55.5. THE PICARD THEOREMS 1753
By Lemma 55.5.1 and 55.5.18,
g(B(0,r)) ⊇ B
(0,
r2 | f ′ (a)|2
6R | f ′ (0)|
)
= B
(0,
r2( 1
2r | f′ (0)|R
)2
6R | f ′ (0)|
)= B
(0,| f ′ (0)|R
24
)Now g(B(0,r)) = f (B(a,r))− f (a) and so this implies
f (B(0,R))⊇ f (B(a,r))⊇ B(
f (a) ,| f ′ (0)|R
24
).
This proves the lemma.Here is a slightly more general version which allows the center of the open set to be
arbitrary.
Lemma 55.5.3 Let f be analytic on an open set containing B(z0,R) and suppose | f ′ (z0)|>0. Then there exists a ∈ B(z0,R) such that
f (B(z0,R))⊇ B(
f (a) ,| f ′ (z0)|R
24
).
Proof: You look at g(z)≡ f (z0 + z)− f (z0) for z ∈ B(0,R) . Then g′ (0) = f ′ (z0) andso by Lemma 55.5.2 there exists a1 ∈ B(0,R) such that
g(B(0,R))⊇ B(
g(a1) ,| f ′ (z0)|R
24
).
Now g(B(0,R)) = f (B(z0,R))− f (z0) and g(a1) = f (a)− f (z0) for some a ∈ B(z0,R)and so
f (B(z0,R))− f (z0) ⊇ B(
g(a1) ,| f ′ (z0)|R
24
)= B
(f (a)− f (z0) ,
| f ′ (z0)|R24
)which implies
f (B(z0,R))⊇ B(
f (a) ,| f ′ (z0)|R
24
)as claimed. This proves the lemma.
No attempt was made to find the best number to multiply by R | f ′ (z0)|. A discussion ofthis is given in Conway [32]. See also [65]. Much larger numbers than 1/24 are availableand there is a conjecture due to Alfors about the best value. The conjecture is that 1/24 canbe replaced with
Γ( 1
3
)Γ( 11
12
)(1+√
3)1/2
Γ( 1
4
) ≈ .47186