Revisited Example 3.4.  We know that      is differentiable and that   .   

Furthermore, the polar coordinate form for    is

                    .  

Use the polar coordinate form of the Cauchy-Riemann equations and prove that    is differentiable for all  .  

Explore Revisited Solution 3.4.

Solution.  It is easy to verify that polar form of the Cauchy-Riemann equations (3-22) are indeed satisfied for all  .  

                    ,     and     
                    
                    .  

Moreover, the partial derivatives    are continuous for all  .  

By Theorem 3.5,   ,  is differentiable for all  .  

Therefore, using Equation (3-23) and (3-24), we have    

                      

and   

                    
                    
as expected.

You might wonder why we required .  

This happens because equations (3-22) do not hold at  .  

Of course, for the function  ,  it is well known that  .   

We are done.

Aside.  Both and can assist us in finding the partial derivatives.  

Aside.  The Mathematica solution uses the commands.  

Mathematica shows that the polar form of the Cauchy-Riemann equations (3-22) are satisfied for all  ,    

                    ,     and     
                    
                    .  

Aside.  The Maple commands are similar.  

          >  

                              

          >  

                              

          >  

                              

          >    

                              

          >    

                              

          >    

                              

          >    

                              

          >    

                              

Maple shows that the polar form of the Cauchy-Riemann equations (3-22) are satisfied for all  ,    

                    ,     and     
                    
                    .  

The Cauchy-Riemann equations hold  all points    in the complex plane,

therefore      is an analytic function, for all  .  

Verify that the derivative can be calculated with either of the formulas:

(3-23)              ,     or    

(3-24)              .  

Aside.  The Mathematica solution uses the commands.  

Aside.  The Maple commands are similar.  

          >    

                              

          >    

                              

          >    

                              

          >    

                              

          >    

                              

Therefore, both Mathematica and Maple have shown that if  

                    ,   

then the derivative is

                    ,   

as expected.

We are really done.

Aside.  Figure R.3.4 a, shows the graphs of      and   .  

The partial derivatives of    are  

                         and     ,   

and the partial derivatives of    are

                         and     .  

        They satisfy the Cauchy-Riemann equations (3-16) because they are the real and imaginary parts of an analytic function.  
        
At the point   ,   we have      and   ,   and these partial

derivatives appear along the edges of the surfaces for    at the points      

and   ,   respectively.   Similarly,  at the point   ,   we have   

  and      and these partial derivatives appear along the edges of the surfaces

for    at the points      and   ,   respectively.

                  .                                                          .  
                                                                        Figure R.3.4 a

                  ,   and                                                 ,   and  

                  .                                                             .  
                                                                        Figure R.3.4 b

                  ,   and                                                 ,   and  

                  .                                                           .  
                            
                            Remark. It is difficult to visualize    because this partial derivative   is

                            taken with respect to changes in the polar angle  ,  and so it cannot be visualized as an "ordinary slope."

                                                                        Figure R.3.4 c

                            For the function      we see that

                            ,     and     
                                                  
                            .  

                                                                        Figure R.3.4

We are really really done.

Aside.  We can let Mathematica check out the calculations given above.  

Remark.  In this book the use of computers is optional.  

Hopefully this text will promote their use and understanding.

This solution is complements of the authors.

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

(c) 2011 John H. Mathews, Russell W. Howell