Exercise 11.  Determine where the following functions are differentiable and where they are analytic.  Explain!  

11 (a).  [Graphics:Images/CauchyRiemannModHome_gr_591.gif].

Solution 11 (a).

See text and/or instructor's solution manual.

Answer.  f(z)  is differentiable only at points on the coordinate axes.    f(z)  is nowhere analytic.

Solution.  [Graphics:../Images/CauchyRiemannModHome_gr_592.gif],   and  

                 [Graphics:../Images/CauchyRiemannModHome_gr_593.gif],     and     [Graphics:../Images/CauchyRiemannModHome_gr_594.gif]    so that  

[Graphics:../Images/CauchyRiemannModHome_gr_595.gif],     [Graphics:../Images/CauchyRiemannModHome_gr_596.gif],     [Graphics:../Images/CauchyRiemannModHome_gr_597.gif],     [Graphics:../Images/CauchyRiemannModHome_gr_598.gif].   

The  Cauchy Riemann equations are  

        [Graphics:../Images/CauchyRiemannModHome_gr_599.gif],  

        [Graphics:../Images/CauchyRiemannModHome_gr_600.gif],

which hold only for the points that satisfy  [Graphics:../Images/CauchyRiemannModHome_gr_601.gif],  which implies that  [Graphics:../Images/CauchyRiemannModHome_gr_602.gif],  and this happens when either  [Graphics:../Images/CauchyRiemannModHome_gr_603.gif].

Hence, f(z) is differentiable only at points on the coordinate axes.    

However, at each point on the x-axis or y-axis there fails to be a neighborhood in which f(z) is differentiable.  

Therefore  f(z)  is nowhere analytic.

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_604.gif].

Recall the identities  [Graphics:../Images/CauchyRiemannModHome_gr_605.gif]  and  [Graphics:../Images/CauchyRiemannModHome_gr_606.gif]  that were used in Section 2.1.

They can be substituted in [Graphics:../Images/CauchyRiemannModHome_gr_607.gif],  and the result is

                    [Graphics:../Images/CauchyRiemannModHome_gr_608.gif]  

So technically there are "hidden" terms involving  [Graphics:../Images/CauchyRiemannModHome_gr_609.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_610.gif].  

We are really done.   

Aside.  We can let Mathematica double check our work.

      [Graphics:../Images/CauchyRiemannModHome_gr_611.gif]     [Graphics:../Images/CauchyRiemannModHome_gr_612.gif]

  

                    The image of this orthogonal grid under  [Graphics:../Images/CauchyRiemannModHome_gr_613.gif]  is not an orthogonal grid.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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