Exercise 1.  Find  [Graphics:Images/ComplexGeometryContinuedModHome_gr_1.gif]  for the following values of  z.  

1 (g).  [Graphics:Images/ComplexGeometryContinuedModHome_gr_104.gif].  

Solution 1 (g).

See text and/or instructor's solution manual.

One solution is obtained by simplifying the quotient  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_105.gif] as follows:

        [Graphics:../Images/ComplexGeometryContinuedModHome_gr_106.gif]

then

    [Graphics:../Images/ComplexGeometryContinuedModHome_gr_107.gif]   

We are done.   

    Another solution is obtained by using the formula  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_108.gif],  where n is an integer.

Since  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_109.gif]  and  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_110.gif] and   [Graphics:../Images/ComplexGeometryContinuedModHome_gr_111.gif]  and  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_112.gif],  an argument of  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_113.gif] is easily found to be

        [Graphics:../Images/ComplexGeometryContinuedModHome_gr_114.gif]  

If we choose  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_115.gif],  then  [Graphics:../Images/ComplexGeometryContinuedModHome_gr_116.gif]  lies in the interval[Graphics:../Images/ComplexGeometryContinuedModHome_gr_117.gif],  

so the principal value of the argument is found to be

        [Graphics:../Images/ComplexGeometryContinuedModHome_gr_118.gif]

We are done.   

Aside.  We can let Mathematica double check our work.

        

[Graphics:../Images/ComplexGeometryContinuedModHome_gr_119.gif]

  

[Graphics:../Images/ComplexGeometryContinuedModHome_gr_120.gif]
[Graphics:../Images/ComplexGeometryContinuedModHome_gr_121.gif]
[Graphics:../Images/ComplexGeometryContinuedModHome_gr_122.gif]




















This solution is complements of the authors.







































 

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