52
VIII – Cauchy Theory
to ω = f (z)dz under a conformal representation z → ϕ(z) = ζ of U on an
open subset V of C. If the inverse of ϕ is the map ψ : V −→ U , a formal
calculation shows that ω is transformed into
= f [ψ(ζ)] ψ
(ζ)dζ .
The function f [ψ(ζ)] ψ
(ζ) is holomorphic on V −ϕ(S), and the more general
result we have in mind is the formula
Res(ω, a) = Res [, ϕ(a)] ,
(5.20)
which holds for all a ∈ S and expresses the invariance of the residue of
a holomorphic differential form under a conformal representation. On the
other hand, r´ esidues at a and b = ϕ(a) of the functions f (z) and f [ψ(ζ)]
are obviously not equal : for f (z) = 1/z, a = 0, ϕ(z) = 2z, ψ(ζ) = ζ/2,
f [ψ(ζ)] = 2/ζ = 1/ζ, but ω = dz/z and = dζ/ζ.
To prove (20), we may assume that a = ϕ(a) = 0 and that U is an open
disc centered at 0 containing no singular points of f other than 0. Then –
use the Laurent series– there is a holomorphic function F on U − {a} such
that f (z) = F
(z) + c/z, where c = Res(f, a). Then,
ω = F
(z)dz + cdz/z = dF + cdz/z ,
and so
= F
[ψ(ζ)] ψ
(ζ)dζ + cψ
(ζ)dζ/ψ(ζ) .
As F
[ψ(ζ)] ψ
(ζ) is the derivative of F [ψ(ζ)], the contribution of the first
term to the residue of at b = 0 is zero; Res(, b) is, therefore, up to a
factor c, the coefficient of 1/ζ in the Laurent series of
ψ
(ζ)/ψ(ζ) = [ψ
(0) + . . .] / [ψ
(0)ζ + . . .] .
As ϕ and ψ are mutually inverse, ψ
(0) = 0, and so Res(ψ
/ψ, b) = 1 and
Res(, b) = c = Res(ω, a), which proves (20).
If a conformal representation leaves the residues of a holomorphic differential form ω = f (z)dz invariant, it may be assumed that the integrals of ω
are also left invariant. To see this, consider a path μ in U − S and its image
t → ν(t) = ϕ [μ(t)] under ϕ; to compute the integral
ν
=
ν
f [ψ(ζ)] ψ
(ζ)dζ ,
(5.21)
by definition, ζ needs to be replaced by ν(t) and dζ by ν
(t)dt; ψ(ζ) is thereby
replaced by ψ {ϕ [μ(t)]} = μ(t) since ψ and ϕ are mutually inverse, f [ψ(ζ)] by
f [μ(t)], and ψ
(ζ)dζ by ψ
[ν(t)] ν
(t)dt; but ψ being holomorphic, definitions
obvious imply what we know, namely that
VIII – Cauchy Theory
to ω = f (z)dz under a conformal representation z → ϕ(z) = ζ of U on an
open subset V of C. If the inverse of ϕ is the map ψ : V −→ U , a formal
calculation shows that ω is transformed into
= f [ψ(ζ)] ψ
(ζ)dζ .
The function f [ψ(ζ)] ψ
(ζ) is holomorphic on V −ϕ(S), and the more general
result we have in mind is the formula
Res(ω, a) = Res [, ϕ(a)] ,
(5.20)
which holds for all a ∈ S and expresses the invariance of the residue of
a holomorphic differential form under a conformal representation. On the
other hand, r´ esidues at a and b = ϕ(a) of the functions f (z) and f [ψ(ζ)]
are obviously not equal : for f (z) = 1/z, a = 0, ϕ(z) = 2z, ψ(ζ) = ζ/2,
f [ψ(ζ)] = 2/ζ = 1/ζ, but ω = dz/z and = dζ/ζ.
To prove (20), we may assume that a = ϕ(a) = 0 and that U is an open
disc centered at 0 containing no singular points of f other than 0. Then –
use the Laurent series– there is a holomorphic function F on U − {a} such
that f (z) = F
(z) + c/z, where c = Res(f, a). Then,
ω = F
(z)dz + cdz/z = dF + cdz/z ,
and so
= F
[ψ(ζ)] ψ
(ζ)dζ + cψ
(ζ)dζ/ψ(ζ) .
As F
[ψ(ζ)] ψ
(ζ) is the derivative of F [ψ(ζ)], the contribution of the first
term to the residue of at b = 0 is zero; Res(, b) is, therefore, up to a
factor c, the coefficient of 1/ζ in the Laurent series of
ψ
(ζ)/ψ(ζ) = [ψ
(0) + . . .] / [ψ
(0)ζ + . . .] .
As ϕ and ψ are mutually inverse, ψ
(0) = 0, and so Res(ψ
/ψ, b) = 1 and
Res(, b) = c = Res(ω, a), which proves (20).
If a conformal representation leaves the residues of a holomorphic differential form ω = f (z)dz invariant, it may be assumed that the integrals of ω
are also left invariant. To see this, consider a path μ in U − S and its image
t → ν(t) = ϕ [μ(t)] under ϕ; to compute the integral
ν
=
ν
f [ψ(ζ)] ψ
(ζ)dζ ,
(5.21)
by definition, ζ needs to be replaced by ν(t) and dζ by ν
(t)dt; ψ(ζ) is thereby
replaced by ψ {ϕ [μ(t)]} = μ(t) since ψ and ϕ are mutually inverse, f [ψ(ζ)] by
f [μ(t)], and ψ
(ζ)dζ by ψ
[ν(t)] ν
(t)dt; but ψ being holomorphic, definitions
obvious imply what we know, namely that