I have a theorem with its proof, but I can't understood some part of proof. Let $\Lambda\in\mathbb{D}$ ($\mathbb{D}$ is open unit desk) with an accumulation point inside $\mathbb{D}$. Then $\{(\lambda^k);\ \lambda\in\Lambda\}$ spans a dense subspace of $\ell^1(\mathbb{N})$.
Proof: Let $u\in\ell^\infty(\mathbb{N})$ which is orthogonal to all $(\lambda^k); \forall \lambda\in\Lambda$ and let $F(\lambda)=\langle u,(\lambda^k)\rangle$. Then $F$ is a holomorphic function in $\mathbb{D}$ (why????) with an accumulation point of zeros inside $\mathbb{D}$. Therefore, $F$ and $u$ are zeroes (why??????), which means that $\{(\lambda^k);\lambda\in\Lambda\}$ spans a dense subspace in $\ell^1(\mathbb{N})$.
 A: For $|\lambda| < 1$, let $f_\lambda =(\lambda^0, \lambda^1,...)  \in l_1$. Let $B = \{ f_\lambda | \lambda \in \Lambda \}$ and $S = \overline{\operatorname{sp}} B$. You want to show $S = l_1$.
Note that since $\Lambda$ has an accumulation point inside the unit ball, it is infinite.
One way is to do this is to show that the only element of $l_1^*$ that satisfies $\phi(x) = 0$ for all $x \in S$ is $\phi=0$ (show this using Hahn Banach).
Since $S$ is the closure of the span of $B$, it is straightforward to showing that this is equivalent to showing that $\phi(x) = 0$ for $x \in B$.
So, suppose $\phi \in l_1^*$ and $\phi(x) = 0$ for all $x \in B$. We want to show that $\phi = 0$.
We can write $\phi(x) = \sum_k \Phi_k x_k $, where $\Phi \in l_\infty$. Define
$F_\phi(z) = \sum_k \Phi_k z^k$, and note that since the $\Phi_k$ are bounded, $f_\phi$ is analytic on the open unit ball. Furthermore, since $F_\phi(\lambda) = 0$ for all $\lambda \in \Lambda$ hence the identity theorem shows that $F_\phi = 0$, or equivalently, $\Phi_k = 0$ for all $k$. Hence $\phi = 0$.
