# coefficients of a laurent series

• Apr 28th 2011, 02:09 PM
hurz
coefficients of a laurent series
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
• Apr 28th 2011, 02:22 PM
TheEmptySet
Quote:

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
• Apr 28th 2011, 02:23 PM
girdav
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.
• Apr 28th 2011, 02:34 PM
hurz
Done. Thanks!