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Thread: coefficients of a laurent series

  1. #1
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    coefficients 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
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by hurz View Post
    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
    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
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  3. #3
    Super Member girdav's Avatar
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    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.
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  4. #4
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    Done. Thanks!
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