Exercise 17.  Let  [Graphics:Images/FunTheoremCalculusModHome_gr_226.gif]  be the principal branch of the square root function.  

17 (b).  Evaluate  [Graphics:Images/FunTheoremCalculusModHome_gr_248.gif],  where C is the right half of the circle  [Graphics:Images/FunTheoremCalculusModHome_gr_249.gif]  joining  [Graphics:Images/FunTheoremCalculusModHome_gr_250.gif].  

Solution 17 (b).

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/FunTheoremCalculusModHome_gr_251.gif][Graphics:../Images/FunTheoremCalculusModHome_gr_252.gif].  

Solution.  The function  [Graphics:../Images/FunTheoremCalculusModHome_gr_253.gif]  is analytic everywhere except points   [Graphics:../Images/FunTheoremCalculusModHome_gr_254.gif],  

along the negative x-axis, where the principal branch of the square root function [Graphics:../Images/FunTheoremCalculusModHome_gr_255.gif] is discontinuous,

and  [Graphics:../Images/FunTheoremCalculusModHome_gr_256.gif]  has  [Graphics:../Images/FunTheoremCalculusModHome_gr_257.gif]  as an antiderivative.  

Letting D be the simply connected domain consisting of the entire complex plane except for the real numbers   [Graphics:../Images/FunTheoremCalculusModHome_gr_258.gif],  

along the negative x-axis, we see that  [Graphics:../Images/FunTheoremCalculusModHome_gr_259.gif]  and its listed antiderivative are analytic in D.  

Since the the right half of the circle  [Graphics:../Images/FunTheoremCalculusModHome_gr_260.gif]  joining  [Graphics:../Images/FunTheoremCalculusModHome_gr_261.gif]  is contained in D, Theorem 6.9 gives

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_262.gif]  

                    [Graphics:../Images/FunTheoremCalculusModHome_gr_263.gif]

                    The path of integration is the right half of the circle  [Graphics:../Images/FunTheoremCalculusModHome_gr_264.gif]  joining  [Graphics:../Images/FunTheoremCalculusModHome_gr_265.gif].  

                    Notice that  [Graphics:../Images/FunTheoremCalculusModHome_gr_266.gif]  is not analytic on the ray   [Graphics:../Images/FunTheoremCalculusModHome_gr_267.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FunTheoremCalculusModHome_gr_268.gif]

[Graphics:../Images/FunTheoremCalculusModHome_gr_269.gif]




















This solution is complements of the authors.



































 

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