# Complex Analysis: Power Series and Convergence

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

Problem

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

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? –  user20266 Dec 7 '11 at 17:55
@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? –  Mathmo Dec 7 '11 at 17:59
no, I'm referring to $1/R = \limsup |c_n|^{(1/n)}$ –  user20266 Dec 7 '11 at 18:02
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! –  Pierre-Yves Gaillard Dec 7 '11 at 19:19
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. –  joriki Dec 7 '11 at 20:21