1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation 345

or

d

dt u[x(t), y(t)] = 0 and

d

dt v[x(t),y(t)] = 0

for a :::; t :::; {3. These equations imply that

u[x(t), y(t)] = C1 and v[x(t),y(t)] = C2

for a:::; t:::; /3, C1 and C2 being constants. Hence f(z) = u(x,y) +iv(x,y)

is a constant along 'Y·

The converse of the preceding theorem is trivial.

Example Let f(z) = zz, -y: z = reit, 0 :::; t:::; 271", r > 0 constant. We

have f 0 (z) = z + ze-^2 i^8 , and since z'(t) = ireit = rei[t+l/^2 11"1, it follows

that () = t +^1 / 2 71". Hence ·

e-2i8 = -e-2it = _ -z

z

so that f 0 (z) = 0 for all z E -y*. Clearly, f(z) = r^2 along 'Y·

Corollary 6.2 If two functions f and g of class 'D(A) have the same

directional derivative along a regular arc -y, then they differ by a constant
along '-y.

Proof It suffices to apply Theorem 6.15 to the function F(z) = f(z)-.g(z).

Theorem 6.16 If f 0 (z) = 0 at a fixed point z for some direction 80 , then
J1(z) = O, and conversely.

Proof We have

fe 0 (z) = fz + fze-^2 ;^80 = 0

so that fz = -fze-^2 i^9 o and lfzl = lfzl· Hence J1(z) = 0 by (6.7-10).
Conversely, if J1(z) = 0, then lfzl = lfzl. If fz = 0, it follows fz = 0

and f 0 (z) = f'(z) = 0. If fz -:f 0, then the equation

e -2i8 =--fz

fz

has real solutions () = -^1 / 2 arg(-fz / fz ).

This property has an obvious geometric interpretation in terms of the
Kasner circle, discussed in Section 6.11.

Theorem 6.17 At a fixed point z, lf 0 (z)I = C (a constant) iff either f

is monogenic or conjugate monogenic at z.

Proof If fz = 0, we have f 0 (z) = fz and lf 0 (z)I = lfzl = a constant at
z. If fz = 0, then f 0 (z) = fze-^2 i^9 and lf 0 (z)I = lfzl = a constant at z.