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Laurent Series

Laurent Series.

The series $\underset{k=- \infty}{\overset{+ \infty}{\sum}} a_k$ is convergent and as as its sum if both $\underset{k=- \infty}{\overset{0}{\sum}} a_k$ and $\underset{k=1}{\overset{+ \infty}{\sum}} a_k$ are convergent and if its sum is equal to the sum of the sums of these series.

Theorem 8.2.1 Take $f(z)= \underset{k=- \infty}{\overset{+ \infty}{\sum}} a_k z^k$ .

The series is convergent in the domain $D=\{ z \in \mathbb{C}; \; r<\vert z\vert<R \}$ , where

$\displaystyle r=\underset{k \rightarrow + \infty}{\text{lim sup}} \vert a_{-k}\... ...rac {1}{\underset{k \rightarrow + \infty}{\text{lim sup}}} \vert a_k\vert^{1/k}$
If , then the function is analytic in the annulus .

Theorem 8.2.2 (Laurent expansion of a function) Suppose that the function is analytic on the annulus $D=\{ z \in \mathbb{C}; \; r<\vert z\vert<R \}$ . Then has an expansion as a Laurent series in .

This series expansion of is unique and its coefficients are given by:

$\displaystyle a_k= \frac {1}{2 \pi i} \int_C \frac {f(z)}{z^{k+1}} \; dz$

where is any simple smooth loop included in and enclosing the inner boundary circle of .

Corollary 8.2.3 (Generalization) Let be analytic in the annulus $D=\{ z \in \mathbb{C}; \; r<\vert z-z_0\vert<R \}$ . Then has a unique Laurent expansion $\underset{k=- \infty}{\overset{+ \infty}{\sum}} a_k (z-z_0)^k$ . The coefficients are given by the formula

$\displaystyle a_k= \frac {1}{2 \pi i} \int_C \frac {f(z)}{(z-z_0)^{k+1}} \; dz$

where is a any simple smooth loop included in and enclosing the inner boundary circle of .

Example 8.2.4

$\displaystyle \forall z \neq 0, \frac {(z+1)^3}{z} = \frac {1}{z} +3 +3z +z^2.$

Yes, this can happen: sometimes the Laurent expansion contains a finite numbers of terms (when?).

Example 8.2.5

$\displaystyle \frac {1}{z^2(1+z)} = \frac {1}{z^2} ( 1 - z + z^2 - z^3 + \dots ) = z^{-2} - z^{-1} + 1 - z + z^2 - \dots,$

This series converges for $\vert z\vert<1$ . The convergence domain in the plane is the open unit disk centered at the origin; cf Fig 4(a).

**Figure 4:** Convergence domains of Laurent series.
$\begin{figure}\mbox{ \subfigure[]{{\epsfig{file=Laurent1.eps,height=3cm}}} \qquad \subfigure[]{{\epsfig{file=Laurent2.eps,height=3cm}}} }\end{figure}$

Example 8.2.6

$\displaystyle \frac {1}{z-2}$	$\displaystyle = \frac {1}{(z-1)-1} = \frac {\frac {1}{z-1}}{1- \frac {1}{z-1}}$
$\displaystyle \quad$	$\displaystyle = (z-1)^{-1} ( 1 + (z-1) + (z-1)^2 + (z-1)^3 + \dots )$
$\displaystyle \quad$	$\displaystyle = (z-1)^{-1} + 1 + (z-1) + (z-1)^2 + \dots.$

This series is convergent when $\left\vert \frac {1}{z-1} \right\vert <1$ , i.e. when $\vert z-1\vert>1$ (cf Fig 4(a).

Example 8.2.7 Let $f(z)=\frac {1}{(z+1)(z-2)}$ . We wish to expand

as a Laurent series convergent on an annulus.

We decompose as a sum of partial fractions:

$\displaystyle f(z)=\frac {1}{(z+1)(z-2)} = \frac 13 \cdot \frac {1}{z-2} - \frac 13 \cdot \frac {1}{z+1}$

Now, we expand both partial fractions as Laurent series about 0:

$\displaystyle \frac {1}{z+1} = 1- z + z^2 - z^3 + \dots + (-1)^n z^n + \dots$

This series is convergent for $\vert z\vert<1$ .
$\displaystyle \frac {1}{z-2}$ $\displaystyle = \frac {1}{z \left( 1-\frac 2z \right)}$

$\displaystyle \quad$ $\displaystyle = \frac 1z \cdot \left[ 1 + \frac 2z + \left( \frac 2z \right)^2 + \left( \frac 2z \right)^3 + \dots \right]$

$\displaystyle \quad$ $\displaystyle = z^{-1}+ 2 + 4z + 8z^2 + 16 z^3 + \dots$

This series is convergent for $\vert \frac 2z \vert<1$ ,i.e. $\vert z\vert>2$ .

The intersection of the two domains of convergence is empty, so we went in a wrong direction. Let us try in another way:

$\displaystyle \frac {1}{z+1}$ $\displaystyle = \frac {1}{z \left( 1 + \frac 1z \right)}$

$\displaystyle \quad$ $\displaystyle =\frac 1z \cdot \left[ 1 + \frac 1z + \left( \frac 1z \right)^2 +\left( \frac 1z \right)^3 +\dots \right]$

$\displaystyle \quad$ $\displaystyle = z^{-1}+1 + z + z^2 + z^3 + \dots$

This series is convergent for $\vert \left( \frac 1z \right) \vert < 1$ , i.e. $\vert z\vert>1$ .
$\displaystyle \frac {1}{z-2}$ $\displaystyle = \frac {-1}{2}\cdot \frac {1}{1- \frac z2}$

$\displaystyle \quad$ $\displaystyle = -\frac 12 \cdot \left[ 1 + \frac z2 + \left( \frac z2 \right)^2 + \left( \frac z2 \right)^3 + \dots \right]$

$\displaystyle \quad$ $\displaystyle = -\frac 12 \left[ 1 + \frac z2 + \frac 14 z^2 + \frac 18 z^3 + \dots \right]$

This series is convergent for $\left\vert \frac z2 \right\vert <2$ , i.e. $\vert z\vert<2$ .

The intersection of these convergence domains is the annulus displayed on Fig 4(b). On this annulus, a Laurent series expansion of

is:

$\displaystyle f(z)$	$\displaystyle = z^{-1}+1 + z + z^2 + z^3 + \dots -\frac 12 \left[ 1 + \frac z2 + \frac 14 z^2 + \frac 18 z^3 + \dots \right]$
$\displaystyle \quad$	$\displaystyle =z^{-1} + \frac 12 +z +\frac 12 z^2 -\frac {15}{16}z^3 + \dots .$

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