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    Laurent expansions of $\frac{z+1}{z(z-1)}$

    P-AZOFF-F01

    Area
    Complex Analysis
    Status
    draft

    Find the Laurent expansions of \(\textstyle { \frac { z + 1 } { z ( z - 1 ) } }\) about

    1. \(z = 0\)

    2. \(z = 1\)

    Solution

    Set \[ F(z)=\frac{z+1}{z(z-1)}. \]

    1. The partial-fraction decomposition is\[ F(z)=-\frac1z+\frac2{z-1}. \]

    One has \[ -\frac1z+\frac2{z-1} = \frac{-(z-1)+2z}{z(z-1)} = \frac{z+1}{z(z-1)}. \]

    2. About \(z=0\), on the annulus \(0<\abs{z}<1\),\[ \boxed{ F(z) = -\frac1z -2\sum_{n=0}^{\infty}z^n. } \]

    1. The factor \(1/z\) requires \(z\neq0\), so theresulting Laurent expansion is valid on \(0<\abs{z}<1\).

    3. About \(z=0\), on the annulus \(\abs{z}>1\),\[ \boxed{ F(z) = -\frac1z +2\sum_{n=0}^{\infty}z^{-n-1}. } \]

    For \(\abs{z}>1\), \[ \frac2{z-1} = \frac2z\frac1{1-z^{-1}} = 2\sum_{n=0}^{\infty}z^{-n-1}. \] Substitution into step 1 gives the displayed expansion.

    4. About \(z=1\), write\[ w=z-1. \]On the annulus \(0<\abs{w}<1\),\[ \boxed{ F(1+w) = \frac2w -\sum_{n=0}^{\infty}(-1)^n w^n. } \]

    Step 1 becomes \[ F(1+w) = -\frac1{1+w} +\frac2w. \] For \(\abs{w}<1\), \[ \frac1{1+w} = \sum_{n=0}^{\infty}(-1)^n w^n. \] Substitution yields the displayed Laurent series, valid away from the center \(w=0\).

    5. About \(z=1\), on the annulus \(\abs{w}>1\),\[ \boxed{ F(1+w) = \frac2w -\sum_{n=0}^{\infty}(-1)^n w^{-n-1}. } \]

    For \(\abs{w}>1\), \[ \frac1{1+w} = \frac1w\frac1{1+w^{-1}} = \sum_{n=0}^{\infty}(-1)^n w^{-n-1}. \] Substitution into \[ F(1+w)=-\frac1{1+w}+\frac2w \] gives the displayed series.

    6. The expansions in steps <1>2–<1>5 exhaust the Laurent expansions onthe maximal annuli centered at \(0\) and \(1\).

    The only singularities of \(F\) are at \(0\) and \(1\). About the center \(0\), the other singularity lies at radius \(1\), producing the maximal annuli \[ 0<\abs{z}<1 \qquad\text{and}\qquad 1<\abs{z}<\infty. \] About the center \(1\), the other singularity \(0\) again lies at distance \(1\), producing \[ 0<\abs{z-1}<1 \qquad\text{and}\qquad 1<\abs{z-1}<\infty. \] Steps <1>2–<1>5 give one Laurent series on each of these maximal annuli.

    7. Q.E.D.

    Steps <1>2–<1>6 give all requested Laurent expansions about both centers.

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