1550251515-Classical_Complex_Analysis__Gonzalez_

246 Chapter 5

From (5.11-5) it follows that ldzl/2y (y > 0) is a differential invari-

ant in Poincare's model. This result can also be obtained directly by

differentiation of (5.11-1) (see Exercises 6.1, problem 38). By adopting

ds = k ldzl = k -jdx2 +dy2 (y > 0) (5.11-7)

y y

as the expression for the differential of arc in the half-plane, where k is a

positive constant to be determined, or

1

t2 Jx'(t)2 + y'(t)2

s=k () dt

t1 y t

(5.11-8)

as the formula for the length of a piecewise smooth arc in parametric form:
x = x(t), y = y(t), ti :::; t:::; t2, we obtain, by applying the formula to the
vertical segment x = 0, y = t with endpoints Yi =ti and Y2 = t2,

s = k - = kln - = kln -

1

t

2

dt I t2 I I Y2 I

t1 t ti Yi

(5.11-9)

In particular, for the points zi = i, z2 = ir, (r > 1), we have

and a comparison with (5.11-6) shows that k = 1 is an appropriate value for
k. However, an arbitrary (but fixed) positive value for k can be introduced
in all distance formulas. This will amount simply to a change of scale.
We note that if in (5.11-9) we keep t 2 fixed and let ti ---+ 0, then s ---+ oo
i.e., as one of the endpoints of the segment tends to the x-axis (the line at
infinity), the length of the segment increases without bound.
Now, to prove the triangle inequality, consider any three distinct points
(i, (2, (a not all on the same line. By a suitable transformation (5.11-1)
of the hyperbolic plane we may carry those three points to the positions
zi = iyi, z2 = iy2 (Y2 > Yi), and Z3 = X3 + iya (xa i= 0), respectively.
Joining zi to z2 via Z3 by an arc 1: x = x(t), y = y(t), ti :::; t:::; t 2 , so that
x(ti) = x(t2) = 0, y(ti) =Yi, y(t2) = y2, we get

L("f)== t2 -/x'(t)2+y'(t)2 dt> t2 y'(t)dt =1Y2 dy =lnlY2 I

lt1 y(t) lt1 y(t) y 1 Y Yi

== d(zi ,z2)

Hence the length of the line segment is the shortest distance between zi
and z2.