Exercise 5.  Use the ratio test to find a disk in which the following geometric series converge.  

5 (a).  [Graphics:Images/ComplexGeometricSeriesModHome_gr_174.gif].

Solution 5 (a).

See text and/or instructor's solution manual.

Answer.   Converges in   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_175.gif].

Solution   Write  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_176.gif]  and use d'Alembert's ratio test and calculate the limit value  L:  

                    [Graphics:../Images/ComplexGeometricSeriesModHome_gr_177.gif]  

When   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_178.gif],   the series  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_179.gif]  will converge.  

Solve  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_180.gif]   and obtain the disk  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_181.gif],  which can be written as  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_182.gif].

We are done.   

Aside.  The sum of this geometric series is  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_183.gif].  

Notice that  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_184.gif]  is not defined at the point  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_185.gif],  where the denominator is zero.  

The distance from  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_186.gif]  to the center  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_187.gif]  of the geometric series is  

                    [Graphics:../Images/ComplexGeometricSeriesModHome_gr_188.gif]

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexGeometricSeriesModHome_gr_189.gif]

[Graphics:../Images/ComplexGeometricSeriesModHome_gr_190.gif]

                                                                                          [Graphics:../Images/ComplexGeometricSeriesModHome_gr_191.gif]

                              The domain set  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_192.gif]  that is used to produce the images under  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_193.gif].

 

[Graphics:../Images/ComplexGeometricSeriesModHome_gr_194.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_195.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_196.gif]


[Graphics:../Images/ComplexGeometricSeriesModHome_gr_197.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_198.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_199.gif]

     Graphs of the mappings  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_200.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_201.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_202.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_203.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_204.gif],   and   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_205.gif].

Remark 1.  In Section 2.1 we saw that the image of a circle under a "linear mapping" is a circle, and in Section 2.5 we saw that the image of a "circle" under a the reciprocal transformation is a "circle."  Here we have  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_206.gif]  and it is the composition of a linear mapping followed by the reciprocal mapping, it too will map "circles" onto "circles".  So it is not surprising that the final graph is a circle.  In Section 10.2 we will see that the bilinear transformation or Möbius transformation also has this property.
Remark 2.  It is worthwhile to compare these images with the ones in 5 (b), (c), (d).

 

Comparison of Geometric Series
Solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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