1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation 347

since (Dou)(Dov)' - (Dov)(Dou)' = J [Exercises 6.2, problem S(c)] and

(Dou)^2 + (Dov)^2 = lf8(z)l^2 = p^2 • Thus we obtain

J = lf8(z)l^2

or

If zl^2 - lfzl^2 = Uz + fze-^2 i^9 )(fz + fze^2 i^8 )

= lfzl^2 + lfzl^2 + fzfze^2 iB + fzfze-^2 iB

Letting A= fzfz, B = 2lfzl^2 , ( = e^2 i^8 , the preceding equation becomes

A(^2 +B(+A = 0

and as in the proof of Theorem 6.17, it follows that

A= fzfz = O, B = 2jfzj^2 = 0

which implies that f z = 0.

Remark If for a value of(), say 00 , we have f8 (z) = O, it follows that

lfzl =I -fze-^2 i^90 j = lfzl· If fz = 0, we have als~ fz = O, and J8(z) = 0

identically at z, in which case Argf8(z) is undefined for all 0. If fz f 0,

then fz f 0 and Arg J8(z) = 'ljJ cannot be a constant. If fact, in this case

J = lfzl^2 - lfzl^2 = 0 and (6.10-10) gives

d'ljJ = -1 or 'ljJ = -0 + c

dO

Theorem 6.19 Suppose that at a fixed point z, f8(z) f 0 for all 0(0 ~

0 ~ 27r). Then

Argf8(z) = 'l/J = w - 2() + 2h

for some appropriate value of k (Fig. 6.5), where w = Arg Jz, iff f is

conjugate monogenic at z.

Fig. 6.5