1550251515-Classical_Complex_Analysis__Gonzalez_

Functions. Limits and Continuity. Arcs and Curves 165

z 0 -:f. z(t) for all t E [a,,8], there is a p > 0 such that lz(t) - zol > p
for all t E [a, ,8] [Exercises 3.3, problem 5(a)]. By the uniform continuity
of z(t) on [a, ,B], the interval [a, ,B] can be divided into a finite number of

subintervals [t;_ 1 ,t;] (i = 1, ... ,n) such that

lz(t) - z(t;-1)1 < p

for all t E [ti-1, ti]· Now letting

we have

z(ti) - zo

ei= z(ti-1)-zo

lei - ll = lz(t;) - z(t;_i)I < 1

lz(t;-1)-zol

(3.16-1)

so Re ei > 0. Hence there exists a unique value for arg e,, say Bi, satisfying

_1/27r < ei < l/27r

Since

n n

Lei= L {Arg[z(ti) - zo] - Arg[z(t;-1) - zo]}

i=l i=l

= Arg[z(,8) - z 0 ] - Arg[z(a) - z 0 ] = 0 (mod27r)

the well-defined sum :L~=l B; is a certain integral multiple of 27r. Then we

proceed to define n,(zo) as follows:

Definition 3.30
n

n,(zo) = c1;27r) L:: ei.

i=l

To justify this definition, we must show that the number D(zo) does not

depend on the chosen partition of the interval [a, ,8]. For this it suffices to

show that if the interval [t;_ 1, ti] is replaced by [ti-1, t'] and [t', t;], where

ti-l < t' <ti, the number S1 1 (z 0 ) remains the same. Denoting byμ' and

μ^11 the arguments of e corresponding to those subdivisions, we have

B; = μ' + μ" + 2k7r

where k is a suitable integer. Since the absolute value of each argument is

less than %7r, we have -7r < μ' + μ^11 < 7r, and (2k - l)7r < B; < (2k + l)7r.

Hence we obtain the inequalities (2k - l)7r <^1 / 2 7r and -^1 / 2 7r < (2k + l)7r,

which yield -^3 / 4 < k < %, so k = 0.

Theorem 3.16 For a fixed 'Y the winding number D 1 (z 0 ) is a contin-

uous function of z 0 on the complement of 'Y*, and hence a constant in