Non-constant analytic function on a compact set

Question

Let $\mathbb{C}$ be the set of complex numbers and $E \subset \mathbb{C}$. Let $f: E \rightarrow \mathbb{C}$ be an analytic function such that $f$ is non-constant on each connected component of $E$. Let $K \subset E$ be a compact subset of $\mathbb{C}$. Let $\lambda_n$ be a sequence of distinct scalars in $K$ such that $f(\lambda_n) \rightarrow f(\lambda_0)$.

Then is it true that $\lambda_n \rightarrow \lambda_0$?

Answer 1

No, take $K$ to be the closed ball of radius 2 and $f(z)=z^2$. Take $\lambda_n=1-1/n$ and $\lambda_0=-1$. This you can do for every even function.

Answer 2

No. Take $E=\mathbb C$, $f(z)=z^2$, $K=S^1=\{z\in\mathbb C\mid\lvert z\rvert=1\}$, $\lambda_n=e^{i/n}+(-1)^n$, and $\lambda_0=1$. Then $\lim_{n\to\infty}f(\lambda_n)=f(\lambda_0)$, but the limit $\lim_{n\to\infty}\lambda_n$ doesn't exist.