1550251515-Classical_Complex_Analysis__Gonzalez_

362 Chapter6

so that lfzl-:/= lfzl for all z. Hence the mapping is one-to-one at least

locally. In fact, the function defines a one-to-one continuous mapping

of the complex plane onto the unit open disk lwl < 1, the inverse

function being given by

w

z=---

1-lwl

The function f(z) = z/(1 + lzl) is monogenic at z = 0 only, with

f1(0) = 1.

Note There are some criteria for a mapping to be globally one-to-one on
a region (see Selected Topics, Section 2.1). Of course, if an explicit formula
for the inverse function can be obtained, and by that formula there is just
one z E D for each given w E f(D), then the mapping is one-to-one, as
in Example 3 above.

6.14 THE DIRECTIONAL DERIVATIVE AND THE

KASNER CIRCLE FOR THE INVERSE FUNCTION

TO w = f(z)

Theorem 6.26 Let w = f(z) E 'D(A), A open, and suppose that f is one-

to-one in N 0 (zo), zo EA. Let Wo = f(zo) and J = J1(zo) t/=-0. Then the
function z = F(w), inverse tow= f(z), has at w 0 the directional derivative

( ~~) ~ = ~ fz - ~ fze-Zi~

where '1j; = arg dw. The corresponding Kasner circle has center at fz/ J

and radius lfz/ JI.

Proof By Exercises 6.2, problem 8(a), we have

1- 1

dz= Jfzdw-Jfzdw

Hence if dw = ld~lei~, we obtain, in view of the statement following
formula (6.10-3),

( ~~) ~ = ~ fz - ~ fze-Zi~ (6.14-1)

It is clear from (6.14-1) that the Kasner circle for the point w 0 indeed has

its center at fz/ J and radius lfz/ JI. Also, it follows from (6.14-1) that

if w = f ( z) is monogenic at zo and f' ( zo) -:/= O, its inverse z = F( w) is
monogenic at wo and F'(wo) = 1/f'(z 0 ), as shown in Section 6.3.