Exercise 9.  Explain why  

9 (a).  [Graphics:Images/ComplexFunExponentialModHome_gr_309.gif]   holds for all  z.  

Solution 9 (a).

See text and/or instructor's solution manual.

Solution  Method I.  Use formula  (5-1)  [Graphics:../Images/ComplexFunExponentialModHome_gr_310.gif].   

                    [Graphics:../Images/ComplexFunExponentialModHome_gr_311.gif]   

We are done.   

Solution  Method II.  Use the Definition 5.1:   [Graphics:../Images/ComplexFunExponentialModHome_gr_312.gif].  

And substitute [Graphics:../Images/ComplexFunExponentialModHome_gr_313.gif] into this and get

                    [Graphics:../Images/ComplexFunExponentialModHome_gr_314.gif].  

Recall Exercise 6 in Section 4.1 where we proved:   If   [Graphics:../Images/ComplexFunExponentialModHome_gr_315.gif] then   [Graphics:../Images/ComplexFunExponentialModHome_gr_316.gif].   

Setting  [Graphics:../Images/ComplexFunExponentialModHome_gr_317.gif]  and  [Graphics:../Images/ComplexFunExponentialModHome_gr_318.gif]  and applying this result yields   

                    [Graphics:../Images/ComplexFunExponentialModHome_gr_319.gif].

Combining the above finishes our proof

                    [Graphics:../Images/ComplexFunExponentialModHome_gr_320.gif].

Detailed solution for  Solution.  Method II.   Let   [Graphics:../Images/ComplexFunExponentialModHome_gr_321.gif],   then   [Graphics:../Images/ComplexFunExponentialModHome_gr_322.gif].   

It follows that  

                    [Graphics:../Images/ComplexFunExponentialModHome_gr_323.gif]   

We are really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexFunExponentialModHome_gr_324.gif]

[Graphics:../Images/ComplexFunExponentialModHome_gr_325.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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