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

5 (c).  [Graphics:Images/ComplexGeometricSeriesModHome_gr_240.gif].

Solution 5 (c).

See text and/or instructor's solution manual.

Answer.   Converges in   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_241.gif].

Solution   Write  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_242.gif]  and use d'Alembert's ratio test and calculate the limit value  L:  

                    [Graphics:../Images/ComplexGeometricSeriesModHome_gr_243.gif]  

When   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_244.gif],   the series  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_245.gif]  will converge.  

Solve  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_246.gif]   and obtain the disk  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_247.gif],  which can be written as  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_248.gif].

We are done.   

Aside.  The sum of this geometric series is  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_249.gif]

Notice that  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_250.gif]  is not defined at the point  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_251.gif],  where the denominator is zero.  

The distance from  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_252.gif]  to the center  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_253.gif]  of the geometric series is  

                    [Graphics:../Images/ComplexGeometricSeriesModHome_gr_254.gif]

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexGeometricSeriesModHome_gr_255.gif]

[Graphics:../Images/ComplexGeometricSeriesModHome_gr_256.gif]

                                                                                          [Graphics:../Images/ComplexGeometricSeriesModHome_gr_257.gif]

                              The domain set  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_258.gif]  that is used to produce the images under  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_259.gif].  

 

[Graphics:../Images/ComplexGeometricSeriesModHome_gr_260.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_261.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_262.gif]


[Graphics:../Images/ComplexGeometricSeriesModHome_gr_263.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_264.gif]  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_265.gif]

     Graphs of the mappings  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_266.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_267.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_268.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_269.gif],   [Graphics:../Images/ComplexGeometricSeriesModHome_gr_270.gif],  and  [Graphics:../Images/ComplexGeometricSeriesModHome_gr_271.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_272.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 (a), (b), (d).

 

Comparison of Geometric Series
Solution.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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