Compute the Laurent series centered at \(z=0\) such that \[ \sum_{n=-\infty}^{\infty} a_nz^n =\frac{1}{z(z-1)(z-2)} \] in the region \(1<|z|<2\).
Solution
For \(1<|z|<2\), \[ \boxed{ \frac{1}{z(z-1)(z-2)} =-\frac1{2z} -\sum_{n=2}^{\infty}z^{-n} -\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}.} \]
1. The rational function has the partial-fraction decomposition\[ \frac{1}{z(z-1)(z-2)} =\frac1{2z}-\frac1{z-1}+\frac1{2(z-2)}. \]
The coefficient of \(1/z\) is \[ \left.\frac1{(z-1)(z-2)}\right|_{z=0}=\frac12. \] The coefficient of \(1/(z-1)\) is \[ \left.\frac1{z(z-2)}\right|_{z=1}=-1, \] and the coefficient of \(1/(z-2)\) is \[ \left.\frac1{z(z-1)}\right|_{z=2}=\frac12. \] This gives the stated decomposition.
2. On the region \(|z|>1\),\[ -\frac1{z-1} =-\sum_{n=1}^{\infty}z^{-n}. \]
Since \(|1/z|<1\), \[ \frac1{z-1} =\frac1z\frac1{1-z^{-1}} =\frac1z\sum_{k=0}^{\infty}z^{-k} =\sum_{n=1}^{\infty}z^{-n}. \] Multiplying by \(-1\) gives the claimed expansion.
3. On the region \(|z|<2\),\[ \frac1{2(z-2)} =-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}. \]
Because \(|z/2|<1\), \[ \frac1{2(z-2)} =-\frac14\frac1{1-z/2} =-\frac14\sum_{n=0}^{\infty}\left(\frac z2\right)^n =-\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}. \]
4. Combining <1>1–<1>3 gives the Laurent series on\(1<|z|<2\).
Substitution into the partial fractions gives \[ \frac{1}{z(z-1)(z-2)} =\frac1{2z} -\sum_{n=1}^{\infty}z^{-n} -\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}. \] The first term combines with the \(n=1\) term of the negative-power sum: \[ \frac1{2z}-\frac1z=-\frac1{2z}. \] Hence \[ \frac{1}{z(z-1)(z-2)} =-\frac1{2z} -\sum_{n=2}^{\infty}z^{-n} -\sum_{n=0}^{\infty}\frac{z^n}{2^{n+2}}, \] and both geometric series converge precisely under the simultaneous conditions \(|z|>1\) and \(|z|<2\).