Example 6.24.  Let  [Graphics:Images/IntegralRepresentationMod_gr_80.gif]  denote a fixed complex value.  Show that, if C is a simple closed positively oriented contour such that  [Graphics:Images/IntegralRepresentationMod_gr_81.gif]  lies interior to C, then  

               [Graphics:Images/IntegralRepresentationMod_gr_82.gif],   and
(6-50)
               [Graphics:Images/IntegralRepresentationMod_gr_83.gif],   for any integer  [Graphics:Images/IntegralRepresentationMod_gr_84.gif]

[Graphics:Images/IntegralRepresentationMod_gr_85.gif]

Explore Solution 6.24 (b).

(b)   Show that  [Graphics:../Images/IntegralRepresentationMod_gr_93.gif],  when  [Graphics:../Images/IntegralRepresentationMod_gr_94.gif]  is an integer.
Remark.  If  m  is an integer and  m < 0,  then  m = -n,  where  n  is a positive integer and the integrand is  [Graphics:../Images/IntegralRepresentationMod_gr_95.gif]
which is a polynomial of degree n.  Hence in this case, f(z) would be analytic and the Cauchy-Goursat theorem implies that the value of the integral is zero.

Now we consider the case when m is a positive integer and  m > 1.
Use the Cauchy's Integral Formulae for Derivatives in the form  [Graphics:../Images/IntegralRepresentationMod_gr_96.gif]

For illustration.  We use m = 5,  and  f(z) = 1.  Then  m = 5 = n + 1 implies that  n = 4.

[Graphics:../Images/IntegralRepresentationMod_gr_97.gif]


[Graphics:../Images/IntegralRepresentationMod_gr_98.gif]

 

 

Thus, we have found the value of the contour integral.

[Graphics:../Images/IntegralRepresentationMod_gr_99.gif]




[Graphics:../Images/IntegralRepresentationMod_gr_100.gif]

[Graphics:../Images/IntegralRepresentationMod_gr_101.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2006 John H. Mathews, Russell W. Howell