1754 CHAPTER 55. COMPLEX MAPPINGS

You can see there is quite a gap between the constant for which this lemma is proved aboveand what is thought to be the best constant.

Bloch’s lemma above gives the existence of a ball of a certain size inside the image ofa ball. By contrast the next lemma leads to conditions under which the values of a functiondo not contain a ball of certain radius. It concerns analytic functions which do not achievethe values 0 and 1.

Lemma 55.5.4 Let F denote the set of functions, f defined on Ω, a simply connectedregion which do not achieve the values 0 and 1. Then for each such function, it is possibleto define a function analytic on Ω, H (z) by the formula

H (z)≡ log

[√log( f (z))

2πi−√

log( f (z))2πi

−1

].

There exists a constant C independent of f ∈F such that H (Ω) does not contain any ballof radius C.

Proof: Let f ∈F . Then since f does not take the value 0, there exists g1 a primitiveof f ′/ f . Thus

ddz

(e−g1 f

)= 0

so there exists a,b such that f (z)e−g1(z) = ea+bi. Letting g(z) = g1 (z)+a+ ib, it followseg(z) = f (z). Let log( f (z)) = g(z). Then for n ∈ Z, the integers,

log( f (z))2πi

,log( f (z))

2πi−1 ̸= n

because if equality held, then f (z) = 1 which does not happen. It follows log( f (z))2πi and

log( f (z))2πi −1 are never equal to zero. Therefore, using the same reasoning, you can define a

logarithm of these two quantities and therefore, a square root. Hence there exists a functionanalytic on Ω, √

log( f (z))2πi

−√

log( f (z))2πi

−1. (55.5.20)

For n a positive integer, this function cannot equal√

n±√

n−1 because if it did, then(√log( f (z))

2πi−√

log( f (z))2πi

−1

)=√

n±√

n−1 (55.5.21)

and you could take reciprocals of both sides to obtain(√log( f (z))

2πi+

√log( f (z))

2πi−1

)=√

n∓√

n−1. (55.5.22)

Then adding 55.5.21 and 55.5.22

2

√log( f (z))

2πi= 2√

n