Exercise 16.  Find the image of the semi-infinite vertical strip   [Graphics:Images/MapTrigonometricFunModHome_gr_1199.gif]   under the mapping   [Graphics:Images/MapTrigonometricFunModHome_gr_1200.gif].

Solution 16.

Answer.   The image of the semi infinite vertical strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_1201.gif]  under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_1202.gif]   

is the horizontal strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_1203.gif].

Hint.   Extend the results in Example 10.13.  You can use the information in Figure 10.17.

Short Solution.   The transformation  [Graphics:../Images/MapTrigonometricFunModHome_gr_1204.gif]  maps the semi-infinite vertical strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_1205.gif]   

onto the upper half plane  [Graphics:../Images/MapTrigonometricFunModHome_gr_1206.gif].   Then the transformation   [Graphics:../Images/MapTrigonometricFunModHome_gr_1207.gif]   maps the upper half plane  

[Graphics:../Images/MapTrigonometricFunModHome_gr_1208.gif]   onto the horizontal strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_1209.gif].

Therefore, the image of the semi infinite vertical strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_1210.gif]  under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_1211.gif]   

is the horizontal strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_1212.gif].

We might be done.   

Solution.   The mapping  [Graphics:../Images/MapTrigonometricFunModHome_gr_1213.gif]  can be written as a composition  

                     [Graphics:../Images/MapTrigonometricFunModHome_gr_1214.gif],    

where   

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1215.gif],    and   

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1216.gif].

        Using Equation (5-33), we write  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1217.gif].  

If  [Graphics:../Images/MapTrigonometricFunModHome_gr_1218.gif],  then the image of the vertical line  [Graphics:../Images/MapTrigonometricFunModHome_gr_1219.gif]  is the curve in the Z-plane given by the parametric equations  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1220.gif]    and   [Graphics:../Images/MapTrigonometricFunModHome_gr_1221.gif],    for    [Graphics:../Images/MapTrigonometricFunModHome_gr_1222.gif].  

The inequalities  [Graphics:../Images/MapTrigonometricFunModHome_gr_1223.gif]  imply that the image points lie in the upper half plane  [Graphics:../Images/MapTrigonometricFunModHome_gr_1224.gif].

Eliminate y from the parametric equations and the result is  [Graphics:../Images/MapTrigonometricFunModHome_gr_1225.gif].  

When [Graphics:../Images/MapTrigonometricFunModHome_gr_1226.gif] the image curve approaches the ray  [Graphics:../Images/MapTrigonometricFunModHome_gr_1227.gif].  

When [Graphics:../Images/MapTrigonometricFunModHome_gr_1228.gif] the image curve approaches the ray  [Graphics:../Images/MapTrigonometricFunModHome_gr_1229.gif].  

        Hence, the image of the semi-infinite strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_1230.gif]   under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_1231.gif]  

the upper half plane  [Graphics:../Images/MapTrigonometricFunModHome_gr_1232.gif].

        The upper half plane  [Graphics:../Images/MapTrigonometricFunModHome_gr_1233.gif]  can be expressed as

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1234.gif].  

Recall that   [Graphics:../Images/MapTrigonometricFunModHome_gr_1235.gif]   which can be written as  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1236.gif]   and   [Graphics:../Images/MapTrigonometricFunModHome_gr_1237.gif].

Then,   [Graphics:../Images/MapTrigonometricFunModHome_gr_1238.gif]   implies that   [Graphics:../Images/MapTrigonometricFunModHome_gr_1239.gif]   which in turn implies that   [Graphics:../Images/MapTrigonometricFunModHome_gr_1240.gif].   

Also,   [Graphics:../Images/MapTrigonometricFunModHome_gr_1241.gif]   implies that   [Graphics:../Images/MapTrigonometricFunModHome_gr_1242.gif].   

        Hence, the image of the upper half-plane   [Graphics:../Images/MapTrigonometricFunModHome_gr_1243.gif],  under the mapping  

[Graphics:../Images/MapTrigonometricFunModHome_gr_1244.gif]  is the horizontal strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_1245.gif].

        Therefore, image of the semi-infinite strip   [Graphics:../Images/MapTrigonometricFunModHome_gr_1246.gif]   under the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_1247.gif]  

is the horizontal strip  [Graphics:../Images/MapTrigonometricFunModHome_gr_1248.gif].

We are done.   

Aside.  We can look at some graphs of the mapping  [Graphics:../Images/MapTrigonometricFunModHome_gr_1249.gif].

 

          The mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_1256.gif],   followed by the mapping   [Graphics:../Images/MapTrigonometricFunModHome_gr_1257.gif].

          Observe the points [Graphics:../Images/MapTrigonometricFunModHome_gr_1258.gif] and their images  [Graphics:../Images/MapTrigonometricFunModHome_gr_1259.gif]  and  [Graphics:../Images/MapTrigonometricFunModHome_gr_1260.gif]  

                    [Graphics:../Images/MapTrigonometricFunModHome_gr_1261.gif]          [Graphics:../Images/MapTrigonometricFunModHome_gr_1262.gif]          [Graphics:../Images/MapTrigonometricFunModHome_gr_1263.gif]  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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