Exercise 10.  Find all values of  z  for which each equation holds.  

10 (a).   .  

Solution 10 (a).

See text and/or instructor's solution manual.

Answer.   ,   where  k  is an integer.  

Solution.   Starting with Identity (5-34)  ,   we write

                    .  

If we equate real and imaginary parts, then we get

                        and    .  

The equation    implies either that  ,  where  n  is an integer, or that  .  

Using    in the equation     leads to the impossible situation  .  

Therefore  ,  where  n  is an integer.  

Since    for all values of  y,  the term    in the equation    must also be positive.  

For this reason we eliminate the odd values of  n  and get  ,  where  k  is an integer.

    Finally, we solve the equation   which can be written as  

and use the fact that    is an even function to conclude that  .  

Therefore the solutions to the equation    are  ,  where  k  is an integer.

We are done.   

                    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 really done.   

Aside.  We can let Mathematica double check our work.

This solution is complements of the authors.

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