Exercise 5.  Find all the roots in both polar and Cartesian form for each expression.

5 (a).  [Graphics:Images/ComplexAlgebraRevisitedModHome_gr_99.gif].    

Solution 5 (a).

See text and/or instructor's solution manual.

Given  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_100.gif].  Then  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_101.gif]  where  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_102.gif],  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_103.gif],  and  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_104.gif].  

The primitive cube root of unity is  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_105.gif],  
and the cube roots of unity are  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_106.gif][Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_107.gif].  

The principal cube root of  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_108.gif]  is  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_109.gif]  

Thus, the cube roots of  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_110.gif]  are  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_111.gif]  which can be written as  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_112.gif]  

                

                     [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_113.gif]

  

                    [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_114.gif]  for  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_115.gif].

We are done.   

The cube roots of  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_116.gif]  can be calculated in the ordinary way.

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_117.gif]  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_118.gif]  

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_119.gif]  

Remark.  The traditional answers for the cube roots of  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_120.gif]  are

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_121.gif],  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_122.gif],  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_123.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

The principal value is

[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_124.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_125.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_126.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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