1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation^359

Examples

1. For f(z) = z^2 we have fz = 2z, fz = 0. Thus lfzl = lf.:I = 0 holds at

z = 0 only. Clearly, the function is not one-to-one in any N6(0) since

(-z)^2 = z^2 = w.

2. Let f(z) = z^2 z. Here fz = 2zz, fz = z^2 , so that lfzl = lfzl = 0 holds
   at z = 0 only. However, the function w =· z^2 z is one-to-one in C since
   w = 0 implies that z = 0, and for w =f 0 we have z = w / lw 1213 • In
   terms of components this mapping is defined by

u = x(x^2 + y^2 ),

with the unique inverse

u

x = (u2 +v2)1/a'

v

Y = (u2 + v2)1/3

for u^2 + v^2 =f 0. For u = v = 0 the inverse image is x = y = 0. In

this case we have

8(u,v) =3(x2+y2)2

8(x,y)

Now suppose that lfzl = lfzl =f 0 and TJ1fz - TJ2fz = 0 at zo. From

(6.12-2) we have

.6.w = (fz + 'T/1).6.z + (!.: + 'T/2).6.z

Let z = rei^8 and chooser > 0 small enough so that

and

Then we have .6.w = O, or

by choosing for () any solution of

e2i8 = _ fz + 'T/2 = _ fz + 'T/ifz - TJ2fz = fz

fz + 'T/1 fz f z(fz + 'T/1) fz

i.e., by taking

Hence in this case the mapping is not one-to-one in Nr(zo).

Example Let f(z) = (z + 1)/(z + 2). In this case

1 .z+2

fz = z+2 = (z+2)^2 '

-(z + 1)

fz= (z+2)^2

(6.13-3)