 Solution

The complex Hahn-Banach theorem says that every bounded complex-linear functional on a linear subspace of a normed complex space extends to the whole space with the same norm.

For the separation assertion choose $x_0\notin M$ and $d=\mathrm{dist}(x_0,M)>0$. On $M+\mathbb{C}{x}_0$ define $g_0(m+\lambda x_0)=\lambda d$. The inequality $\Vert m+\lambda x_0\Vert \ge |\lambda |d$ makes its norm at most one; approximating the distance $d$ shows the norm is exactly one. Hahn-Banach gives an extension $g$ of norm one annihilating $M$.

Each geometric sequence satisfies $\Vert x_k{\Vert }_1=(1-|{\beta }_k|{)}^{-1}\le (1-\delta {)}^{-1}$. Let $M$ be their closed linear span. If $M\ne {\ell }^1$, the separation just proved gives a nonzero bounded functional annihilating all $x_k$. Represent it by $b\in {\ell }^{\infty }$ and formThe series converges locally uniformly and defines a holomorphic function. It vanishes at every ${\beta }_k$. These distinct points lie in the compact disk $|z|\le \delta <1$, so have an accumulation point inside the domain. The identity theorem gives $F=0$, hence every Taylor coefficient $b_j=0$, contradicting the functional's nonzero norm. Therefore