§ 2. Cauchy’s Integral Formulas
51
f (z) = . . . + c p−1 z
p−1 + c p z
p
with c p = 0 ,
(5.18”)
in other words, if f (z) z
p large large z. Otherwise, the expression essential singular point at infinity is used, for example in the case of e
z . Then,
expansion (18”) leads to setting
Res(f, ∞) = −c −1
as above, whence
|z|=R
f (z)dz = −2πi Res(f, ∞) for large R
(5.19)
follows, attention needing to be paid to the sign on the right hand side. . .
A rational function only has polar singularities in ˆ
C = C ∪ {∞}, in other
words is meromorphic on ˆ
C, and no other function satisfies this property.
First of all, such a function f can only have finitely many poles because,
even if it has a pole at infinity, it is holomorphic outside a compact disc
D. However, it can only have finitely many poles in D. Multiplying f by a
polynomial chosen so as to remove the poles of f in D, we get a function
whose only possible singularity in ˆ
C is a pole at infinity; it is, therefore, an
entire function on C whose order of magnitude at infinity is that of a power
of z, and hence, by Liouville’s theorem (Chapter VII, § 4, n
◦ 18, theorem 15),
is a polynomial. The result follows.
(v) The conformal invariance of a residue. At first sight, the definition
of the residue of f at infinity seems strange; apart from the sign chosen, in
the series expansion of f (z), the residue is the coefficient of a power of z
approaching 0 as z approaches infinity, whereas the exact opposite holds for
residues at points = ∞. This requires some explanation, found by replacing
the variable z by 1/z.
Indeed, computing ` a la Leibniz, the change of variable z = 1/ζ transforms
the expression
34 ω = f (z)dz into = f (1/ζ)d(1/ζ) = g(ζ)dζ, where
g(ζ) = −f (1/ζ)ζ
−2 = − (. . . + c −1 ζ + . . .) ζ
−2 = . . . − c −1 /ζ + . . . .
Therefore,
Res(f, ∞) = −c −1 = Res(g, 0) .
This suggests that the residue of a function f at a point a in fact involves
the differential form ω = f (z)dz rather than the function f itself; it would
therefore be better to write it Res(ω, a). To justify this, let us generalize the
situation by considering a holomorphic function f on U −S, where S is closed
and discrete in an open subset U of C, and let us investigate what happens
34 It is some sort of differential in the sense of Chapter IX; the transformation it is
made to undergo here consists in computing its “ inverse image ” under z → 1/z.
51
f (z) = . . . + c p−1 z
p−1 + c p z
p
with c p = 0 ,
(5.18”)
in other words, if f (z) z
p large large z. Otherwise, the expression essential singular point at infinity is used, for example in the case of e
z . Then,
expansion (18”) leads to setting
Res(f, ∞) = −c −1
as above, whence
|z|=R
f (z)dz = −2πi Res(f, ∞) for large R
(5.19)
follows, attention needing to be paid to the sign on the right hand side. . .
A rational function only has polar singularities in ˆ
C = C ∪ {∞}, in other
words is meromorphic on ˆ
C, and no other function satisfies this property.
First of all, such a function f can only have finitely many poles because,
even if it has a pole at infinity, it is holomorphic outside a compact disc
D. However, it can only have finitely many poles in D. Multiplying f by a
polynomial chosen so as to remove the poles of f in D, we get a function
whose only possible singularity in ˆ
C is a pole at infinity; it is, therefore, an
entire function on C whose order of magnitude at infinity is that of a power
of z, and hence, by Liouville’s theorem (Chapter VII, § 4, n
◦ 18, theorem 15),
is a polynomial. The result follows.
(v) The conformal invariance of a residue. At first sight, the definition
of the residue of f at infinity seems strange; apart from the sign chosen, in
the series expansion of f (z), the residue is the coefficient of a power of z
approaching 0 as z approaches infinity, whereas the exact opposite holds for
residues at points = ∞. This requires some explanation, found by replacing
the variable z by 1/z.
Indeed, computing ` a la Leibniz, the change of variable z = 1/ζ transforms
the expression
34 ω = f (z)dz into = f (1/ζ)d(1/ζ) = g(ζ)dζ, where
g(ζ) = −f (1/ζ)ζ
−2 = − (. . . + c −1 ζ + . . .) ζ
−2 = . . . − c −1 /ζ + . . . .
Therefore,
Res(f, ∞) = −c −1 = Res(g, 0) .
This suggests that the residue of a function f at a point a in fact involves
the differential form ω = f (z)dz rather than the function f itself; it would
therefore be better to write it Res(ω, a). To justify this, let us generalize the
situation by considering a holomorphic function f on U −S, where S is closed
and discrete in an open subset U of C, and let us investigate what happens
34 It is some sort of differential in the sense of Chapter IX; the transformation it is
made to undergo here consists in computing its “ inverse image ” under z → 1/z.