1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation 329

Example If f(z) = lzl^4 = z^2 z^2 , we have 8f /8z = 2zz^2 and 8f /az =
2z^2 z.
We now proceed to establish some of the basic properties of the operators
az and az as defined by (6.7-6).

Theorem 6. 7 Let I( z) = z be the identity function, and let J ( z) = z be

the conjugation function. Then we have

Proof It follows at once by applying (6.7-8) and (6.7-9).

(6.8-1)

(6.8-2)

For convenience, we shall write properties (6.8-1) and (6.8-2) in the form

az(z) = 1,

az(z) = 0,

az(z) = 0

az(z) = 1

(6.8-3)

(6.8-4)

Theorem 6.8 Let f and g be functions of class 'D(A). Then we have

azu + g) = azf + azg,

az(fg) = f8zg + g8zf,

az (L) = gazf -1azg,

g g2

az(f + g) = 8zf + azg

az(fg) = f8zg + g8zf

az (L) = gazJ - Jazg

g g2

(6.8-5)

(6.8-6)

(6.8-7)

for all z in A excepting, in the case of (6.8-7), those values of z, if any,
for which g(z, z) = 0.

Proof We shall prove only the first formula in (6.8-6). The other results
can be established in a similar manner.
From the definition of az we have

Now, it is easy to check that

and similarly,

so that

(fg)x = fgx + gfx

az(fg) = 1/2[! gx + gfx -ifgy -igfy]

= J · %[gx -igy] + g •^1 /2[f x -ify]

= fozg+ gozf