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I do know that if $f$ is entire, one can expand about $0$ by taylor series, and we can write $f(z) = \sum_{k=0}^{\infty}a_kz^k$ for all $z \in \mathbb{C}$. Hence implying that $\overline{f(\overline{z})}= \sum_{k=0}^{\infty}\overline{a_k}z^k$

However, how does one ensure that $\sum_{k=0}^{\infty}\overline{a_k}z^k$ does have a radius of convergence of infinity, same as $f(z)$ to guarantee the usage of "if a power series $\sum_{k=0}^{\infty}a_k(z-z_0)^k$ has a radius of convergence $R$, then it is analytic on the ball $B(z_0,R)$"

I think i lack some understanding on converging, please highlight to me where i overlooked so i can improve.

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    $\begingroup$ The radius of convergence can be determined from the absolute values of the coefficients in the power series and $|\overline{a}|=|a|$. $\endgroup$ – Mark McClure Apr 4 '17 at 13:57
  • $\begingroup$ If $\sum_{k=0}^\infty a_k z^k$ converges for every $z$ then for every $R$ : $\lim_{k \to \infty} |a_k| R^k = 0$ $\endgroup$ – reuns Apr 4 '17 at 13:57
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Hint:

if $f(z)=\sum_0^\infty a_n z^n$ and the radius of convergence is infinite (i.e. $f(z)$ is an entire function), than we have :

$\lim_{n\to \infty}|a_n|^{\frac{1}{n}}=0$.

And $|\bar a_n|=|a_n|$.

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