55.5. THE PICARD THEOREMS 1751

Proof: By assumption,

f (z) =∞

∑k=0

akzk, |z| ≤ r. (55.5.16)

Then by the Cauchy integral formula for the derivative,

ak =1

2πi

∫∂B(0,r)

f (w)wk+1 dw

where the integral is in the counter clockwise direction. Therefore,

|ak| ≤1

2π

∫ 2π

0

∣∣ f (reiθ)∣∣

rk dθ ≤ Mrk .

In particular, br ≤M. Therefore, from 55.5.16

| f (z)| ≥ b |z|−∞

∑k=2

Mrk |z|

k = b |z|−M(|z|r

)2

1− |z|r

= b |z|− M |z|2

r2− r |z|

Suppose |z|= r2b4M < r. Then this is no larger than

14

b2r2 3M−brM (4M−br)

≥ 14

b2r2 3M−MM (4M−M)

=r2b2

6M.

Let |w|< r2b4M . Then for |z|= r2b

4M and the above,

|w|= |( f (z)−w)− f (z)|< r2b4M≤ | f (z)|

and so by Rouche’s theorem, z→ f (z)−w and z→ f (z) have the same number of zerosin B

(0, r2b

4M

). But f has at least one zero in this ball and so this shows there exists at least

one z ∈ B(

0, r2b4M

)such that f (z)−w = 0. This proves the lemma.

55.5.1 Two Competing LemmasLemma 55.5.1 is a really nice lemma but there is something even better, Bloch’s lemma.This lemma does not depend on the bound of f . Like the above two lemmas it is interestingfor its own sake and in addition is the key to a fairly short proof of Picard’s theorem. Itfeatures the number 1

24 . The best constant is not currently known.

Lemma 55.5.2 Let f be analytic on an open set containing B(0,R) and suppose | f ′ (0)|>0. Then there exists a ∈ B(0,R) such that

f (B(0,R))⊇ B(

f (a) ,| f ′ (0)|R

24

).