55.5. THE PICARD THEOREMS 1755

which contradicts the above observation that log( f (z))2πi is not equal to an integer.

Also, the function of 55.5.20 is never equal to zero. Therefore, you can define thelogarithm of this function also. It follows

H (z)≡ log

(√log( f (z))

2πi−√

log( f (z))2πi

−1

)̸= ln

(√n±√

n−1)+2mπi

where m is an arbitrary integer and n is a positive integer. Now

limn→∞

ln(√

n+√

n−1)= ∞

and limn→∞ ln(√

n−√

n−1)= −∞ and so C is covered by rectangles having vertices at

points ln(√

n±√

n−1)+ 2mπi as described above. Each of these rectangles has height

equal to 2π and a short computation shows their widths are bounded. Therefore, thereexists C independent of f ∈F such that C is larger than the diameter of all these rectangles.Hence H (Ω) cannot contain any ball of radius larger than C.

55.5.2 The Little Picard TheoremNow here is the little Picard theorem. It is easy to prove from the above.

Theorem 55.5.5 If h is an entire function which omits two values then h is a constant.

Proof: Suppose the two values omitted are a and b and that h is not constant. Letf (z) = (h(z)−a)/(b−a). Then f omits the two values 0 and 1. Let H be defined inLemma 55.5.4. Then H (z) is clearly not of the form az+ b because then it would havevalues equal to the vertices ln

(√n±√

n−1)+2mπi or else be constant neither of which

happen if h is not constant. Therefore, by Liouville’s theorem, H ′ must be unbounded. Pickξ such that |H ′ (ξ )| > 24C where C is such that H (C) contains no balls of radius larger

than C. But by Lemma 55.5.3 H (B(ξ ,1)) must contain a ball of radius |H′(ξ )|24 > 24C

24 =C,a contradiction. This proves Picard’s theorem.

The following is another formulation of this theorem.

Corollary 55.5.6 If f is a meromophic function defined on C which omits three distinctvalues, a,b,c, then f is a constant.

Proof: Let φ (z) ≡ z−az−c

b−cb−a . Then φ (c) = ∞,φ (a) = 0, and φ (b) = 1. Now consider

the function, h = φ ◦ f . Then h misses the three points ∞,0, and 1. Since h is meromorphicand does not have ∞ in its values, it must actually be analytic. Thus h is an entire functionwhich misses the two values 0 and 1. Therefore, h is constant by Theorem 55.5.5.

55.5.3 Schottky’s Theorem

Lemma 55.5.7 Let f be analytic on an open set containing B(0,R) and suppose that fdoes not take on either of the two values 0 or 1. Also suppose | f (0)| ≤ β . Then lettingθ ∈ (0,1) , it follows

| f (z)| ≤M (β ,θ)