1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation

has argument () ± 71", so that

f8±7r(z) = fz + fze-^2 i(fJ±7r) = f8(z)

From formula (6.7-7), namely,

we obtain

df = f z dz+ fz dz

df dz

dz = fz + fz dz

343

By taking d~ i:n the direction () we have dz = ldzlei^6 , dz = ldzle-i^6 , so

that dz/ dz == e-^2 ie. Hence

! = fz + fze.-^2 ;e = f8(z) (6.10-3)

Thus the directional derivative can be expressed as the ratio of the total

di;fferential df to the differential dz with argument eq~al tQ. the argument

of the tangent to the arc 'Y at z. We recall that if z == z(t) is the equati,0n

of the arc, then dz= z'(t)dt has argument (), and ldzl = ds is the so-call~d

differential of the arc.

Equation (6.10-3) can, be written in the form

dw = df = f8(z) dz (6X0-4)

which shows that the total differential of w = f(z) is the product -Of the

directional derivative times the differential of the independent V~lj'ia'ble with

argument ().

Let 'Y': w = w(t) =-f(z(t)), a :::; t :::; /3, be the image 0£· 1 under j.

Then we hav{:).

Therefore,

and

w'(t) = fzz'(t) + fzz'(t)

= (fz + fte-^2 i^6 )z'(t)

= f8(z)z'(t)

du= Jdwl = lfHz)lldzl = lfHz)I ds

argw'(t) = Argf8(z) + Argz'(t)

assuming that f8(z) i= 0. Thus we have proved the following:

t6.10-5)

(6.ib-7)

Theorem 6.12 Under the mapping defined by f E 1'( A) the magliifica-

tion ratio of corresponding differential of arcs is given by lf8(z)I, ana the
direction of the original arc is rotated by the amount Arg f8( z) if f8( z) # 0.