Page 101 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Hence,

f(z)   f(s)  j
 ds =   j 1 ds (z z )



0





0
C 2 s z j 0 C  2 (s z ) 

  f(s) 
=   j 1 ds (z z ) j



0

j 0  C  2 (s z )   
0

For the second of these two integrals, note that for s  C we have |s – z | < |z – z |, and so
0
1
0
 
1  1  1   1  
s z = (z z ) (s z )  z z    





s z 
0
0
0
 1   0  

   z z 0   
 1    j  1
s z 
=   0     (s z ) j

0
z z 0 j 0 z z 0  j 0 (z z ) j 1



 
0

 1   1  1
=   (s z ) j 1 j      j 1  j


0



 
j 0 (z z ) j 1 (s z ) 0  (z z )
0

0
As before,
f(s)   f(s)  1
 ds =     j 1 ds  j






C 1 s z j 1 C  1 (s z )  (z z )
0
0

  f(s)  1
=    j 1 ds j





j 1  C  1 (s z )   (z z )

0
0
Putting this altogether, we have the Laurent series:
1 f(s) 1 f(s)
f(z) =  ds   ds
2 i C 2 s z 2 i C 1 s z




  1 f(s)    1 f(s)  1
=    ds (z z )     ds  .
j


0





j 0 2 i (s z ) C  0 j 1   j 1 2 i (s z0) C   j 1   (z z ) j
 
 
0
Let f be defined by
1
f(z)  .
z(z 1)

First, observe that f is analytic in the region 0 < |z| < 1. Let’s find the Laurent series for f valid
in this region. First,
1 1 1
f(z)     .
z(z 1) z z 1


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