1550251515-Classical_Complex_Analysis__Gonzalez_

460 Chapter^7

Proof Choose a 0 r/. C* and € > 0. We must show that there exists 8 > 0

such that la -ao I < 8 implies that

1nc(a)-nc(ao)I = _!__ j (-

1

1

-) dz < i:

2?T z - a z - ao

G

Let p = d(ao, C*) = mina<t<,s lao - z(t)I. For 81 =^1 / 2 p we have

lz(t) -al = l(z(t) - ao) - (a - ao)I > p - %P =^1 / 2 p

and it follows that

I

1 1 I I a - ao I la - a^01

z -a - z - ao = (z -a)(z - ao) < %P^2

for z E C. Hence
1

lf2c(a)-nc(ao)I:::; - 2 la - aolL(C) < i:

7rp

whenever la-a 0 I < ?T p^2 i:/ L( C) = 82. Thus it suffices to take ii = min( 81 , 82)
to satisfy the required inequality.

Since the connected set M has no point in common with C*, nc( a) is

a continuous function of a on M. By Theorem 7.16, f2c(a) takes only

integral values. Hence nc( a) must be a constant on any arc whose graph

is in M, and so constant on M. If M is unbounded it contains a sequence

of points {an} such that an -t oo as n -t oo. Because C* is compact, there

is a constant h such that lz(t)I < h for all t E [a,,B]. Choosing n large

enough so that Ian!> h + L(C)/27r for n > N, say, we have, for z EC,

Thus
1 1
0:::; lf2c(an)I:::; 2?T Ian!-h L(C) < 1

which shows that f2c(an) = 0 for n > N, since the winding number is an

integer. If Mis also connected, then f2 0 (a) = 0 everywhere on M.

Theorem 7 .18 For any piecewise regular closed contours C1, C2 and a <f.
Ci, c; we have

and

Proof Follows at once from (7.13-1) and the known properties of the
integral.
In view of the preceding properties of the winding number we may
characterize analytically the orientation of a simple closed contour in the