Solution 1 (b).

See text and/or instructor's solution manual.

Answer.     is harmonic for  .

Solution.  For  ,  we have

          ,   and  

          ,   and  

          ,

which holds for all  .  Hence    is harmonic for  .

We are done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

Aside.  In Exercise 8 (a) in Section 3.2 we saw that    is analytic in the domain  

.   When f(z) is expressed in Cartesian coordinates we see that

                    ,

is analytic in the domain    and from Theorem 3.8 it follows that it's real and imaginary parts

                    ,      are harmonic functions for all  .  

For this exercise we can conclude that    is a harmonic function too.  

Notice that Laplace's equation holds for all  ,  so that and are actually harmonic for  all  .  

Remark.  In Section 5.2 we will learn that this is the complex logarithm function, i.e.  

                    ,

  is analytic in the domain  .  

          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 inverse function  .  

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

          Remark.  In Section 5.1 we will introduce the formulas for the complex exponential function  .   

This solution is complements of the authors.

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