Exercise 3.  Find the radius of convergence of the following.  

3 (a).  [Graphics:Images/ComplexPowerSeriesModHome_gr_164.gif].  

Solution 3 (a).

See text and/or instructor's solution manual.

Answer   The radius of convergence of  [Graphics:../Images/ComplexPowerSeriesModHome_gr_165.gif]  is   [Graphics:../Images/ComplexPowerSeriesModHome_gr_166.gif].

Solution.  For   [Graphics:../Images/ComplexPowerSeriesModHome_gr_167.gif],  we have  [Graphics:../Images/ComplexPowerSeriesModHome_gr_168.gif].  

Use Theorem 4.16 and calculate the limit in in d'Alembert's Ratio Test:    

                    [Graphics:../Images/ComplexPowerSeriesModHome_gr_169.gif]   
                    
Since  [Graphics:../Images/ComplexPowerSeriesModHome_gr_170.gif]  the radius of convergence of the series  [Graphics:../Images/ComplexPowerSeriesModHome_gr_171.gif]  is   [Graphics:../Images/ComplexPowerSeriesModHome_gr_172.gif].  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexPowerSeriesModHome_gr_173.gif]

[Graphics:../Images/ComplexPowerSeriesModHome_gr_174.gif]

We are really done.   

Aside.  The Taylor series for  [Graphics:../Images/ComplexPowerSeriesModHome_gr_175.gif]  was studied in calculus, and we have  

                    [Graphics:../Images/ComplexPowerSeriesModHome_gr_176.gif],  

which converges for all  x.

In Section 5.4 we will find that complex series are extensions of real series and we will derive the formula

                    [Graphics:../Images/ComplexPowerSeriesModHome_gr_177.gif],

which converges for all  z.  Using the substitution  [Graphics:../Images/ComplexPowerSeriesModHome_gr_178.gif] will establish that the series at hand is really

                    [Graphics:../Images/ComplexPowerSeriesModHome_gr_179.gif].

Remark.  At first glance, a first reaction might be:  [Graphics:../Images/ComplexPowerSeriesModHome_gr_180.gif]  could not have a power series expansion about [Graphics:../Images/ComplexPowerSeriesModHome_gr_181.gif]  because [Graphics:../Images/ComplexPowerSeriesModHome_gr_182.gif] does not.

However, the series [Graphics:../Images/ComplexPowerSeriesModHome_gr_183.gif] involves powers of   [Graphics:../Images/ComplexPowerSeriesModHome_gr_184.gif].

We are really done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexPowerSeriesModHome_gr_185.gif]

[Graphics:../Images/ComplexPowerSeriesModHome_gr_186.gif]

                                                                                [Graphics:../Images/ComplexPowerSeriesModHome_gr_187.gif]

                              The domain set  [Graphics:../Images/ComplexPowerSeriesModHome_gr_188.gif]  that is used to produce the images under  [Graphics:../Images/ComplexPowerSeriesModHome_gr_189.gif]

 

[Graphics:../Images/ComplexPowerSeriesModHome_gr_190.gif]          [Graphics:../Images/ComplexPowerSeriesModHome_gr_191.gif]          [Graphics:../Images/ComplexPowerSeriesModHome_gr_192.gif]


[Graphics:../Images/ComplexPowerSeriesModHome_gr_193.gif]          [Graphics:../Images/ComplexPowerSeriesModHome_gr_194.gif]          [Graphics:../Images/ComplexPowerSeriesModHome_gr_195.gif]

     Graphs of the mappings  [Graphics:../Images/ComplexPowerSeriesModHome_gr_196.gif],   [Graphics:../Images/ComplexPowerSeriesModHome_gr_197.gif],   [Graphics:../Images/ComplexPowerSeriesModHome_gr_198.gif],   [Graphics:../Images/ComplexPowerSeriesModHome_gr_199.gif],   [Graphics:../Images/ComplexPowerSeriesModHome_gr_200.gif],   and   [Graphics:../Images/ComplexPowerSeriesModHome_gr_201.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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