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Let $f(z)=\displaystyle \sum_{n=0}^{\infty}c_nz^n$ have radius of convergence $R$.

Problem

Prove that $\displaystyle \sum_{n=0}^{\infty}\overline{c_n}z^n$ has radius of convergence $R$ and that $\overline{f(\overline{z})}=\displaystyle \sum_{n=0}^{\infty}{\overline{c_n}}z^n$ for $|z|

Progress

I have previously proven that $\sup\{|z|:c_nz^n\rightarrow 0 \quad as\quad n\rightarrow\infty\}$ is equal to $R$ for such an $f$.

Can I make the following argument?

$c_nz^n\rightarrow 0\Longleftrightarrow c_n\rightarrow 0 \Longleftrightarrow Re(c_n) \rightarrow 0$ and $Im(c_n)\rightarrow 0 \Longleftrightarrow \overline{c_n} \rightarrow 0 \Longleftrightarrow \overline{c_n}z^n \rightarrow 0$

If so, I think this demonstrates that the radii of convergence are equal.

The next part suggests making use of the fact that $z\mapsto \overline{z}$ is continuous, but I'm not sure how that can be applied.

Any help with determining if my working thus far is correct, and assistance with the remainder of the question would be greatly appreciated.

TJO

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    There are formulas for the radius of convergence, which will yield the same result when you replace the coefficients by their conjugates, do you know them?2011-12-07
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    @Thomas Are you referring to the standard definition i.e. replacing $c_nz^n \rightarrow 0$ with $\sum{|c_nz^n|} converges$? If so, can we just observe that $|cn|=|\overline{c_n}|$ and conclude the $R$ values are equal?2011-12-07
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    no, I'm referring to $1/R = \limsup |c_n|^{(1/n)}$2011-12-07
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    Dear TJO. The equality $$\overline{f(\overline z)}=\sum_{n=0}^\infty\ \overline{c_n}z^n$$ for $|z|$$\overline{f(z)}=\sum_{n=0}^\infty\ \overline{c_n}\ \overline z ^n$$ for $|z|2011-12-07
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    @Pierre-YvesGaillard: If you don't mind explaining, how do we arrive at $\overline{f(\overline{z})}=\sum_{n=0}^{\infty}\overline{c_n}z^n$? Can we simply state that for each $n$, $\quad \overline{c_n\overline{z^n}}=\overline{c_n}z^n$?2011-12-07
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    Dear TJO: Yes! (We're also using the fact that $z\mapsto\overline z$ is continuous.) But your argument using the absolute convergence is even simpler. Your other argument with $c_n\,z^n\to0$ is also very nice! (My argument isn't as nice...)2011-12-07
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    @Pierre-YvesGaillard: Excellent; I understand. And thanks!2011-12-07
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    @Pierre-YvesGaillard: Does this mean then that the function $g$ defined as $g(z)=\overline{f(\overline{z})}$ is holomorphic in $D(0;R)$? Is there any way of demonstrating this more explicitly?2011-12-07
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    Dear TJO: Yes! If $f$ is holomorphic at $a$, then $z\mapsto\overline{f(\overline z)}$ is holomorphic at $\overline a$ and its derivative at $\overline a$ is $\overline{f'(\overline a)}$: $$\frac{\overline{f(\overline z)}-\overline{f(\overline a)}}{z-a}=\left(\frac{f(\overline z)-f(\overline a)}{\overline z-\overline a}\right)^-.$$ Other notation: denote $\overline z$ by $z^*$. Then $$\frac{f(z^*)^*-f(a^*)^*}{z-a}=\left(\frac{f(z^*)-f(a^*)}{z^*-a^*}\right)^*.$$ Very good point!2011-12-07
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    The outer equivalences in your chain of equivalences are wrong. Whether $c_nz^n$ tends to zero depends on $|z|$, whereas whether $c_n$ tends to zero doesn't.2011-12-07

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