530 Chapter^8
that lz --al = p < R. For the case R = oo, let z be an arbitrary point of
the complex plane, i.e., such that lz - al = p < oo.
Next, choose r satisfying p < r < R (R finite or infinite) and consider
the circle C': ( - a = reit, 0 :::; t :::; 271' (Fig.8.2). For ( ¥= z, ( ¥= a we
have the algebraic identity
1 1 1
(-z =(-a l-(z-a)/((-a)~ (z -a)k (z - a)n+l
= f::o (( a)k+l + (( z)(( _ a)n+1
Multiplying by J(()/27l'i and integrating both sides along C', we get
_1 j J(()d( = ~ (z -a)k j f(()d(
27l'i (-z L.J 27l'i ((-a)k+l
~ k=O ~(z-a)n+1 f J(()d(
+ 27l'i (( - z)((-a)n+l
(8.3-2)
C'Cauchy's formula (7.17-3) and the formula for the derivatives yield
n J(k)( )f(z) = L Ti (z -a)k + Rn(z)
k=O(8.3-3)where
(z-a)n+i f f(()d(
Rn z =
() 27l'i ((-z)((-a)n+l
(8.3-4)
C'is called the remainder after n + 1 terms. Since f( () is continuous along
C', we have lf(()I :::; M (a constant) for all (EC'. Also, I( -al= rand
Fig. 8.2