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

Regards

Printable View

- Apr 28th 2011, 01:09 PMhurzcoefficients of a laurent series
A function $\displaystyle f(z)$ does have a laurent series about the origin, with real $\displaystyle a_n$ coefficients. Show that $\displaystyle \bar{f}(z) = f(\bar{z})$

Regards - Apr 28th 2011, 01:22 PMTheEmptySet
So that means that f can be written as

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

Now write out

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

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

Use some properties of complex numbers and simplify - Apr 28th 2011, 01:23 PMgirdav
So, what did you try ? Write the Laurent series and use the conjugate to compute $\displaystyle \overline{f(z)}$. You can put the conjugate into the sum because $\displaystyle z\mapsto \overline z$ is continuous.

- Apr 28th 2011, 01:34 PMhurz
Done. Thanks!