{C}$ is holomorphic such that $f(\mathbb{S}^1)\subset B_r(z_0)$ then $f(\mathbb{B}^2)\subset B_r(z_0)$
Where:
$B_r(z_{0})=\{z \in \mathbb{C} : |z-z_0|<r \}$
$\mathbb{B}^2=\{z \in \mathbb{C} : |z|< 1 \}$
$\mathbb{S}^1=\{z \in \mathbb{C} : |z|= 1 \}$
I don't have a lot of resources to do the proof (I've seen few complex analysis theorems) but we saw Cauchy's theorem and it mentioned the following:
"The values of a function within a disk are determined by its boundary"
So, what I think is that this problem is a consequence of the Cauchy integral formula, but I don't know how to express that this result follows from this.