1550251515-Classical_Complex_Analysis__Gonzalez_

376 Chapter 6

Thus, with definitions (6.17-7) and (6.17-8) the equation (6.17-6) holds

without exception. Note that %(Argfz + Argfz) = Arg(fz + fz) under

the assumption lfzl = lfzl.

Example For the function w = f(z) = zz we have, at z = 1,

f~(l) = 1 + e-2i8

All the rays z = 1 + tei^8 (t ;::: 0) map into

w = 1 + t^2 + 2t cos 8 = (1 - t)^2 + 2t(l +cos 8) 2:: 0

In this example, 8' = %(Argfz + Argf-z) = 0.

Case III. If J < 0, from equation (6.17-2) we have that d'ljJ/d8 is always

negative, so that 'ljJ is a strictly decreasing function of 8. In this case the

Kasner circle encloses the origin (Fig. 6.18), and the construction shows

that if 8 is increased by a, then 'ljJ decreases by >. =a+ S, where S;::: 0.

A particular case occurs when f z = 0. In this case the center of the circle

is at the origin (Fig. 6.19), and if 8 is increased by a, then 'ljJ decreases by

2a. Hence if 8~ is the direction corresponding to 8 +a, we have

and

8~ = ( 'ljJ - 2a) + ( 8 + a)

so that

8~ - 8' =-a

while (8 +a) - 8 =a. Thus the mapping defined by f is then indirectly

isogonal, which, of course, agrees with our result in Section 6.15.

Fig. 6.18