Exercise 11.  Show that the zeros of    are at    where n is an integer.  

Solution 11.

See text and/or instructor's solution manual.

Answer.   By identity (5-33),   ,  if and only if  .  

Equate real and imaginary parts to show this occurs if and only if     where n is an integer.   

Solution.  Starting with Identity (5-33)    we get:

                    ,    if and only if    

                    .   

Equating the real and imaginary parts of this equation gives

                        and    .  

The real-valued function    is never zero, so the equation    implies that  ,  

from which we obtain    for any integer  n .  

Using the values    in the equation    yields  

            .  

which implies that  ,  so the only zeros for    are the values   for  n  an integer.

Aside.  We can let Mathematica double check our work.

This solution is complements of the authors.

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