complex metric space, algebra that doesn't separate points, extension of baby Rudin

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Let the metric space be $(C([-\pi, \pi]), \mathbb{C})$ with the uniform metric. Define $f_k=e^{ikx}$ where $-\pi\le x\le\pi$. Let $A$ be the algebra of $\{f_k\}$ where $k$ is a nonnegative integer, and let $B=\{g\in (C([-\pi, \pi]), \mathbb{C}))\}$ such that $g(-\pi)=g(\pi)$. Show that $B\neq\bar{A}$ in the metric space defined.

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A uniform limit of analytic functions is analytic, so all you need to do is construct a continuous function $g$ with $g(-\pi)=g(\pi)$, not analytic (for instance, not everywhere differentiable).