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    Laurent series of $1/((z-1)(z-2))$ in the annulus $1<|z|<2$

    P-8CA28

    Area
    Complex Analysis
    Topics
    Laurent Series, Partial Fractions
    Seen at
    UGA
    Years
    2018
    Status
    draft

    Find the Laurent series expansion of \[ f(z)=\frac{1}{(z-1)(z-2)} \] valid in the annulus \(1<|z|<2\).

    Solution

    Use partial fractions: \[ \frac1{(z-1)(z-2)}=-\frac1{z-1}+\frac1{z-2}. \] For \(|z|>1\), \[ -\frac1{z-1} =-\frac1z\frac1{1-1/z} =-\sum_{n=0}^\infty z^{-n-1}. \] For \(|z|<2\), \[ \frac1{z-2} =-\frac12\frac1{1-z/2} =-\sum_{n=0}^\infty \frac{z^n}{2^{n+1}}. \] Hence, on the maximal common annulus \(1<|z|<2\), \[ \boxed{ \frac1{(z-1)(z-2)} =-\sum_{n=0}^\infty z^{-n-1} -\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}.} \]

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