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21.2 The Laurent Expansion 725

21.2 The Laurent Expansion

If f is differentiable in some disk about z 0 , then f (z) has a Taylor series representation about z 0 .
If f is not differentiable at z 0 , then f (z) may have a different kind of series expansion about z 0 ,
which is a Laurent expansion. This will have important applications in evaluating integrals. First
we need some terminology.
The open set of points between two concentric circles is called an annulus. Typically, an
annulus with center z 0 is described by inequalities

r < |z − z 0 | < R,
in which r is the radius of the inner circle and R is the radius of the outer circle. We allow
r = 0, in which case the annulus 0 < |z − z 0 | < R is the open disk of radius R about z 0 with the
center removed. Such an annulus is called a punctured disk. We also allow R =∞, in which
case the annulus r < |z − z 0 | < ∞ consists of all points outside the inner circle. An annulus
0 < |z − z 0 | < ∞ is the entire plane with z 0 removed.
We can now state the fundamental result on Laurent series.

THEOREM 21.5 The Laurent Expansion

Let f be differentiable in the annulus r < |z − z 0 | < R where 0 ≤r < R ≤∞. Then, for each z in
this annulus,

∞

n
f (z) = c n (z − z 0 ) ,
n=−∞
where for each integer n,
1 f (z)
c n = dz
2πi γ (z − z 0 ) n+1
with γ as any closed path in the annulus enclosing z 0 .

A proof is outlined in Problem 11.
The Laurent expansion about z 0 enables us to write, in some annulus about z 0 ,
−1 ∞

n n
f (z) = c n (z − z 0 ) + c n (z − z 0 ) = h(z) + g(z),
n=−∞ n=0
where
−1

h(z) = c n (z − z 0 ) n
n=−∞
c −3 c −2 c −1
= ··· + + +
(z − z 0 ) 3 (z − z 0 ) 2 (z − z 0 )
contains all of the terms in the expansion with negative powers of z − z 0 and
∞
n
g(z) = c n (z − z 0 )
n=0
2
= c 0 + c 1 (z − z 0 ) + c 2 (z − z 0 ) + ···

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October 14, 2010 15:35 THM/NEIL Page-725 27410_21_ch21_p715-728