1550251515-Classical_Complex_Analysis__Gonzalez_

348 Chapter6

Proof If fz = 0, we have f 0 (z) = fze-^2 i^8 '!-0, so that lf 0 (z)I = lf.zl and

Argf 0 (z) ='I/;= w - 28 + 2k7r (6.10-11)

for some integer k. Next, suppose that (6.10-11) holds. Then we have

d'lj; =-2

d(J

and from (6.10-10) we have

d'l/; = !.._ -1

d(J p2

Hence we conclude that

P^2 = IJ8(z)l2 = -J (6.10-12)

Thus, our assumption implies that J < 0. Equation (6.10-12) yields

(fz + f.ze-^2 i^8 )(fz + f.ze^2 i^8 ) = -lfzl^2 + Jf.zJ^2

and

(6.10-13)

where ( = e^2 i^8 • Since (6.10-13) must be satisfied for all (such that l(J = 1,

we obtain f z = 0, so f must be conjugate monogenic at z.

Theorem 6.20 Let z = rei.P, w = f(z) = u(r,'1/;) + iv(r,'lj;) E 'D(A), 1:

z = z(t) = r(t)eit/J(t), a :St :S /3, a regular arc with graph contained in A,

and z'(t) = lz'(t)lei^8. Then for any z E 1*,

fMz) = fz + fze-^2 ;^8

where

and

fz= ~e;,μ(8! + ~ 8f)

2 8r r 8'1/;

(see Exercises 6.1, problem 34).

Proof As in (6.10-1) we have

f,'( (^8) z ) = e -i8 [c Ur+ ivr. ) lz'(t)I r'(t) + ( u,μ +iv,μ. ) '1/;'(t)] Jz'(t)J (6.10-14)