Solution 4.

See text and/or instructor's solution manual.

Solution.  For  ,  we have

                    ,   and  ,  

                    ,   and  .

Substitute these values into  Laplace's equation:

                    ,  

which holds for all  .  

This also shows that    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.  

Caveat.  We must use caution when evaluating the function  ,  which is never defined at  .  Also, you cannot substitute  ,  so that it must be defined separately for points that lie on the y-axis.  Furthermore, its value for point    in quadrant III might compute as if they were in quadrant I.  (and point    in quadrant IV might compute as if they were in quadrant II.)   For this reason, the software Mathematica has the version of the arctangent called  ,  and it takes into consideration these difficulties, the following values are computed correctly:

                   ,    ,    ,    

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

                    ,

  is analytic in the domain  .  

This solution is complements of the authors.

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