Exercise 7.  Find all the roots of the equation  [Graphics:Images/ComplexAlgebraRevisitedModHome_gr_258.gif]  given that  [Graphics:Images/ComplexAlgebraRevisitedModHome_gr_259.gif]  is a root.  

Solution 7.

See text and/or instructor's solution manual.

Since the coefficients are real, roots come in conjugates, so that [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_260.gif]  is also a root.  

Thus,  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_261.gif],  and  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_262.gif]  are factors.  Hence  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_263.gif]  is a factor.  

Dividing the polynomial by  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_264.gif]  yields  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_265.gif],  which can be solved with the quadratic formula.

Here  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_266.gif],  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_267.gif],  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_268.gif],  and  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_269.gif].  Thus

        [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_270.gif]  

Therefore, the roots of  [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_271.gif]  are [Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_272.gif].

We are done.   

Aside.  We can let Mathematica double check our work.

                

[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_273.gif]

  

[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_274.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_275.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_276.gif]
[Graphics:../Images/ComplexAlgebraRevisitedModHome_gr_277.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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