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Unit 6: Cauchy’s Integral Formula

where Notes

cosz
f(z)  .
z 5

Observe that f is analytic on and inside C, and so,

cosz f(z)
2  dz =  dz  2 if(1)

z  6z 5 z 1


C C
cos1 i
= 2 i   cos1
1 5 2

6.2 Functions defined by Integrals
Suppose C is a curve (not necessarily a simple closed curve, just a curve) and suppose the
function g is continuous on C (not necessarily analytic, just continuous). Let the function G be
defined by :

g(s)
G(z) =  ds

C s z
for all z  C. We shall show that G is analytic. Here we go.
Consider,

G(z   z) G(z) = 1  1  1  g(s)ds

 z  z C    s z   z s z  


=  g(s) ds.


C (s z   z)(s z)
Next,


G(z   z) G(z) – g(s) ds =    1  1  g(s)ds

 z  (s z) 2 (s z Dz)(s z) (s z) 2 




C C  



 (s z) (s z   z)
=   2  g(s)ds


C  (s z   z)(s z) 
=  z  g(s0 2 ds.


C (s z   z)(s z)
Now we want to show that
 g(s) 
 
lim  z ds  0.



 z 0  C (s z   z)(s z) 2 
To that end, let M = max {|g(s)| : s  C}, and let d be the shortest distance from z to C. Thus, for
s  C, we have |s – z|  d > 0 and also
|s – z – z|  |s – z| – |z|  d – |z|.
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