§ 2. Cauchy’s Integral Formulas
49
Supp(μ), neither does f (z)−c
for c
sufficiently near c; when c
approaches c,
f
/(f − c
) obviously converges uniformly to f
/(f − c) on Supp(μ), implying
the result. This also means that the number v μ (f − c) of solutions of f (z) = c
in the interior of μ only depends on the connected component of C − Supp(μ)
containing c.
If for example f has a zero of order p at a and does not vanish elsewhere
on a disc |z − a| < r, then, for any sufficiently small c ∈ C, the equation
f (z) = c has p solutions in the disc. This shows that the image under f of a
disc centered at a contains a disc centered at f (a) and, as a is arbitrary, that
f maps every open subset of G to an open subset of C. Moreover, the roots
of f (z) = c = 0 in |z − a| < r are pairwise distinct for sufficiently small r
since even if f
(a) = 0, f
(z) = 0 for 0 < |z − a| < r if r is sufficiently small,
which prevents f (z) = c from having multiple roots in this disc.
For p = 1, this argument shows that f is injective in the neighbourhood
of a if and only if f
(a) = 0. If f is globally injective on G, then f
(z) = 0
everywhere and as shown above (and in Chapter III, § 5, n
◦ 24, by using
the real-variable version of the local inversion theorem), f is a conformal
representation on an open subset as remarked earlier. In conclusion :
Theorem 7. Any holomorphic and non-constant function f on G maps open
subsets of G to open subsets of C. It is a conformal representation of G
on f (G) if and only if f is injective on G.
(iv) Residues at infinity. The simplest functions the residue formula can
be applied to are rational functions f (z) = p(z)/q(z), where p and q are
polynomials without common roots. As will be seen later, using this method,
the integral over R of any function of this type can be computed, at least when
it converges. But an important theoretical result about residues of rational
functions can now be proved.
Indeed, integrate f over a circle |z| = R going around it once counterclockwise. If R is sufficiently large, we get, up to a factor of 2πi, the sum
of the residues of f at all its poles, which are the roots of q. Now, for large
z, there is an asymptotic estimate of the form f (z) ∼ cz
n , with c = 0 and
n = d
◦ (p) − d
◦ (q), and in particular |f (z)| ≤ M |z|
n , where M is a constant.
The integral over the circle, equal to
1
0
2πif [Re(t)] Re(t)dt ,
(5.12)
where e(t) = exp(2πit), is, therefore, O(R
n+1 ). So it approaches 0 if n ≤ −2,
i.e. if
d
◦ (q) ≥ d
◦ (p) + 2 .
(5.13)
Hence assumption (13) implies the relation
Res(f, a) = 0 .
(5.14)
49
Supp(μ), neither does f (z)−c
for c
sufficiently near c; when c
approaches c,
f
/(f − c
) obviously converges uniformly to f
/(f − c) on Supp(μ), implying
the result. This also means that the number v μ (f − c) of solutions of f (z) = c
in the interior of μ only depends on the connected component of C − Supp(μ)
containing c.
If for example f has a zero of order p at a and does not vanish elsewhere
on a disc |z − a| < r, then, for any sufficiently small c ∈ C, the equation
f (z) = c has p solutions in the disc. This shows that the image under f of a
disc centered at a contains a disc centered at f (a) and, as a is arbitrary, that
f maps every open subset of G to an open subset of C. Moreover, the roots
of f (z) = c = 0 in |z − a| < r are pairwise distinct for sufficiently small r
since even if f
(a) = 0, f
(z) = 0 for 0 < |z − a| < r if r is sufficiently small,
which prevents f (z) = c from having multiple roots in this disc.
For p = 1, this argument shows that f is injective in the neighbourhood
of a if and only if f
(a) = 0. If f is globally injective on G, then f
(z) = 0
everywhere and as shown above (and in Chapter III, § 5, n
◦ 24, by using
the real-variable version of the local inversion theorem), f is a conformal
representation on an open subset as remarked earlier. In conclusion :
Theorem 7. Any holomorphic and non-constant function f on G maps open
subsets of G to open subsets of C. It is a conformal representation of G
on f (G) if and only if f is injective on G.
(iv) Residues at infinity. The simplest functions the residue formula can
be applied to are rational functions f (z) = p(z)/q(z), where p and q are
polynomials without common roots. As will be seen later, using this method,
the integral over R of any function of this type can be computed, at least when
it converges. But an important theoretical result about residues of rational
functions can now be proved.
Indeed, integrate f over a circle |z| = R going around it once counterclockwise. If R is sufficiently large, we get, up to a factor of 2πi, the sum
of the residues of f at all its poles, which are the roots of q. Now, for large
z, there is an asymptotic estimate of the form f (z) ∼ cz
n , with c = 0 and
n = d
◦ (p) − d
◦ (q), and in particular |f (z)| ≤ M |z|
n , where M is a constant.
The integral over the circle, equal to
1
0
2πif [Re(t)] Re(t)dt ,
(5.12)
where e(t) = exp(2πit), is, therefore, O(R
n+1 ). So it approaches 0 if n ≤ −2,
i.e. if
d
◦ (q) ≥ d
◦ (p) + 2 .
(5.13)
Hence assumption (13) implies the relation
Res(f, a) = 0 .
(5.14)