1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 443

is satisfied, although it may be satisfied by more than one rectangle. If

there is just one, call it Ri, and if there is more than one, call Ri that with
the smallest superscript. Repeating the process with Ri, then with R2,
and so on indefinitely, there results a nested sequence of closed rectangles
(Fig. 7. 7) R :::> Ri :::> R2 :::> • • • :::> Rn :::> Rn+i :::> • • • such that

· JJ(Rn)I :'.'.'.^1 /4JI(Rn-1)l

giving

JJ(Rn)I :'.'.'.

4

1

n JJ(R)J (7.10-2)

Clearly, the diameter dn of the rectangle Rn tends to zero as n -+ oo,
since dn =^1 / 2 dn-1 =^1 / 4 dn-2 = · · · = (1/2n)d, where d is the diameter
(length of the diagonal) of R. Hence, by Theorem 2.24 and Corollary 2.2,
there is just one point z 0 = (x 0 ,yo) that belongs to all rectangles Rn, and
also to R.

Since f is analytic at z 0 , given € > 0 there is a ti > 0 such that f is

analytic in lz - zo I < ti and

I

f(z) - f(zo) _ f'(zo)

Z - Zo

provided that 0 < lz - zo I < ti. Letting

11(z) = J(z)-J(zo). -J'(zo)

Z - Zo
and

ry(zo)=O

<€

for z E N~(zo)

we have limz-+zo ry(z) = ry(zo), so ry(z) is continuous on No(zo).

From (7.10-3), (7.10-4), and (7.10-5) it follows that

l11(z)I < €

Fig. 7.7

(7.10-3)

(7.10-4)

(7.10-5)

(7.10-6)