1549055259-Ubiquitous_Quasidisk__The__Gehring_

54 3. CONFORMAL INVARIANTS

The square of the length l of the diagonal (z 1 , z3) is then given by

l^2 = c^2 + d^2 - 2cdcose = a^2 + b^2 + 2abcose,

where e is the angle formed by the segments ( z4, z1) and ( z4, z3), and we obtain

c^2 - a^2 ::::: c^2 - a^2 + d^2 - b^2 = 2 ( cd + ab) cos e ::::: 2 ( c + a) d cos e.

Thus

Hence

0 ::::: c - a ::::: 2d cos e

1

x-2cos8<- -x'

and

c a c a

• 
  
  
  
  • 
    
    
    
    • < - - - < 2 cos e.
      d b-d d-
    
    
    
    
  
  
  
  

x::::: J1 + cos^2 e + cose

where x = c/d, and we obtain

~ = x < v'2+1,

a 1 r;:;

-=->v2-l.

b x

Letting e --+ 0 in the above calculations shows that the constants J2 + 1 and J2- 1
cannot be improved. D

LEMMA 3.12.9. Suppose b is a constant with b 2'. l. If

(3.12.10)

(3.12.11)

whenever [z 1 , z 2 , z 3 , z 4 ] = 2 where a= a(b) 2'. 2.

PROOF. We begin by showing that if Zi, Zj, Zk is a triple of points in 8B with

(3. 12 .12)

then

(3. 12 .13) c=b+l.

For this let / denote the arc in 8B with endpoints Zi, z k which contains Zj and let
( denote the midpoint off. Next we may assume that ¢(zk) is on the larger arc
of 8B connecting ¢ (zi) and ¢ (zj ) since otherwise (3 .12.13) follows trivially. Then
¢(()lies on the smaller arc connecting ¢ (zi ) and ¢(zj)· Hence

1¢(zi) - ¢(zk)I :S: 1 ¢(zi) - ¢(()1+1¢(() - ¢(zk)I :S: cl¢(zi) - ¢(() 1

:S: cl¢(zi) -¢(zj)I

by (3. 12 .10).
Suppose now that z1, z2, z3, z4 is an ordered quadruple of points in 8B with
[z1, z2, z3, z4] = 2. Then

lz1 - z2llz3 - z4I = lz1 - z3llz2 - z4l - lz2 - z3llz4 - z1I

= [z1, z2, z3, z4] lz2 - z3ilz4 - z1l - lz2 - z3llz4 - z1I

= lz2 - z3llz4 - z1I