52.4. EXTENSIONS OF MAXIMUM MODULUS THEOREM 1657

while for a ∈ A,limsup

z→a|F (z)| ≤M.

Therefore, for α ∈ ∂∞Ω, limsupz→α |F (z)| ≤ max(M,Mη−η) and so by Theorem 52.3.5,

| f (z)| ≤(

mη

|φ(z)|η)

max(M,Mη−η) . Now let η → 0 to obtain | f (z)| ≤M.

In applications, it is often the case that B = {∞}.Now here is an interesting case of this theorem. It involves a particular form for Ω, in

this case Ω ={

z ∈ C : |arg(z)|< π

2a

}where a≥ 1

2 .

Ω

Then ∂Ω equals the two slanted lines. Also on Ω you can define a logarithm, log(z) =ln |z|+ iarg(z) where arg(z) is the angle associated with z between −π and π. Therefore,if c is a real number you can define zc for such z in the usual way:

zc ≡ exp(c log(z)) = exp(c [ln |z|+ iarg(z)])= |z|c exp(icarg(z)) = |z|c (cos(carg(z))+ isin(carg(z))) .

If |c|< a, then |carg(z)|< π

2 and so cos(carg(z))> 0. Therefore, for such c,

|exp(−(zc))| = |exp(−|z|c (cos(carg(z))+ isin(carg(z))))|= |exp(−|z|c (cos(carg(z))))|

which is bounded since cos(carg(z))> 0.

Corollary 52.4.2 Let Ω ={

z ∈ C : |arg(z)|< π

2a

}where a≥ 1

2 and suppose f is analyticon Ω and satisfies limsupz→a | f (z)| ≤M on ∂Ω and suppose there are positive constants,P,b where b < a and

| f (z)| ≤ Pexp(|z|b)

for all |z| large enough. Then | f (z)| ≤M for all z ∈Ω.

Proof: Let b < c < a and let φ (z) ≡ exp(−(zc)) . Then as discussed above, φ (z) ̸= 0on Ω and |φ (z)| is bounded on Ω. Now

|φ (z)|η = |exp(−|z|c η (cos(carg(z))))|

lim supz→∞

| f (z)| |φ (z)|η ≤ lim supz→∞

Pexp(|z|b)

|exp(|z|c η (cos(carg(z))))|= 0≤M