Exercise 1.  Use the Cauchy Riemann conditions to determine where the following functions are differentiable, and evaluate the derivatives at those points where they exist.  

1 (e).  [Graphics:Images/CauchyRiemannModHome_gr_137.gif].  

Solution 1 (e).

See text and/or instructor's solution manual.

Answer.  [Graphics:../Images/CauchyRiemannModHome_gr_138.gif]  is differentiable only at the point  [Graphics:../Images/CauchyRiemannModHome_gr_139.gif],  and  [Graphics:../Images/CauchyRiemannModHome_gr_140.gif].  

Solution.  [Graphics:../Images/CauchyRiemannModHome_gr_141.gif],     and      

                 [Graphics:../Images/CauchyRiemannModHome_gr_142.gif],      [Graphics:../Images/CauchyRiemannModHome_gr_143.gif]     so that  

[Graphics:../Images/CauchyRiemannModHome_gr_144.gif],     [Graphics:../Images/CauchyRiemannModHome_gr_145.gif],     [Graphics:../Images/CauchyRiemannModHome_gr_146.gif],     [Graphics:../Images/CauchyRiemannModHome_gr_147.gif].   

The  Cauchy Riemann relations are  

        [Graphics:../Images/CauchyRiemannModHome_gr_148.gif],  

        [Graphics:../Images/CauchyRiemannModHome_gr_149.gif],

which hold when [Graphics:../Images/CauchyRiemannModHome_gr_150.gif],  which occurs only at the point  [Graphics:../Images/CauchyRiemannModHome_gr_151.gif].   

The partials are continuous everywhere, so at the point  [Graphics:../Images/CauchyRiemannModHome_gr_152.gif]  we find that    

                    [Graphics:../Images/CauchyRiemannModHome_gr_153.gif].  

Caveat.  It would be false to use the formula  [Graphics:../Images/CauchyRiemannModHome_gr_154.gif]  for values other than  [Graphics:../Images/CauchyRiemannModHome_gr_155.gif].    

We are done.   

Aside.  A reason why f(z) is not analytic is given in Exercise 16.  

     Loosely speaking, if f(z) is analytic then there cannot be any occurrence of the variable  [Graphics:../Images/CauchyRiemannModHome_gr_156.gif].

Recall the identities  [Graphics:../Images/CauchyRiemannModHome_gr_157.gif]  and  [Graphics:../Images/CauchyRiemannModHome_gr_158.gif]  that were used in Section 2.1.

They can be substituted in [Graphics:../Images/CauchyRiemannModHome_gr_159.gif],  and the result is

                    [Graphics:../Images/CauchyRiemannModHome_gr_160.gif]  

So technically there are "hidden" terms involving  [Graphics:../Images/CauchyRiemannModHome_gr_161.gif]  in the formula for  f(z),  and this precludes the possibility that  f(z)  can be analytic.

Indeed, the complex form of the Cauchy-Riemann equations fails to hold because  [Graphics:../Images/CauchyRiemannModHome_gr_162.gif].  

We are really done.   

Aside.  We can let Mathematica double check our work.

          [Graphics:../Images/CauchyRiemannModHome_gr_163.gif]     [Graphics:../Images/CauchyRiemannModHome_gr_164.gif]

  

                    The image of this orthogonal grid under  [Graphics:../Images/CauchyRiemannModHome_gr_165.gif]  is by luck an orthogonal grid.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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