10 1. PRELIMINARIES
FIGURE 1.5THEOREM 1.3.4. If f : R^2 -+ R^2 is K-quasiconformal and if
lz2 - zol :<::::: lz1 - zol,then
(1.3.5)
where c = e^8 K.
PROOF. By means of preliminary similarity transformations, we may assume
that zo = f(zo) = 0 and that lz1I = 1, whence lz2I :<::::: 1. We may also assume that
lf(z1)I < lf(z2)I since otherwise there is nothing to prove.
Let r' be the family of circles {w: lwl = t} where lf(zi)I < t < lf(z2)I. Then
2_ log lf(z^2 )1 < mod(f')
27l' lf(zi)I -by Lemma 1.3.2.
To estimate the modulus of r = f-^1 (f'), let ¢ denote the stereographic projec-
tion of R
2
onto the Riemann sphere S^2 = {x E R^3 : lxl = l}. If"'/ Er= f-^1 (f'),
then"'/ separates the points 0 and z 1 from oo and z 2 ; hence ¢("'!) is a closed curve
on S^2 which separates the points ¢(0) and ¢(z 1 ) from ¢(00) and ¢(z 2 ). Since each
arc on S^2 which joins ¢(0) to ¢(z 1 ) or ¢(00) to ¢ (z 2 ) has length at least 7r/2,
!, 1 +
2
lzl 2 ldz l =length(¢("'!)) :'.'.'. 7rand hence the density
1 2
p(z) =:; 1 + lzl2
is admissible for r. Thus
1
mod(r) :<::::: p(z)^2 dm =^11 4 4
R2^2 7l' R2 ( 1 + I Z^12 )^2 dm = -7l'and we obtain
1 lf(z2)I 4K- log --< mod(f') < K mod(f) < -
27!' lf(z1)I - - - 7l'
from which (1.3.5) follows. 0