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726 CHAPTER 21 Series Representations of Functions

contains all the nonnegative powers of z − z 0 . The series defining g(z) is a power series about z 0
and so is a differentiable function in the open disk |z − z 0 |< R. Any “bad” behavior of f (z) near
z 0 is contained in h(z).
It can be shown (Problem 12) that, if
∞ ∞

n n
f (z) = c n (z − z 0 ) = d n (z − z 0 )
n=−∞ n=−∞
in some annulus about z 0 , then c n = d n for each integer n. This means that a Laurent expansion is
unique to f and z 0 and will be the same no matter how it is derived. This is important, because
usually, we obtain a Laurent expansion by manipulating known series and very rarely use the
integral formula to compute the coefficients.

EXAMPLE 21.7

The Laurent expansion of e 1/z about 0 is
∞ n
1 1 1 1 1
= 1 + + + ··· ,
n! z z 2! z 2
n=0
z
which is obtained by replacing z with 1/z in the Maclaurin expansion of e . This Laurent
expansion is valid for 0 < |z < ∞, hence, in the entire plane with the origin removed.

EXAMPLE 21.8

5
f (z) = cos(z)/z is differentiable in the annulus 0 < |z| < ∞, which is the entire plane with
the origin removed. We know the Taylor expansion of cos(z) about 0. Therefore, we know the
Laurent expansion of f (z) in 0 < |z| < ∞ is
∞
1 1 (−1) n
f (z) = cos(z) = z 2n
z 5 z 5 (2n)!
n=0
∞ n
(−1)

= z 2n−5
(2n)!
n=0
1 1 1 1 1 1 1 3
= − + − z + z − ··· for z = 0.
z 5 2 z 3 24 z 720 40,320
We can think of
cos(z)
= h(z) + g(z),
z 5
where
1 1 1 1 1
h(z) = − + ,
z 5 2! z 3 4! z
and
1 1
3
g(z) =− z + z − ··· .
6! 8!
g(z) is differentiable for all z and h(z) is differentiable on the plane with the origin removed, so
h(z) + g(z) is differentiable on the plane with the origin removed. It is the behavior of h(z) near
the origin that determines the behavior of cos(z)/z there.
5

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October 14, 2010 15:35 THM/NEIL Page-726 27410_21_ch21_p715-728