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I know this question had already been asked and answered here.

If $f: \mathbb{C} \to \mathbb{C}$ is continuous and analytic off $[-1,1]$ then is entire.

However I was trying power series approach. By analyticity, $f$ can be expanded in power series around $\iota$ in radius of convergence $1$. Take the sequence $\frac{\iota}{n}\to0$. Then $f(\frac{\iota}{n})$ is given by power series $\sum a_k(z-\iota)^k$. Can we take limit inside power series? Then $f$ will be analytic at $0$. Similarly on other points of $[-1,1]$.

Problem I am facing is uniform convergence of power series is guaranteed inside open ball. Also Abel's theorem does not help much as I need to go to reverse direction, i.e., using continuity of $f$ to ensure convergence of power series.

Anirban Sarkar
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    There's a simple argument that the Laurent Series at $0$ for $1\lt \vert z \vert \lt \infty$ is in fact a power series, giving the result. But I don't see a way to progress your argument. – user8675309 Jan 28 '24 at 18:15

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