Find the Laurent expansions of \(\frac { z + 1 } { z ( z - 1 ) ^ { 2 } }\) about
\(z = 0\)
\(z = 1\)
Hint: Recall that power series can be differentiated.
Solution
Set \[ F(z)=\frac{z+1}{z(z-1)^2}. \]
1. One has the partial-fraction decomposition\[ F(z) = \frac1z -\frac1{z-1} +\frac2{(z-1)^2}. \]
Multiplying the proposed identity by \(z(z-1)^2\) gives \[ (z-1)^2-z(z-1)+2z = z+1. \] Thus the two rational functions agree.
2. About \(z=0\), on \(0<\abs{z}<1\),\[ \boxed{ F(z) = \frac1z +\sum_{n=0}^{\infty}(2n+3)z^n. } \]
For \(\abs{z}<1\), \[ \frac1{1-z} = \sum_{n=0}^{\infty}z^n. \] Differentiating term by term gives \[ \frac1{(1-z)^2} = \sum_{n=0}^{\infty}(n+1)z^n. \] Since \[ -\frac1{z-1}=\frac1{1-z}, \qquad \frac2{(z-1)^2}=\frac2{(1-z)^2}, \] step 1 becomes \[ \begin{aligned} F(z) &= \frac1z +\sum_{n=0}^{\infty}z^n +2\sum_{n=0}^{\infty}(n+1)z^n\\ &= \frac1z +\sum_{n=0}^{\infty}(2n+3)z^n. \end{aligned} \] The pole at the center excludes \(z=0\), so this is valid on \(0<\abs{z}<1\).
3. About \(z=0\), on \(\abs{z}>1\),\[ \boxed{ F(z) = \sum_{k=2}^{\infty}(2k-3)z^{-k}. } \]
For \(\abs{z}>1\), \[ -\frac1{z-1} = -\frac1z\frac1{1-z^{-1}} = -\sum_{n=0}^{\infty}z^{-n-1}, \] and differentiation of the geometric series gives \[ \frac2{(z-1)^2} = \frac2{z^2}\frac1{(1-z^{-1})^2} = 2\sum_{n=0}^{\infty}(n+1)z^{-n-2}. \] Adding these to \(1/z\) as in step 1, the \(z^{-1}\) terms cancel. For \(k\geq2\), the coefficient of \(z^{-k}\) is \[ -1+2(k-1)=2k-3. \] This yields the displayed Laurent series.
4. About \(z=1\), write\[ w=z-1. \]On \(0<\abs{w}<1\),\[ \boxed{ F(1+w) = \frac2{w^2} -\frac1w +\sum_{n=0}^{\infty}(-1)^n w^n. } \]
Substituting \(z=1+w\) into step 1 gives \[ F(1+w) = \frac1{1+w} -\frac1w +\frac2{w^2}. \] For \(\abs{w}<1\), \[ \frac1{1+w} = \sum_{n=0}^{\infty}(-1)^n w^n. \] Substitution proves the formula.
5. About \(z=1\), on \(\abs{w}>1\),\[ \boxed{ F(1+w) = -\frac1w +\frac2{w^2} +\sum_{n=0}^{\infty}(-1)^n w^{-n-1}. } \]
For \(\abs{w}>1\), \[ \frac1{1+w} = \frac1w\frac1{1+w^{-1}} = \sum_{n=0}^{\infty}(-1)^n w^{-n-1}. \] Insert this into the recentered expression from step 4.
6. Steps <1>2–<1>5 give all Laurent expansions on the maximal annuliabout \(0\) and \(1\).
The only singularities of \(F\) are \(0\) and \(1\). Their mutual distance is \(1\). Hence the maximal annuli centered at \(0\) are \[ 0<\abs{z}<1 \qquad\text{and}\qquad 1<\abs{z}<\infty, \] while those centered at \(1\) are \[ 0<\abs{z-1}<1 \qquad\text{and}\qquad 1<\abs{z-1}<\infty. \] The preceding four steps supply one Laurent series on each such annulus.
7. Q.E.D.
Step 6 gives the complete collection of requested expansions.