Conway J. B. - Functions of One Complex Variable II (1995) Errata

Corrections
for
Functions of One Complex Variable, II
by
John B Conway
This is a list of corrections for my book Functions of One Complex Vari-

able, II. This is also available from my WWW page (http://www.math.utk.
edu/∼conway).
Thanks to R B Burckel.
I would appreciate any further corrections or comments you
wish to make.
Minor corrections
Page Line From To
ix 10 built
Pm in built-in
Pm j
3 22 j=1 n(γ; a) = 0 j=1 n(γ ; a) = 0

∂f 1 ∂f ∂f ∂f 1 ∂f ∂f
7 11 = −i . = +i .
∂z 2 ∂x ∂y ∂ z̄ 2 ∂x ∂y
13 18 Area (Dn ∩ Gk ) Area (f (Dn ) ∩ Λk )
15 16 ux y ux dy
22 -2 analtyic analytic
23 7 approach a approach a
24 18 ε
24 -2 ε
25 23 approaches approach
25 -13 free analyticity free analytic
25 -10 analytically analyticity
25 -7 h(0) = 0 h(0) = a
25 -5 π2 < arg(z − r) < π2 + α} π/2 − α < arg(z − t) < π/2 + α}
32 7 f (B(b; ε)) f (B(a; ε))
33 11 bound region bounded region
34 -7 \Xn \τ (Xn )
37 17 Schwartz Schwarz
51 15 Jordan region simple Jordan region
51 17 Jordan simple Jordan
51 -4 Jordan simple Jordan
53 12 Jordan simple Jordan
Z
θ2 Z θ2
0
54 11 = iθ iθ
τ (−1 + re )rie dθ. ≤ |τ 0 (−1 + reiθ )|r dθ.

θ1 θ1
54 12 the angle the largest angle

54 12 1 + reiθ −1 + reiθ
54 14 Schwartz Schwarz
55 -18 Jordan simple Jordan
56 -10 polynimally polynomially
67 14 ≤ ≥
67 15 ≤ ≥
69 -18 (7.5) (1.4)
69 -2 an ≥ g (n) (0)/n! |an | ≥ |g (n) (0)|/n!
71 -9 G G (italics)
71 -7 the j’s and k’s here should
be slanted.
77 3 φ1 (∂ D), φ1 (∂ D) = φ1 (∂ K00 ) φ1 (∂ D) = φ1 (∂ K00 )
81 -7 {Φ0 , Φ1 , . . . , Φn } Φ0 , Φ 1 , . . . , Φn
82 -3 Thus there are Jordan arcs Since Cj is an analytic curve, there
are Jordan arcs
83 1 vlaue value
82 -2 η1 (0) 6= η2 (0), η1 (0) 6= η2 (0),|ηi (t) − ai | < ε,
83 5 and ins C ⊆ G and C ⊆ {z ∈ G : dist (z, Cj ) < ε},
so ins C ⊆ G
83 8 such that φ(ins C) = ins γ such that φ(ins C) ⊆ ins γ
84 3 (z − a)−1 (z − α)−1
84 -1 multiplicites multiplicities
85 9 C C
85 18 set is non-empty set contains 0
91 11 that is insistent on that requires
91 -8 inner circle of Ω. inner circle of Ω and orient γ1
so that n(φ(γ1 ); 0) = −1.
92 13 with ψ. with ψ. Since f is a conformal
equivalence we have
n(ψ(γ1 ); 0) = n(f (φ(γ1 )); 0) = −1.
92 -5 So n(γ1 ; 0) = −1, n(γ0 ; 0) = 1, So n(γj ; 0) = 0 for 2 ≤ j ≤ n
and n(γj ; 0) = 0 for 2 ≤ j ≤ n. and we can orient γ0 and γ1
such that n(γ1 ; 0) = −1 and
n(φ0 ; 0) = 1.
93 11 anlytic Jordan analytic n-Jordan
94 3 Argument Principle Argument Principle, if ζ ∈ / φ(γj )
for 0 ≤ j ≤ n, then
94 8 that 0 ≤ j ≤ n and |ζ| =
6 rj , that for 0 ≤ j ≤ n,
97 -11 f :G→Λ f :Ω→Λ
98 -2 n(γj , a) n(γj ; a)
rj2 rj2
101 -12
z̄ − āj z̄ − āj
102 -8 (14.7.14) (14.7.16)
X n X ∞
103 -13
k=m k=m
106 10 It follows that g(D) = D and It follows from Proposition 7.5
so g(z) − λ(z − a)(1 − āz)−1 . that g is a Möbius transformation.
107 16 thereom. theorem.
107 -14 equivalence equivalences
110 20 z ∈ T (w) z ∈ T (W )
110 -4 bet be
113 -2 (G, τ ) (G1 , τ1 )
114 -5 But the only way such a But according to Proposition 14.1.1,
conformal equivalence can G = C and h(z) = az + b for complex
exist is if G = C. But numbers a and b with a 6= 0.
then Proposition 14.1.1 implies
that h(z) = az + b for
complex numbers a and b
with a 6= 0.
116 10 dentoed denoted
119 10 Im z = Im M −1 M (z) < Im z = Im M M −1 (z) < Im M −1 (z)
≤ Im z
119 -10 c, d ∈ Z} c, d ∈ Z and c, d occur in some
M in G}
119 -7 G G
121 10 an and
122 5 no common divisor, no common divisor,
there is an odd integer there is an odd integer a
d such that b and d have no
common divisor and there
is an odd integer a
122 -14 a neighborhood of z+ and an open neighborhood
of z+ (Verify!) and
122 -4 First λ First, λ
124 6 Proposition 2.1 Theorem 1.3 and Proposition 2.1
124 15 F|B F|B
124 -7 w in B z in B
126 15 continuation continuation along γ
126 18 continuation and continuation, gi (∆i ) ⊆ G, and
126 20 (Gn , ∆0 ), (Gn , ∆0 ) and g(∆) ⊆ G,
126 -9 path γ path γ such that gt (∆t ) ⊆ G for all t
126 -8 z in ∆t z (italics) in ∆t
127 2 neighborhood of α0 neighborhood of α0 that is
contained in Ω
127 5 g(α0 ) and g(α0 ) = 0 and
127 7 continuation continuation along γ
127 8 that gt (∆t ) that ∆t ⊆ Ω and gt (∆t )
127 10 continuation and continuation with ∆i ⊆ Ω and
127 15 Since h0 ∈ F, F =6 ∅. Hence h0 ∈ F and F = 6 ∅.
127 -16 B(αO ; δ) B(α0 ; δ)
128 8 the function a function
128 12 continuation of h ever continuation of h in G
with values in D ever
128 -8 g 0 (h(αO ))κ g 0 (h(α0 ))κ
129 4 the function a function
129 19 Thus the Thus (there is something
extra to do here) the
129 -7 appraches approches
131 -7 f (a) = 0 f (a) = a

1 + w̄ 1 + z w̄
205 -10 Re Re
1 − z w̄ 1 − z w̄
More substantial corrections

Page Line From To
13 14 Because f (∂Dn ) is a The set G\ ∪n Dn can be wr
smooth curve, countable union of compact s
∪j Kj (Why?). Since f is ana
locally Lipschitz. Thus Area
for each j ≥ 1. Thus
82 20 Theorem 3.4 should also show
that if G is on the “left”
of a boundary curve γ
of G, then Ω is on
the“left” of φ(γ).
136 -8 Add the following sentence as a separate paragraph.
The treatment in this section and the next are based on Duren [1983].
385 In the appropriate place, add the following reference.
P L Duren [1983], Univalent Functions, Springer-Verlag, New York.

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Corrections

Functions of One Complex Variable, II

This is a list of corrections for my book Functions of One Complex Vari-

54 12 the angle the largest angle

More substantial corrections

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