Solution 5.

See text and/or instructor's solution manual.

Answer.   

Solution.   The transformation  [Graphics:../Images/JoukowskiTransModHome_gr_247.gif]  maps the point   [Graphics:../Images/JoukowskiTransModHome_gr_248.gif]   onto the point   [Graphics:../Images/JoukowskiTransModHome_gr_249.gif].

For the function  [Graphics:../Images/JoukowskiTransModHome_gr_250.gif]   we have   [Graphics:../Images/JoukowskiTransModHome_gr_251.gif].   Since   [Graphics:../Images/JoukowskiTransModHome_gr_252.gif],   

by Theorem 10.1 in Section 10.1 , the mapping is conformal at  [Graphics:../Images/JoukowskiTransModHome_gr_253.gif]  and "angles are preserved."

        The ray  [Graphics:../Images/JoukowskiTransModHome_gr_254.gif]  at  [Graphics:../Images/JoukowskiTransModHome_gr_255.gif]  make an angle  [Graphics:../Images/JoukowskiTransModHome_gr_256.gif].   In Section 10.2, we proved that a bilinear transformation maps the class of half-planes and disks onto itself.  

Hence, the image of a line through the origin under the Möbius transformation   [Graphics:../Images/JoukowskiTransModHome_gr_257.gif]   is a "circle."    Since   [Graphics:../Images/JoukowskiTransModHome_gr_258.gif],   and   

[Graphics:../Images/JoukowskiTransModHome_gr_259.gif]  the image of the ray  [Graphics:../Images/JoukowskiTransModHome_gr_260.gif] will be an arc   [Graphics:../Images/JoukowskiTransModHome_gr_261.gif]  of a circle that passes through the points  [Graphics:../Images/JoukowskiTransModHome_gr_262.gif].  

        Therefore, the arc   [Graphics:../Images/JoukowskiTransModHome_gr_263.gif]   is inclined at the angle  [Graphics:../Images/JoukowskiTransModHome_gr_264.gif]  at the point   [Graphics:../Images/JoukowskiTransModHome_gr_265.gif].  

 

We are done.   

 

Aside.  The image of the point [Graphics:../Images/JoukowskiTransModHome_gr_266.gif] will be the point   [Graphics:../Images/JoukowskiTransModHome_gr_267.gif]   on the arc   [Graphics:../Images/JoukowskiTransModHome_gr_268.gif]  as shown by the calculation

                    [Graphics:../Images/JoukowskiTransModHome_gr_269.gif]  

 

We are really done.   

 

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/JoukowskiTransModHome_gr_270.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_271.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_272.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_273.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_274.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_275.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_276.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_277.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_278.gif]


We are really really done.   

 

Aside.  We can explore some graphs.

          [Graphics:../Images/JoukowskiTransModHome_gr_279.gif]          [Graphics:../Images/JoukowskiTransModHome_gr_280.gif]

                    The image of the ray  [Graphics:../Images/JoukowskiTransModHome_gr_281.gif]  under   [Graphics:../Images/JoukowskiTransModHome_gr_282.gif]   is an arc  [Graphics:../Images/JoukowskiTransModHome_gr_283.gif]  of a circle that passes through  [Graphics:../Images/JoukowskiTransModHome_gr_284.gif]  and  [Graphics:../Images/JoukowskiTransModHome_gr_285.gif].  

                    If the ray  [Graphics:../Images/JoukowskiTransModHome_gr_286.gif] make an angle  [Graphics:../Images/JoukowskiTransModHome_gr_287.gif]  at  [Graphics:../Images/JoukowskiTransModHome_gr_288.gif]  then the arc  [Graphics:../Images/JoukowskiTransModHome_gr_289.gif]  is inclined at the angle  [Graphics:../Images/JoukowskiTransModHome_gr_290.gif]  at the point  [Graphics:../Images/JoukowskiTransModHome_gr_291.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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