Singularities/Residues/ Applications 667
Thus the residue of f at a is equal to the coefficient A_ 1 in the Laurent
expansion of f(z) valid in a deleted neighborhood of a.t
Examples
- Resz=O ~ = Resz=o( * + ~ + ~ + · · ·) = 1.
- Resz=l (z!l) 2 = 2 ~i fa+ ((~}) 2 = 0, by 7.8-13.
Alternatively, we may observe that 1/(z - 1)^2 is just the Laurent ~x
pansion of the same function in the vicinity of z = 1. Hence A_ 1 = 0 and
the residue at z = 1 is zero.
Definition 9.12 ;Let a = oo be either a regular point or an isolated sin-
gularity off, and let c-: (=Re-it, 0:::; t:::; 27r, be a circle contained in
a deleted neighborhood N' ( oo) of oo where f is analytic. Then the residue
of f at oo is defined by
~~f(z) = 2 ~i j J(()d( (9.7-3)
a-
The notation c-is used to emphasize that in this case the circle. C
is described once in the negative (clockwise) direction. Again, the value
obtained in (9.7-3) is independent of Ras long as C* remains in a region
Ro < JzJ < oo where f is analytic.
Let ( = 1/z, d( = -dz/z^2 • If ( = Re-it we have z = (1/R)eit
(0:::; t:::; 27r), so that c-maps into the circle 1+: z = 1/Reit, which is de-
scribed once in the positive direction, as the notation indicated (Fig. 9.5).
Then (9.7-3) becomes
Resf(z)=--^1 j dz 1 ( 1)
2
. f(l/z)2 =-Res2f -
z=oo 7l"Z Z z=O Z Z
(9.7-4)
-y+
If f(z) = E~: Bmzm is the Laurent expansion of f(z) valid for JzJ >
R 0 , and we take R > R 0 , we get
B = _1 J f(()d(
m 27l"i (m+l
a+
tThis fact was used by A. L. Cauchy to define the residue of a function at a pole
(Sur un nouveau genre de calcul analogue au calcul infinitesimal, Exer. Math.,
Paris, 1826).