coefficients of a laurent series

**hurz** · April 28th 2011, 01:09 PM

A function $f(z)$ does have a laurent series about the origin, with real $a_n$ coefficients. Show that $\bar{f}(z) = f(\bar{z})$

Regards

**TheEmptySet** · April 28th 2011, 01:22 PM

Originally Posted by hurz

A function $f(z)$ does have a laurent series about the origin, with real $a_n$ coefficients. Show that $\bar{f}(z) = f(\bar{z})$

Regards

So that means that f can be written as

$f(z)=\sum_{n=-\infty}^{\infty}a_n z^n,a_n \in \mathbb{R}$

Now write out

$\overline{f(z)}=\overline{\sum_{n=-\infty}^{\infty}a_n z^n}=...$

$f(\bar{z})=\sum_{n=-\infty}^{\infty}a_n \bar{z}^n=...$

Use some properties of complex numbers and simplify

**girdav** · April 28th 2011, 01:23 PM

So, what did you try ? Write the Laurent series and use the conjugate to compute $\overline{f(z)}$ . You can put the conjugate into the sum because $z\mapsto \overline z$ is continuous.

**hurz** · April 28th 2011, 01:34 PM

Done. Thanks!

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