Solution 1 (a).

See text and/or instructor's solution manual.

Answer.     and    are harmonic for all values of  .  

Solution.  For  ,  we have

          ,   and  ,  

          ,   and  ,  

          ,  

which holds for all z.  Hence    is harmonic for all values of  (x,y).

Similarly, for  ,  we have

          ,   and  ,  

          ,   and  ,  

          ,  

which holds for all z.  Hence    is harmonic for all values of  (x,y).

We are done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

Aside.  In Exercise 5 in Section 3.2 we saw that    is differentiable for all  z.  

Therefore    is analytic for all  z  and from Theorem 3.8 it follows that it's real and imaginary parts

                    ,      are harmonic functions for all  .    

Remark. In Section 5.1 we will learn that    is the complex exponential function.  

            The level curves    and  .  

          The orthogonal grid formed with    and  .  

We are really really done.   

                    In Section 11.4 we will prove that the image of an orthogonal grid under an analytic function is an orthogonal grid.  

          It is best to worry about these concepts when we get there because this example involves the inverse transformation  .

          The orthogonal grid formed by the image of a rectangular grid under the multivalued inverse function  .  

          The spacing between curves is not the same as in the previous figures because lines in the domain grid are equally spaced.

          Remark.  In Section 5.2 we will introduce the formulas for the complex logarithm function.   

This solution is complements of the authors.

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