Solution 1 (a).

See text and/or instructor's solution manual.

Solution.  Factor the denominator to see that  [Graphics:../Images/CauchyGoursatModHome_gr_5.gif].  

The integrand  [Graphics:../Images/CauchyGoursatModHome_gr_6.gif]  is analytic everywhere except at the points  [Graphics:../Images/CauchyGoursatModHome_gr_7.gif]   which lie outside the contour  [Graphics:../Images/CauchyGoursatModHome_gr_8.gif].

Hence, by the Cauchy-Goursat Theorem we have   [Graphics:../Images/CauchyGoursatModHome_gr_9.gif].  

                              [Graphics:../Images/CauchyGoursatModHome_gr_10.gif]

                    The points  [Graphics:../Images/CauchyGoursatModHome_gr_11.gif]  and  [Graphics:../Images/CauchyGoursatModHome_gr_12.gif]  which lie outside the contour  [Graphics:../Images/CauchyGoursatModHome_gr_13.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/CauchyGoursatModHome_gr_14.gif]

[Graphics:../Images/CauchyGoursatModHome_gr_15.gif]



[Graphics:../Images/CauchyGoursatModHome_gr_16.gif]

[Graphics:../Images/CauchyGoursatModHome_gr_17.gif]

















This solution is complements of the authors.















 















(c) 2008 John H. Mathews, Russell W. Howell