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.

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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.

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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.

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

Hopefully this text will promote their use and understanding.

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