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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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    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

0 Answers 0