1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 429

Putting back this value in (7.8-8) and letting z = z 1 , we obtain

('Y) 1:

1

f(()d(=1/;(z1) -1/J(zo) (7.8-9)

Note From (7.8-9) it does not follow that the value of the integral depends

only on the initial and terminal points of the arc "(, because the function

1/J itself depends in general on 'Y· If the initial point coincides with the

terminal point while L('Y) =f. 0, it does not follow either that the integral

along the closed path 'Y is necessarily zero, because the function 1fJ is in

general multiple-valued.

Examples 1. Let f(z) = z, "(: z = z(t) = t +it, 0 :::; t:::; 1, and consider

the integral

1+;

('Y) 1 zdz

For the given arc we have ei^8 = (1 + i)/./2, so that e-^2 i^8 = -i, and we
may choose 1/;(z) =^1 / 4 (z + z)^2 since

1/J~(z) = %(z + z) +^1 / 2 (z + z)(-i) = t -it= z

along 'Y. Hence

1+i

('Y) 1 z dz = 1/;(1 + i) -1/;(0) = 1

2. Consider the integral of the same function along the unit circle C: z =
   z(t) = eit, 0 :::; t:::; 27f from z 0 = 1 to z = eit. We have

(C) 1z zdz= lte-iteitidt=it=logz

where we must take the value of the logarithm that corresponds to the

domain of t. Thus for the integral around the circle from z 0 = 1 to z 1 =

e^2 71"i = 1, we obtain

e211"i 21r'

[ z dz = log z ]

1

= it ]

0

= 27fi

Theorem 7 .8 If f is continuous in a region R, the integral

('Y) 1:1 !(() d(

where 'Y is a smooth arc with graph contained in R, depends only on the
endpoints of 'Y iff there exists a single-valued analytic function F in R such