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How do I find the region of convergence in the following case:

Let $f(z)=\sum_{i=0}^\infty a_iz^i$ be a power series with a positive radius of convergence. Determine the region of convergence of the Laurent series $\sum_{i=-\infty}^\infty a_{\vert i\vert}z^i$ and identify the function which it represents there.

The question in the book (Theory of complex functions, Remmert, page 360)

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  • $\begingroup$ \begin{align} \sum_{-\infty}^\infty a_{|k|}z^k &= \sum_1^\infty a_{|k|}z^k + \sum_{-\infty}^0 a_{|k|}z^k \\&= \sum_1^\infty a_kz^k + \sum_0^\infty a_k z^{-k} \\&= -a_0 + \sum_0^\infty a_kz^k + \sum_0^\infty a_k z^{-k} \\&= -a_0 + \sum_0^\infty a_k \big(z^k + z^{-k} \big) \\&= a_0 + \sum_1^\infty a_k \big(z^k + z^{-k} \big) \end{align} $\endgroup$ – Brevan Ellefsen Apr 23 at 5:24
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The series is nothing but $f(z)+ \sum\limits_{n=1}^{\infty} a_n\frac 1 {z^{n}}$. It converges if $|z| <R$ and $|\frac 1 z|<R$ ie, if $\frac 1 R <|z|<R$. The sum is $f(0)+g(z)+g(\frac 1 z)=f(z)+f(\frac 1 z)-f(0)$ (where $g(z)=f(z)-f(0)$).

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