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   .

Explore Solution 3.10.

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.

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  except at points that lie on the negative -axis and   .

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.

>

>

>

>

>

Both Mathematica and Maple have shown that if

,

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

,

where   .

We are really done.

Aside.  Figure E.3.10 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 (3-22) 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 E.3.10 a

,   and                                               ,   and

.                                                             .
Figure E.3.10 b

,   and                                             ,   and

.                                                          .

Remark. It is difficult to visualize    and    because these partial derivatives

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

Figure E.3.10 c

For the function      we see that

,     and

.

Figure E.3.10

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