Page 57 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Putting this all together, we can estimate the integrand above:

g(s) M
(s z   z)(s z) 2  (d | z|)d 2




for all s  C. Finally,
g(s) M
 z  2 ds  | z| (d | z|)d 2 length(C),


C (s z   z)(s z)  
and it is clear that
 g(s) 
 
lim  z ds  0,



 z 0 (s z   z)(s z) 2
 C 
just as we set out to show. Hence G has a derivative at z, and
g(s)
G'(z)   2 ds

C (s z)
Next we see that G’ has a derivative and it is just what you think it should be. Consider
G'(z   z) G'(z) = 1    1  1  g(s)ds



 z  z C  (s z   z) 2 (s z) 2  
1  (s z)  (s z   z) 
2
2


=   2  g(s)ds
2
 z C  (s z   z) (s z) 


1  2(s z) z ( z) 
2

  
=   2  g(s)ds
2
 z C  (s z   z) (s z) 



 2(s z)   z 
=   2 2  g(s)ds


C  (s z   z) (s z) 
Next,

G'(z   z) G'(z) g(s)
=  z  2  C (s z) 3 ds

 2(s z)   z 2 

=   2 2  3  g(s)ds



C  (s z   z) (s z) (s z) 
2
2
 2(s z)   z(s z) 2(s z   z) 




=   2 3  g(s)ds


C  (s z   z) (s z) 
2
2
4 z(s z) 2( z) 
 2(s z)   z(s z) 2(s z)      2




=   2 3  g(s)ds


C  (s z   z) (s z) 
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