1752 CHAPTER 55. COMPLEX MAPPINGS

Proof: Let K (ρ)≡max{| f ′ (z)| : |z|= ρ} . For simplicity, let Cρ ≡ {z : |z|= ρ}.Claim: K is continuous from the left.Proof of claim: Let zρ ∈Cρ such that

∣∣ f ′ (zρ

)∣∣= K (ρ) . Then by the maximum mod-ulus theorem, if λ ∈ (0,1) ,∣∣ f ′ (λ zρ

)∣∣≤ K (λρ)≤ K (ρ) =∣∣ f ′ (zρ

)∣∣ .Letting λ → 1 yields the claim.

Let ρ0 be the largest such that

(R−ρ0)K (ρ0) = R∣∣ f ′ (0)∣∣ .

(Note (R−0)K (0) =R | f ′ (0)| .) Thus ρ0 <R because (R−R)K (R) = 0. Let |a|= ρ0 suchthat | f ′ (a)|= K (ρ0). Thus ∣∣ f ′ (a)∣∣(R−ρ0) =

∣∣ f ′ (0)∣∣R (55.5.17)

Now let r = R−ρ02 . From 55.5.17,

∣∣ f ′ (a)∣∣r = 12

∣∣ f ′ (0)∣∣R, B(a,r)⊆ B(0,ρ0 + r)⊆ B(0,R) . (55.5.18)

0

a

Therefore, if z ∈ B(a,r) , it follows from the maximum modulus theorem and the defi-nition of ρ0 that

∣∣ f ′ (z)∣∣ ≤ K (ρ0 + r)<R | f ′ (0)|

R−ρ0− r=

2R | f ′ (0)|R−ρ0

=2R | f ′ (0)|

2r=

R | f ′ (0)|r

(55.5.19)

Let g(z) = f (a+ z)− f (a) where z ∈ B(0,r) . Then |g′ (0)| = | f ′ (a)| > 0 and for z ∈B(0,r),

|g(z)| ≤∣∣∣∣∫

γ(a,z)g′ (w)dw

∣∣∣∣≤ |z−a| R | f′ (0)|r

= R∣∣ f ′ (0)∣∣ .