Exercise 1.  Find all values of the following.  

1 (a).   .  

Solution 1 (a).

Answer.       where  n  is an integer.

Solution.   Use the Identity in (5-45)      and get:    

                      

There are two choices   

                       
and
                      

Therefore,       where  n  is an integer.

                    The points   

                    and  .  

        You may wonder why the solutions look the way they do.  Recall the identity    which holds for all z.  

Hence, if  z  is a value such that      then so is     because   .  

We are done.   

Aside.  We can let Mathematica double check our work.

This can be accomplished by performing the computation    .  

We are really done.

Aside.  The principal value of    can be calculated with Mathematica.

                    The point    the principal value image point  .  

Also, we can substitute  n = 0  into the general formula    and get   

We are really really done.

                    The point    and the principal value image point  .  

Remark.  We will study mappings with the trigonometric functions in Section 10.4 and in some of the applications that follow in Chapter 11.

This solution is complements of the authors.

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