6.5 Laurent Expansion
355
and expanded by the binomial theorem, which now follows from the Taylor
series [Eq. (6.61)].
Noting that for C1 , |z − z0 | > |z − z0 |, whereas for C2 , |z − z0 | < |z − z0 |,
we find
∞
1 f (z )dz f (z) =
(z − z0 )n
n+1
2πi n=0
C1 (z − z0 )
∞
1 +
(z − z0 )−n
(z − z0 )n−1 f (z )dz .
(6.69)
2πi n=1
C2
The minus sign of Eq. (6.68) has been absorbed by the binomial expansion.
Labeling the first series S1 and the second S2 ,
∞
f (z )dz 1 S1 =
(z − z0 )n
,
(6.70)
n+1
2πi n=0
C1 (z − z0 )
which is the regular Taylor expansion, convergent for |z − z0 | < |z − z0 | = r1 ,
that is, for all z interior to the larger circle, C1 . For the second series in
Eq. (6.68), we have
∞
1 S2 =
(z − z0 )−n
(z − z0 )n−1 f (z )dz (6.71)
2πi n=1
C2
convergent for |z− z0 | > |z − z0 | = r2 , that is, for all z exterior to the smaller
circle C2 . Remember, C2 goes counterclockwise.
These two series are combined into one series13 (a Laurent series) by
f (z) =
∞
an(z − z0 )n,
(6.72)
f (z)dz
.
(z − z0 )n+1
(6.73)
n=−∞
where
an =
1
2πi
C
Since, in Eq. (6.72), convergence of a binomial expansion is no problem, C may
be any contour within the annular region r < |z− z0 | < R encircling z0 once in
a counterclockwise sense. The integrals are independent of the contour, and
Eq. (6.72) is the Laurent series or Laurent expansion of f (z).
The use of the contour line (Fig. 6.16) is convenient in converting the
annular region into a simply connected region. Since our function is analytic
in this annular region (and therefore single-valued), the contour line is not
essential and, indeed, does not appear in the final result [Eq. (6.72)]. For n ≥ 0,
the integrand f (z)/(z − z0 )n+1 is singular at z = z0 if f (z0 ) =
0. The integrand
has a pole of order n + 1 at z = z0 . If f has a first-order zero at z = z0 , then
f (z)/(z − z0 )n+1 has a pole of order n, etc. The presence of poles is essential
for the validity of the Laurent formula.
13 Replace
n by −n in S 2 and add.