Page 55 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Let  > 0 be any positive number. We know that f is continuous at z and so there is a number 
0
such that |f(z) – f(z )| <  whenever |z – z | < . Now let  > 0 be a number such that  <  and the
0 0
f(z)
circle C = {z : |z – z | = } is also inside C. Now, the function z z 0 is analytic in the region
0
0

between C and C ; thus,
0
f(z) f(z)
 dz   dz.


C z z 0 C 0 z z 0
We know that  1 dz  2 i, so we can write


C z z 0
f(z) f(z) 1
0 
 z z dz – 2 if(z ) =  z z dz – f(z ) z z dz

0
C  0 C  0 C 0  0

=  f(z) f(z ) dz.
0

C 0 z z 0
For z  C we have
0
f(z) f(z ) f(z) f(z )


0
0
z z 0 = z z 0



 .

Thus,
f(z) f(z) f(z )

 dz 2 if(z ) =  0 dz


0


C 0 z z 0 C 0 z z 0

 2  2 .

which is exactly what we set out to show.
Look at this result. It says that if f is analytic on and inside a simple closed curve and we know the
values f (z) for every z on the simple closed curve, then we know the value for the function at
every point inside the curve—quite remarkable indeed.
Example:
Let C be the circle |z| = 4 traversed once in the counterclockwise direction. Let’s evaluate the
integral

cosz
2  dz.

C z  6z 5
We simply write the integrand as

cosz cosz f(z) ,
2

z  6z 5  (z 5)(z 1)  z 1



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