Page 100 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 9: Taylor and Laurent Series

Okay, now let’s derive the above formula. First, let r and r be so that R < r  |z – z |  r < R 2 Notes
0
2
1
1
2
1
and so that the point z and the curve C are included in the region r  |z – z |  r . Also, let  be
0
1
2
a circle centered at z and such that  is included in this region.

f(z)
Then is an analytic function (of s) on the region bounded by C , C , and , where C is the
s z 1 2 1

circle |z| = r and C is the circle |z| = r . Thus,
2
2
1
f(z) f(z) f(z)
 s z ds   s z ds   s z ds.
C 2  C 1   
f(z)
(All three circles are positively oriented, of course.) But  ds  2 if(z), and so we have


 s z
f(z) f(z)

2 if(z)   ds   ds


C 2 s z C 1 s z
Look at the first of the two integrals on the right-hand side of this equation. For s  C , we have
2
|z – z | < |s – z |, and so
0
0
1 1
s z = (s z ) (z z )




0
0
 
1  1 
=  
s z   1     0   

z z 
0

   s z 0   
1    j
z z 
=   0 
s z 0 j 0 s z 0 

 
 1
=  (z z ) . j

0

j 0 (s z ) 0 j 1


LOVELY PROFESSIONAL UNIVERSITY 93