Section 3.2  The Cauchy-Riemann Equations

    In Section 3.1 we showed that computing the derivative of complex functions written in a form such as      

is a rather simple task.  But life isn't always so easy.   Many times we encounter complex functions written as  

(3-13)              .   

For example, suppose we had  

(3-13)              .  

Is there some criterion - perhaps involving the partial derivatives of      and   ,  

that we can use to determine whether    is differentiable, and if so, to find the value of  ?

    The answer to this question is yes, thanks to the independent discovery of two important equations by the French

mathematician Augustin Louis Cauchy (1789-1857) and the German mathematician Georg Friedrich Bernhard Riemann (1826-1866).

    First, let's reconsider the derivative of   .    As we have stated, the limit given in Equation (3-1) must not depend

on how approaches  ,  and a calculation similar to Example 3.1 (in Section 3.1), will prove that   .

              

We can drop the subscript on    to obtain      as a general formula.

The Special Cartesian Limits.

        For the Cartesian coordinate form of a complex function

                    ,

it is important to determine how the function values change as we move along the horizontal grid line  

                        at the point    ,  

and how the function values change as we move along the vertical grid line  

                        at the point    .  

        We investigate these two approaches: a horizontal approach and a vertical approach to  .  Recall from our graphical

analysis of      in Example 2.12, in Section 2.2, that the image of a square is a "curvilinear quadrilateral" and the

images of the horizontal and vertical edges are portions of parabolas in the  -plane.   For convenience, we let the square have

vertices   ,    ,    ,   and   .    Then the image

points are  ,    ,    ,   and   ,  

as shown in Figure 3.1.  

                  Figure 3.1  The image of a small square under the mapping    ,  

                                     the vertex  vertex   ,   is mapped onto the point   .  

        We know that      is differentiable, so the limit of the difference quotient      exists

no matter how we approach   .   Let us investigate the two special Cartesian limits.

First, we can numerically approximate      by using a horizontal increment in .   

Use      and      where   

to compute the difference quotient.

                       

Second, we can numerically approximate      by using a vertical increment in .  

Use      and      where   

to compute the difference quotient.

                          

Comparing these two numerical approximations we see that

                    ,     

                    and  

                    ,  

which leads us to speculate that   .  

    These numerical approximations lead to the idea of taking limits along the horizontal and vertical directions.  

First, we can taking the limit along the horizontal direction.  

                       

Second, we can taking the limit along the vertical direction.  

                      

Comparing these two limits we see that

                    ,     

                    and  

                    .  

Since the above two limits were not taken along all possible approaches to   ,   they alone

are not sufficient to prove that   ,   but they prepare our thinking for Theorem 3.3.

Exploration

    We now generalize this idea by taking limits of an arbitrary differentiable complex function and obtain an important result.

Theorem 3.3 (Cauchy-Riemann Equations).  Suppose that  

(3-14)              ,  

is differentiable at the point  .   Then the partial derivatives of    exist at the point  ,  

and can be used to calculate the derivative at .  That is,    

(3-14)              ,     

                        and also  

(3-15)              .  

Equating the real and imaginary parts of Equations (3-14) and (3-15) gives the so-called Cauchy-Riemann Equations:

(3-16)                   and     .  

Proof.

Exploration for the Cauchy-Riemann Equations.

        Note some of the important implications of Theorem 3.3.
    
(i).    If f is differentiable at  ,  then the Cauchy-Riemann Equations (3-16) will be satisfied at  ,

         and we can use either either Equation (3-14) or (3-15) to evaluate  .  
    
(ii).   Taking the contrapositive, if Equations (3-16) are not satisfied at  ,  

          then we know automatically that    is not differentiable at  .  

(iii).  Even if Equations (3-16) are satisfied at  ,  we cannot necessarily conclude that    is differentiable at  .

    We now illustrate each of these points.

Example 3.4.  We know that      is differentiable and that   .   

Furthermore, the Cartesian coordinate form for    is

                    .  

Use the Cartesian coordinate form of the Cauchy-Riemann equations and find  .  

Solution.  It is easy to verify that Cauchy-Riemann equations (3-16) are indeed satisfied:

                         and     .  

Using Equations (3-14) and (3-15), respectively, to compute    gives

                    ,     

                    and   

                    ,
                    
as expected.

Explore Solution 3.4.

Example 3.5.  Show that      is nowhere differentiable.

Solution.  We have   ,   where  

                         and     .  

Thus, for any point  ,  

                         and     .  

The Cauchy-Riemann equations (3-16) are not satisfied at any point  ,  so we conclude that  

   is nowhere differentiable.

Explore Solution 3.5.

Example 3.6.  Show that the function defined by

                      

is not differentiable at the point    even though the Cauchy-Riemann equations (3-16) are satisfied at the point  .  

Solution.  We must use limits to calculate the partial derivatives at  .  

                    ,

                    ,
                    
                    ,   

                    .

Thus, we can see that   

                    ,     and     .  

Hence the Cauchy-Riemann equations (3-16) hold at the point  .  

        We now use Equation (3-1),   ,   from Section 3.1,

and show that    is not differentiable at the point  .   We do this by choosing two paths

that go through the origin and compute the limit of the difference quotient along each path.

First, let approach    along the -axis, given by the parametric equations  ,  then  

                      

Second, let approach    along the line  ,  given by the parametric equations  ,  then  

                      

        The limits along the two paths are different, so there is no possible value for the right side of Equation (3-1).  

Therefore,     is not differentiable at the point   .

Explore Solution 3.6.

        Example 3.6 reiterates that the mere satisfaction of the Cauchy-Riemann equations is not sufficient to guarantee the differentiability

of a function.  The following theorem, however, gives conditions that guarantee the differentiability of    at  ,  so that which we

can use Equation (3-14) or (3-15) to compute  .  They are referred to as the Cauchy-Riemann conditions for differentiability.

Theorem 3.4 (Cauchy-Riemann conditions for differentiability).  Assume that     

is a continuous function that is defined in some neighborhood of the point  .  If all the partial derivatives  

  are continuous at the point    and if the Cauchy-Riemann equations  

(3-14)                  and    

hold at   ,   then    is differentiable at  ,  and the derivative     

can be computed with either formula (3-14) or (3-15),  i. e.  

(3-14)              ,     

                        or  

(3-15)              .  

Proof.

Example 3.7.  At the beginning of this section (Equation (3-13)) we defined the function  

                    .   

Show that this function is differentiable for all , and find its derivative.

Solution.  We compute      and   ,   so the

Cauchy-Riemann Equations (3-16), are satisfied.  Moreover, the partial derivatives

  are continuous everywhere.   

By Theorem 3.4,      is differentiable everywhere, and, from Equation (3-14),  

                      

Alternatively, from Equation (3-15),

                    

This result isn't surprising because   ,    

and so the function    is really our old friend   .  

Explore Solution 3.7.

Extra Example 1.  Given  .   

Show that this function is differentiable for all , and find its derivative.

Solution.  We compute the partial derivatives and get  

                    ,     and   

                    ,  

so that the Cauchy-Riemann Equations (3-16), are satisfied.  Moreover, the partial derivatives

  are continuous everywhere.   

By Theorem 3.4,   ,   

is differentiable everywhere, and, from Equation (3-14),  

                      

Alternatively, from Equation (3-15),

                    

This result isn't surprising because  ,    

and so the function    is really our old friend   .  

Explore Extra Solution 1.

Example 3.8.  Show that the function     

is differentiable for all    and find its derivative.

Solution.  We first observe that   

                         and     .   

Then compute the partial derivatives and get  

                    ,     and
                    
                    .  

Moreover, the partial derivatives    are continuous everywhere.   

By Theorem 3.4,   ,  is differentiable everywhere.  

Therefore, using Equation (3-14) and (3-15), we have    

                    ,     and  

                    .

Aside.  Can you guess the "complex" form of    ?

Explore Solution 3.8.

Extra Example 2.  Show that the function  

is differentiable for all    and find its derivative.

Solution.  We first observe that   

                         and     .   

Then compute the partial derivatives and get  

                    ,     and

                    .  

Moreover, the partial derivatives    are continuous everywhere.   

By Theorem 3.4,   ,  is differentiable everywhere.  

Therefore, using Equation (3-14) and (3-15), we have    

                         
                    
                    and  

                    

Aside.  Can you guess the "complex" form of    ?

Explore Extra Solution 2.

    The Cauchy-Riemann conditions are particularly useful in determining the set of points for which a function f is differentiable.

Example 3.9.  Show that the function    

is differentiable at points that lie on the -axis, and at points that lie on the -axis, but    is nowhere analytic.

Solution.  Recall Definition 3.1 (from Section 3.1):  when we say a function is analytic at a point we mean that the function

is differentiable not only at , but also at every point in some  neighborhood of .  With this in mind, we proceed

to determine where the Cauchy-Riemann equations (3-16) are satisfied.  We write  

                         and     ,  

and compute the partial derivatives:

                    ,    ,     and  

                    ,    .  

Here    are continuous, and  

                      

holds for all points in the complex plane.   

But      if and only if   ,   which is equivalent to  

                    .  

Hence, the Cauchy-Riemann equations hold only at the points where   .  

According to Theorem 3.4,      is differentiable only when   ,  

which occurs only at points that lie on the coordinate axes.  Furthermore, for any point on the coordinate axes,

there contains an -neighborhood about it, in which there exist points where    is not differentiable.

Applying Definition 3.1 (from Section 3.1) , we see that the function      

is not analytic on either of the coordinate axes.

Therefore,      is nowhere analytic.

Explore Solution 3.9.

The Special Polar Limits.

    When polar coordinates are used to locate points in the plane, we use Expression (2-2) for a complex function.

That is, the Cartesian coordinate form

                    ,  

can be rewritten in the polar coordinate form

                      

where    are real functions of the real variables .  

        For the Polar coordinate form of a complex function

                    ,

it is important to determine how the function values change as we move along the radial grid line  

                       at the point ,  

and how the function values change as we move along the circular grid line  

                       at the point  .  

The polar form of the Cauchy-Riemann equations and a formula for finding    in terms of the partial derivatives of

,  are given in Theorem 3.5, which we ask you to prove in Exercise 10.  This theorem makes use

of the validity of the Cauchy-Riemann equations for the functions  ,  so the relation between them

and the functions  ,  namely,  

                    ,     and  

                    ,

is important.

Theorem 3.5  (Polar Form of the Cauchy-Riemann equations).  Let    

be a continuous function that is defined in some neighborhood of the point  .   If all the partial derivatives  

  are continuous at the point ,  

and if the polar form of the Cauchy-Riemann equations,

(3-22)                 and     ,    

hold, then     is differentiable at  ,  and we can compute the derivative    by using either  

(3-23)            ,     or    

(3-24)            .  

Proof.

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  .  

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  .   

Explore Revisited Solution 3.4.

Example 3.10.  Show that, if    is is the principal square root function given by  

                    ,     

where the domain is restricted to be   ,   then the derivative is given by  

                    ,    

for every point in the domain   .  

Solution. We write  

                    ,     

                    and   
            
                    .  
                    
Thus,

                    ,     

                    and   

                    .  

Moreover, the partial derivatives      are continuous in the domain   

    (note the strict inequality in  ).   

By Theorem 3.5,   ,   is differentiable in the domain   

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

                      

And an alternative calculation is  

                      

Note that      is discontinuous on the negative real axis and is undefined at the origin.  

Using the terminology of Section 2.4, the negative real axis is a branch cut, and the origin is a branch point for this function.  

Explore Solution 3.10.

    Two important consequences of the Cauchy-Riemann equations close this section.

Theorem 3.6.  Let      be an analytic function on the domain  .  

Suppose for all      that   ,   where    is a constant.   Then      is constant on  .  

Proof.

Theorem 3.7.   Let      be an analytic function on the domain  .   

If      for all   ,   then      is constant on  .  

Proof.

Optional Material

Theorem  (Complex form of the Cauchy-Riemann Equations).  Suppose the formula for    involves  .  

We can view    as a function of    and write:  

            .  

The complex form of the Cauchy-Riemann equations is  .

Proof.

Revisited Example 3.7.  Given   .   

Show that this function is differentiable for all , and find its derivative.

Solution.  Recall the identities    and    that were used in Section 2.1.

They can be substituted in   ,   and the result is

                       

When we view    as a function of the two variables ,  we see that

                    .  

Therefore, the complex form of the Cauchy-Riemann equations hold for all and      is analytic for all  .

Indeed,      is the revealed formula of    alone, and we are permitted to use the rules for differentiation in Section 3.1, and we find that

                    .

Explore Revisited Solution 3.7.

Revisited Example 3.9.  Given     

is differentiable at points that lie on the    axes but    is nowhere analytic.  

Solution.  Recall the identities    and    that were used in Section 2.1.

They can be substituted in   ,   and the result is

                       

When we view    as a function of the two variables  ,  we see that

                    .

Therefore, the complex form of the Cauchy-Riemann equations do not hold

and      is not analytic.

To determine where    has a derivative we must solve the equation  .

First expand the quantity    as follows.

                    

Hence, the equivalent equation we need to solve is   .  

      So we find that the complex form of the Cauchy-Riemann equations hold only when  ,  

and according to Theorem 3.4,    is differentiable only at points that lie on the coordinate axes.  

But this means that    is nowhere analytic because any -neighborhood about a point on either axis

contains points that are not on those axes.

Therefore      is only differentiable at points on the and axes.

Explore Revisited Solution 3.9.

Exercises for Section 3.2.  The Cauchy-Riemann Equations  

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

(c) 2012 John H. Mathews, Russell W. Howell