1550251515-Classical_Complex_Analysis__Gonzalez_

360

Hence we have If z I = If z I -# 0 for every z such that

lz+21 = lz+ll

which gives

z = -3/ 2 +iy,. -oo < y < +oo

Also, we have

Chapter.6

(6.13-4)

'TJi= (z+2)(z+6z+2)' "'^2 = (z+2)^2 (z+6z+2)

so that ry 1 fz -ry2fz = 0 identically. From (6.13-3) we obtain B = %7r + k7r.

For any of the values given by (6.13-4) we get w = -1. Therefore, the

mapping is not one-to-one (Fig. 6.8).

Remark The condition lfzl = lfzl at a point zo implies that J = 0 at

that point. Geometrically, this condition means that the Kasner circle for

the point z 0 passes through the origin, or else it reduces to the point z 0 •

Thus that condition implies either that some directional derivatives vanish

at z 0 (in fact, two of them in opposite directions), or that the ordinary

derivative vanishes at z 0 •

Theorem 6.25 Let w = f(z) = u+iv E '.D(A), A open. If lfzl = lfzl and

Tfifz - 'f/dz = 0 in some region A' CA, then the mapping f: A~ f(A)

is not one-to-one. If, in addition, the partial derivatives u.,, uy, v.,, Vy are

continuous in N 0 (z 0 ) C A'(8 > 0), then for some 8' such that 0 < 81 ::; 8

the image of N 0 1(zo) by f is the graph of an arc in f(A').

·Proof If lfzl = lfzl = 0 in A' then f is analytic, as well as conjugate
analytic, in A', and f' ( z) = 0 identically in A'. Hence f ( z) = c (a constant)

in A', so that f(A') = {c} is in this case a point arc. If lfzl = lf.:I -f. 0

and Tfifz = Tf2fz = 0 in A' it follows from Theorem 6.24 that f is not

z = -% + iy y v

x 0 u

Fig. 6.8