Page 98 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 9: Taylor and Laurent Series

Notes
f(s)   f(s)  j

 dz    j 1 ds (z z ) , or


0




C s z j 0  C (s z ) 
0

1 f(s)   1 f(s) 

f(z)   ds     ds (z z ) . j


2 i s z j 0 2 i (s z ) C  0 j 1   0



 
C
We have, thus, produced a power series having the given analytic function as a limit:

j

f(z)   c (z z ) , z z  r,

0
j
0
j 0

where
1 f(s)

c = 2 i (s z ) j 1 ds



j
0
C
This is the celebrated Taylor Series for f at z = z .
0
We know we may differentiate the series to get


f'(z)   jc (z z ) j 1

j
0
j 0

and this one converges uniformly where the series for f does. We can, thus, differentiate again
and again to obtain

(n)



f (z)   j(j 1)(j 2)...(j n 1)c (z z ) j n .



0
j
j n

Hence,
f (z ) = n!c , or
(n)
0 n
f(n)(z )
c  n! 0 .
n
But we also know that,
1 f(s)

c  2 i (s z ) n 1 ds.
n



0
C
This gives us,
n! f(s)
(n)

f (z )  2 i (s z ) n 1 ds,for n  0,1,2,.....
0



0
C
This is the famous Generalized Cauchy Integral Formula. Recall that we previously derived this
formula for n = 0 and 1.
What does all this tell us about the radius of convergence of a power series? Suppose we have,

j
f(z)   c (z z ) ,

j
0
j 0

LOVELY PROFESSIONAL UNIVERSITY 91