Exercise 18.  Show the following concerning the exponential map  [Graphics:Images/ComplexFunExponentialModHome_gr_533.gif].  

18 (d).  The image of the vertical line segment  [Graphics:Images/ComplexFunExponentialModHome_gr_599.gif],  where  [Graphics:Images/ComplexFunExponentialModHome_gr_600.gif]  is half a circle.

Solution 18 (d).

See text and/or instructor's solution manual.

Solution.    [Graphics:../Images/ComplexFunExponentialModHome_gr_601.gif].  

Substitute  [Graphics:../Images/ComplexFunExponentialModHome_gr_602.gif]   into the equations  [Graphics:../Images/ComplexFunExponentialModHome_gr_603.gif]  and  [Graphics:../Images/ComplexFunExponentialModHome_gr_604.gif]  and obtain

                    [Graphics:../Images/ComplexFunExponentialModHome_gr_605.gif].

Notice that   [Graphics:../Images/ComplexFunExponentialModHome_gr_606.gif]   and   [Graphics:../Images/ComplexFunExponentialModHome_gr_607.gif]   and   [Graphics:../Images/ComplexFunExponentialModHome_gr_608.gif].  

This set is the parametric representation of half a circle of radius  [Graphics:../Images/ComplexFunExponentialModHome_gr_609.gif]  subtended by the angles  [Graphics:../Images/ComplexFunExponentialModHome_gr_610.gif]

We are done.   

Aside.  We can let Mathematica double check our work.

                              [Graphics:../Images/ComplexFunExponentialModHome_gr_611.gif]          [Graphics:../Images/ComplexFunExponentialModHome_gr_612.gif]

                                                            The image the horizontal strip  [Graphics:../Images/ComplexFunExponentialModHome_gr_613.gif],  
                                                            is the half circle   [Graphics:../Images/ComplexFunExponentialModHome_gr_614.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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