## Laurent Series Example

Problem Find all Laurent series centered at $z=0$ for $f(z) = \frac{1}{z(z-1)(z-2)}$.

Solution First, we will find the possible Laurent expansions having $0$ as the center. These are determined by the singularities of $f$, which are at $0$, $1$, and $2$. These determine two circles, wtih center $z=0$ and radii $|z|=1$ and $|z|=2$, and three regions: I, $\{ 0 < |z| <1 \}$; II, $\{ 1 < |z| <2 \}$; and, III, $\{ 2 < |z| \}$. Each of these three regions has a unique Laurent expansion associated with it. There are no other Laurent expansions, centered at $z=0$, for $f$.

To obtain the Laurent expansions for $f$, put $f$ in partial fraction form: $f(z) = \frac{1}{2z} - \frac{1}{z-1} + \frac{1}{2(z-2)}$ In each region, the Laurent expansion is obtained by expnading the terms in powers of $z$. Obviously the first term is already in that form.

Region I. $-\frac{1}{z-1} = \frac{1}{1-z}$. Since $|z|<1$, we may expand this term using an appropriate geometric series: $\frac{1}{1-z}=\sum_{k=0}^\infty z^k$. Similarly, $\text{(1)}\quad \frac{1}{2(z-2)} = -\frac{1}{4}\frac{1}{1-z/2} = - \frac{1}{4} \sum_{k=0}^\infty \frac{z^k}{2^k} = \sum_{k=0}^\infty (-2^{-k-4})z^z,$ which converges in $|z|<2$. Combining the three terms gives us the Laurent expansion valid in region I: $\text{Region I:}\quad f(z) = \frac{1}{2}z^{-1} + \sum_{k=0}^\infty (1 - 2^{-k-4})z^k.$

Region II. Since (1) is an expansion for $\frac{1}{2(z-2)}$ valid in $|z|<2$, we only need to get an expansion for $- \frac{1}{z-1}$ valid in $1 <|z| <2$. This is easy to do: $\text{(2)}\quad - \frac{1}{z-1} = -z^{-1}\frac{1}{1-z^{-1}} = -z^{-1}\sum_{k=0}^\infty z^{-k}= -z^{-1} + \sum_{n=-\infty}^{-2} (-1)z^n,$ which converges for all $|z|>1$. It follows that the expansion valid in region II is $\text{Region II:}\quad f(z) = \sum_{n=-\infty}^{-2} (-1)z^n - \frac{1}{2}z^{-1} +\sum_{n=0}^\infty (-2^{-n-4})z^n$

Region III. This region is exterior to the circle $|z|=2$. We need to expand all three terms in powers of $z^{-1}$. Equation (2) gives us that for $- \frac{1}{z-1}$. The only term left is $\frac{1}{2(z-2)}$. Again, this is easy: $\text{(3)}\quad \frac{1}{2(z-2)}=\frac{1}{2} z^{-1}\frac{1}{1-\frac{2}{z}} = \frac{1}{2} z^{-1}\sum_{k=0}^\infty 2^k z^{-k} = \frac{1}{2} z^{-1}+\sum_{k=1}^\infty 2^{k-1} z^{-k-1} = \frac{1}{2}z^{-1} + \sum_{n=-\infty}^{-2}2^{-n-2}z^n$ Finally, using (2) and (3), we see that the Laurent expansion for region III is given by $\text{Region III:}\quad f(z) = \frac{1}{2}z^{-1} -z^{-1} - z^{-2} + \sum_{n=-\infty}^{-3}(-1)z^n + \frac{1}{2}z^{-1} + z^{-2} +\sum_{n=-\infty}^{-3}2^{-n-2}z^n = \sum_{n=-\infty}^{-3}(2^{-n-2}-1)z^n$

Updated 3/3/2018 (fjn)