Page 102 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 9: Taylor and Laurent Series

From our vast knowledge of the Geometric series, we have Notes

1 
f(z)     z . j
z j 0

Now let’s find another Laurent series for f, the one valid for the region 1 < |z| < .
First,

 
1 1  1  .
z 1  z  1   1  

 z 

1
Now since  1, we have
z

 
1 1  1  1  j   z , j 
z 1  z  1   1    z  z  

j 1
j 0


 z 
and so
1 1 1 
f(z)        z j 
z z 1 z j 1



f(z) =  z . j 
j 2

9.3 Summary

Suppose f is analytic on the open disk |z – z | < r. Let z be any point in this disk and choose C to
0
be the positively oriented circle of radius , where |z – z | <  < r. Then for s  C, we have,
0
 
1 1 1  1   (z z ) j

0
s z  (s z ) (z z )  (s z )  1  z z 0     0 j 1







j 0 (s z )
0 
0
0

  s z 

0 
z z

since 0 < 1. The convergence is uniform, so we may integrate
s z 0

f(s)   f(s) 
j
 dz    j 1 ds (z z ) , or



0




0
C s z j 0  C (s z ) 

1 f(s)   1 f(s)  j

f(z)   ds     ds (z z ) .

0
2 i s z j 0 2 i (s z ) C  0 j 1  




 
C
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