Description

d1(fn,f) ≤ d2(fn,f) ≤ … ≤ d∞(fn,f).
We write d(fn,f) as a finite sum (denoted by An) plus its tail (denoted by Bn), with N some natural number:
.
Since limn→∞ dk(fn,f) = 0 we have limn→∞ An = 0. For Bn (we have omitted absolute value signs in these calculations since all terms are nonnegative),
.
Thus limn→∞ d(fn,f) = 0.
(b) =⇒ (a): Suppose that
.
As a series of nonnegative terms, we have

for any N ∈ N. Thus,
. (1)
The function f : [0,∞) → R defined by is only zero at x = 0, and limx→∞ f(x) = 1. Therefore, (1) implies that limn→∞ dN(fn,f) = 0, ie. fn → f uniformly on KN. Since this is true for any N ∈ N, and for any compact K ⊂ Ω we have KN ⊃ K for some N, we conclude that fn → f uniformly on compact subsets of Ω.
2. Since g is nonconstant and holomorphic, z0 is an isolated zero of g, so we can write
g(z) = (z − z0)mh(z) (2)
for all z in an open neighbourhood U1 of z0, and where h : U1 → C is holomorphic and nonvanishing. Take an open U2 ⊂ U1 such that z0 ∈ U2 and U2 is simply connected, then we can define a holomorphic function H : U2 → C such that eH(z) = h(z) for all z ∈ U2. We then define ϕ : U2 → C by
ϕ(z) = (z − z0)eH(z)/m
so that ϕm(z) = g(z) by (2). We then have
.
Then ϕ′(z0) ̸= 0, so by problem 4 of homework 4 we find an open disk U3 ⊂ U2 centred at z0 such that ϕ|U3 is injective. By the open mapping theorem im(ϕ|U3) contains a disk Dr(z0) for some r > 0, so taking V to be (ϕ|U3)−1(Dr(z0)) we see that ϕ : V → Dr(z0) satisfies properties (a) and (b).
3. (a) Taking the hint, define Ψ : D → D by Ψ = ψf(w) ◦ f ◦ ψw−1. As a
composition of holomorphic functions Ψ is holomorphic, and
.
By the Schwarz lemma, |Ψ(z)| ≤ |z| for all z ∈ D, ie.
. (3)
By substituting z with ψw(z) in (3) we obtain ρ(f(z),f(w)) ≤ ρ(z,w). If φ : D → D is an automorphism, we apply the inequality twice:
ρ(z1,z2) = ρ((φ−1 ◦ φ)(z1),(φ−1 ◦ φ)(z2)) ≤ ρ(φ(z1),φ(z2)) ≤ ρ(z1,z2), so ρ(φ(z1),φ(z2)) = ρ(z1,z2).
(b) By (a), for any w,z ∈ D we have
.
Rearrange to get
, (4)
so letting w → z in (4) we see that
.
4. (a) Let γ be a curve from z1 to z2. By the chain rule, we have
Z 1 Z 1
∥(f ◦ γ)′(t)∥(f◦γ)(t)dt ≤ ∥γ′(t)∥γ(t)dt.
0 0
(b) Let φ : D → D be an automorphism. Using part (a), d(z1,z2) = d((φ−1 ◦ φ)(z1),(φ−1 ◦ φ)(z2)) ≤ d(φ(z1),φ(z2)) ≤ d(z1,z2), so d(φ(z1),φ(z2)) = d(z1,z2).
Conversely, suppose φ : D → D is a function such that d(φ(z1),φ(z2)) = d(z1,z2).
(c) We construct a family of automorphisms D → D of the form
φz1,z2 = e−iθψz1,
where θ = arg(ψz1(z2). We see that φz1,z2(z1) = 0 and φz1,z2(z2) = |ψz1(z2)|. Since |ψα| is continuous in α (homework 1, composed with z → |7 z|), by the intermediate value theorem each s ∈ [0,1) is of the form |ψz1(z2)| for some z1,z2 ∈ D.
(d) We first show that by calculating the integral for

γ(t) = st, the straight line from 0 to s.

Next we show . We write γ(t) = x(t)+iy(t) and note that the integral is minimized if x (t) is monotone and does not change sign. Thus
(5) We substitute η = x(t) so that (5) equals

by the same calculation as above. Thus .
(e) By (b), d is preserved by automorphisms, so
.
5. By proposition 1.1 in chapter 8 of the textbook, f′(z) ̸= 0 for all z ∈ C. If f is a polynomial, then f must have degree 1, otherwise f′ has a zero by the fundamental theorem of algebra. Therefore it suffices to show f is a polynomial of any degree.
Denote the image of f by U. Since f is holomorphic and injective, U is simply connected, and by the open mapping theorem U is open, so if U is a proper subset of C then by the Riemann mapping theorem there exists a conformal map ϕ : U → D. Then we obtain a holomorphic injective map ϕ ◦ f : C → D, a contradiction to Liouville’s theorem. Thus U = C. In particular, there exists a unique zero of f, and so we assume without loss of generality that f(0) = 0, denote the multiplicity of this zero by n ∈ N. Denote the positively oriented unit circle by C and the negatively oriented unit circle by C←. By the argument principle we have
(6)
and so a short calculation shows, where ), that
where by (2).
Since g has no zeros inside or on C, by the argument principle g has a pole of order n at 0, so by problem 5 (b) of homework 3, f is a polynomial of degree n, completing the proof.
6. If Ω = C then by problem 5 we have f1(z) = α1z + β1 and f2 = α2z + β2 for some α1,α2,β1,β2 ∈ C{0}. Then f1(a) = f2(a) and f1(b) = f2(b) imply
, a contradiction to a and b being distinct. Thus Ω is a proper
subset of C.
As Ω is proper and simply connected, by the Riemann mapping theorem there exists a conformal map F : Ω → D such that F(f1(a)) = 0. We define two maps φ1,φ2 : D → D by
φj = F ◦ fj ◦ F −1 ◦ ψF(a),
where j = 1,2. As a composition of conformal maps, φj is an automorphism, so
φj = eiθjψζj
for some θj ∈ [0,2π) and ζj ∈ D. However, we calculate φj(0) = 0, so φj is a rotation. Then, because φ1 and φ2 agree at (ψF(a) ◦ F)(b), and are nonzero at that point, we must have φ1 = φ2. Composing both sides of this inequality with ψF(a) ◦ F on the right and F −1 on the left, we obtain f1 = f2.
7. By Montel’s theorem it suffices to show G is uniformly bounded on compact subsets of U. Let K ⊂ U be compact, and let γ be a rectifiable loop in UK such that any homotopy from γ to a point contains all of K in its image (ie. K is in the region bounded by γ), and such that there exists an a > 0 such that dist(z,K) ≥ a for all z ∈ γ. Denote the length of γ by L. Since f ∈ F and F is a normal family, by Montel’s theorem there exists an M > 0, independent of f, such that supz∈U |f(z)| ≤ M. Let z0 ∈ K, then by the Cauchy integral
5
formula and ML-estimate we have

Since this bound is independent of f or z0, f′ is uniformly bounded on K, completing the proof.
8. Let fn : D → C be defined by fn = z + nz2 for each n ∈ N. Evidently we have fn holomorphic, fn(0) = 0, and f′(0) = 1 for each n ∈ N, so {fn}n∈N is a sequence in F. Applying the results of problem 7 twice, we see that if H = {f′′ | f ∈ F} is not a normal family then neither is F. Indeed, the sequence has no convergent subsequence, since every subsequence is unbounded. Thus F is not a normal family.
9. We imitate the proof in the textbook of the Hurwitz theorem. Supposethat there is a z ∈ Ω such that f(z) = 0. If f is not identically zero then z is an isolated zero, so choose a circle γ (positively oriented) with z in its interior, small enough so that f(ζ) ̸= 0 for all ζ ̸= z in an open set containing γ and its interior. Then and uniformly on γ so

However, this is a contradiction since the argument principle implies
= 0 for all n ∈ N, and .
Therefore f is either nonvanishing or identically zero on Ω.
6