Exercise 7.   Let  [Graphics:Images/RoucheTheoremModHome_gr_468.gif].  

7 (b).   Show that there are three zeros in  [Graphics:Images/RoucheTheoremModHome_gr_490.gif].  

Solution 7 (b).

Solution.   Let  [Graphics:../Images/RoucheTheoremModHome_gr_491.gif].  Then for  [Graphics:../Images/RoucheTheoremModHome_gr_492.gif]  we have

        [Graphics:../Images/RoucheTheoremModHome_gr_493.gif]  

Since  [Graphics:../Images/RoucheTheoremModHome_gr_494.gif]  has three roots in  [Graphics:../Images/RoucheTheoremModHome_gr_495.gif],  so does  [Graphics:../Images/RoucheTheoremModHome_gr_496.gif],  by Remark 8.5 to Rouché's Theorem.

We are done.   

Aside.  We can use Mathematica to verify the inequality mentioned above.

                    [Graphics:../Images/RoucheTheoremModHome_gr_497.gif]

                    For  [Graphics:../Images/RoucheTheoremModHome_gr_498.gif]  on the circle  [Graphics:../Images/RoucheTheoremModHome_gr_499.gif]  we have the inequality
                    [Graphics:../Images/RoucheTheoremModHome_gr_500.gif]

We are really done.   

Aside.  We can use Mathematica to plot the zeros of  g(z).

                    [Graphics:../Images/RoucheTheoremModHome_gr_501.gif]

                    Three zeros of  [Graphics:../Images/RoucheTheoremModHome_gr_502.gif]  lie in the disk [Graphics:../Images/RoucheTheoremModHome_gr_503.gif].  

Aside.  We can let Mathematica double check our work.

However, the zeros of  [Graphics:../Images/RoucheTheoremModHome_gr_504.gif]  have a complicated algebraic representation.  

So we might prefer numerical approximations:

[Graphics:../Images/RoucheTheoremModHome_gr_505.gif]

                    [Graphics:../Images/RoucheTheoremModHome_gr_506.gif]    

[Graphics:../Images/RoucheTheoremModHome_gr_507.gif]

                    [Graphics:../Images/RoucheTheoremModHome_gr_508.gif]   

The moduli of these roots are:  

                    [Graphics:../Images/RoucheTheoremModHome_gr_509.gif]  

Remark.  A factorization of the polynomial using numerical approximations for the coefficients is

          [Graphics:../Images/RoucheTheoremModHome_gr_510.gif]  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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