1550251515-Classical_Complex_Analysis__Gonzalez_

432 Chapter^7

From (7.8-10) we have

F'((k-1)((k - (k-1) = F((k) - F((k-1) -11k((k - (k-1)

so that
n n n
l:F'((k-1)((k - (k-1) = L[F((k) - F((k-1)] - L 1Jk((k - (k-1)
k=l k=l k=l
n
= F(z1) - F(zo) - L 1Jk((k - (k-1)
k=l
Hence
n n
lim -... F'((k-1)((k - (k-1) = F(zi) - F(zo) - lim L 1Jk((k - (k-1)

IPl-->O ~ IPJ-->0 k=l

(7.8-11)

However, for IPI < 8' we have

1~1Jk((k -(k-1)1 :S.E ~ i(k -(k-11 :S EL(7)

Therefore, the last term in (7.8-11) vanishes, and we obtain

1

Z1

('¥) F'(z)dz=F(z 1 )-F(zo)

zo

(7.8-12)

which shows that the integral is independent of the path.

Corollary 7.4 If a single-valued analytic function exists such that F' ( z) =

f(z) in a region R, then

j f(z)dz = 0

c

for every closed rectifiable curve C such that C* C R.

Proof H follows at once from Theorems 7.5 and 7.8.

Example If m is a nonnegative integer, we have

j(z-ardz = o

r

(7.8-13)

for any closed rectifiable curve r with graph contained in <C, since the

integrand is the derivative of F(z) = (z - a)m+l /m + 1, a function that is

single-valued and analytic in <C. The result is also valid for m a negative

integer other than -1, provided that I'* does not pass through a, since