Exercise 7.  Consider the conformal mapping   [Graphics:Images/ElectrostaticsModHome_gr_279.gif].  

7 (a).   Show that  [Graphics:Images/ElectrostaticsModHome_gr_280.gif]  maps the domain  D  that is the portion of the right half-plane   [Graphics:Images/ElectrostaticsModHome_gr_281.gif],  

that lies exterior to the circle   [Graphics:Images/ElectrostaticsModHome_gr_282.gif]   onto the annulus   [Graphics:Images/ElectrostaticsModHome_gr_283.gif].  

Solution 7 (a).

See text and/or instructor's solution manual.

Solution.   First, the boundary of the right half-plane  [Graphics:../Images/ElectrostaticsModHome_gr_284.gif]  can be give a positive orientation by using the points [Graphics:../Images/ElectrostaticsModHome_gr_285.gif],  [Graphics:../Images/ElectrostaticsModHome_gr_286.gif],  and  [Graphics:../Images/ElectrostaticsModHome_gr_287.gif].  

The image points are  [Graphics:../Images/ElectrostaticsModHome_gr_288.gif],  [Graphics:../Images/ElectrostaticsModHome_gr_289.gif],  and  [Graphics:../Images/ElectrostaticsModHome_gr_290.gif],  

and give the disk [Graphics:../Images/ElectrostaticsModHome_gr_291.gif]  a positive orientation.   Next, consider region that lies exterior to the circle   [Graphics:../Images/ElectrostaticsModHome_gr_292.gif].    

The boundary  [Graphics:../Images/ElectrostaticsModHome_gr_293.gif]  can be give a positive orientation by using the points [Graphics:../Images/ElectrostaticsModHome_gr_294.gif],  [Graphics:../Images/ElectrostaticsModHome_gr_295.gif],  and  [Graphics:../Images/ElectrostaticsModHome_gr_296.gif].  

The image points are  [Graphics:../Images/ElectrostaticsModHome_gr_297.gif],  [Graphics:../Images/ElectrostaticsModHome_gr_298.gif],  (where  [Graphics:../Images/ElectrostaticsModHome_gr_299.gif]),  and  [Graphics:../Images/ElectrostaticsModHome_gr_300.gif]  

and give the region [Graphics:../Images/ElectrostaticsModHome_gr_301.gif]  a positive orientation.  

        Therefore   [Graphics:../Images/ElectrostaticsModHome_gr_302.gif]   maps the portion of the right half-plane   [Graphics:../Images/ElectrostaticsModHome_gr_303.gif]  

that lies exterior to the circle   [Graphics:../Images/ElectrostaticsModHome_gr_304.gif]    onto the annulus  [Graphics:../Images/ElectrostaticsModHome_gr_305.gif].  

Furthermore,  the imaginary axis   [Graphics:../Images/ElectrostaticsModHome_gr_306.gif]   is mapped onto the circle   [Graphics:../Images/ElectrostaticsModHome_gr_307.gif],  

and the  circle   [Graphics:../Images/ElectrostaticsModHome_gr_308.gif]   is mapped onto the unit circle   [Graphics:../Images/ElectrostaticsModHome_gr_309.gif].  

We are done.   

Aside.  For illustration purposes we can graph the mapping   [Graphics:../Images/ElectrostaticsModHome_gr_310.gif].   

                     [Graphics:../Images/ElectrostaticsModHome_gr_311.gif]          [Graphics:../Images/ElectrostaticsModHome_gr_312.gif]

  

                    The mapping   [Graphics:../Images/ElectrostaticsModHome_gr_313.gif].  

 

We are really done.   

 

        We can use Mathematica to explore some of the computations.

        The function   [Graphics:../Images/ElectrostaticsModHome_gr_314.gif]   maps  [Graphics:../Images/ElectrostaticsModHome_gr_315.gif]

onto   [Graphics:../Images/ElectrostaticsModHome_gr_316.gif],   respectively.  

[Graphics:../Images/ElectrostaticsModHome_gr_317.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_318.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_319.gif]


[Graphics:../Images/ElectrostaticsModHome_gr_320.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_321.gif]


[Graphics:../Images/ElectrostaticsModHome_gr_322.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_323.gif]

Hence, the right half-plane  [Graphics:../Images/ElectrostaticsModHome_gr_324.gif]  is mapped onto the circle  [Graphics:../Images/ElectrostaticsModHome_gr_325.gif].   

        The function   [Graphics:../Images/ElectrostaticsModHome_gr_326.gif]   maps  [Graphics:../Images/ElectrostaticsModHome_gr_327.gif]

onto   [Graphics:../Images/ElectrostaticsModHome_gr_328.gif] ,   respectively.  

[Graphics:../Images/ElectrostaticsModHome_gr_329.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_330.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_331.gif]


[Graphics:../Images/ElectrostaticsModHome_gr_332.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_333.gif]


[Graphics:../Images/ElectrostaticsModHome_gr_334.gif]

[Graphics:../Images/ElectrostaticsModHome_gr_335.gif]

Hence, the circle   [Graphics:../Images/ElectrostaticsModHome_gr_336.gif]   is mapped onto the circle  [Graphics:../Images/ElectrostaticsModHome_gr_337.gif].   

        Therefore   [Graphics:../Images/ElectrostaticsModHome_gr_338.gif]   maps the portion of the right half-plane   [Graphics:../Images/ElectrostaticsModHome_gr_339.gif]  

that lies exterior to the circle   [Graphics:../Images/ElectrostaticsModHome_gr_340.gif]    onto the annulus  [Graphics:../Images/ElectrostaticsModHome_gr_341.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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