1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions 521

By hypothesis 2 to every f > 0 there corresponds an integer N, such

that n > N, implies that

JFn(z) - F(z)I < f

for all z E C. Therefore, for n > N, we have

j Fn(z)dz-j F(z)dz = j[Fn(z)dz-F(z)]dz:::; EL(C)

c c c

where L( C) denotes the length of C. This shows that

n->oo lim f Fn(z)dz = JF(z)dz

c c

Corollary 8.1 Suppose that:

1. The functions f n(z) are continuous on C (n = 1, 2, ... ).

2. F(z) = E~=l fn(z) uniformly for all z E C.
   Then

j F(z)dz= j fi(z)dz+···+ j fn(z)dz+···

c c c

Proof Let Fn(z) = fi(z) + · · · + fn(z). Then the functions Fn(z) (n =

1, 2, ... ) are continuous on C, and Fn(z) ~ F(z) for all z E C, so that the

conditions of Theorem 8.1 are satisfied. Hence,

J

F(z)dz= n-+oo lim JFn(z)dz= n-+oo lim jU1(z)+···+fn(z)]dz

c c c

= nli_.~ [! fi(z) dz+···+ j f n(z) dz] = f j f n(z) dz

C C n=lc

Corollary 8.2 Suppose that:

1. The functions fn(z) are analytic on a simply connected region R.


2. F(z) = E~=l fn(z), the convergence being uniform on compact subsets
   of R.

Then F( z) is analytic in R.

Proof Let C be any closed contour in R. By Corollary 8.1 we have

j F(z) dz= f j fn(z) dz

c n=lc