COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpub.com, http://www.jbpub.com/


 CHAPTER 9 CONFORMAL MAPPING

Section 9.1 Basic Properties of Conformal Mappings

Let f be an analytic function in the domain D , and let z[0] be a point in D . If `f '(`*z[0]*`)` <> 0 , then we can express f in the form

 

`f(z)` = `f(`*z[0]*`)` + `f '(`*z[0]*`)(` z-z[0] `)` + eta `(`*z*`)(` z-z[0] `)` ,

 

where Limit(eta*`(`*z*`)`,z = z[0]) = 0 .

 

If z is near z[0] , then the transformation w = `f(z)` has the linear approximation

 

f(z) = A+B(z-z[0]) = B*z+A-z[0] ,

 

where A = `f(`*z[0]*`)` and B = `f '(`*z[0]*`)` .

 

Because Limit(eta*`(`*z*`)`,z = z[0]) = 0 for points near z[0] the transformation w = `f(z)` has an effect much like the linear mapping w = `S(z)` . The effect of the linear mapping S is a rotation of the plane through the angle alpha = Arg(`f '(`*z[0]*`)`) , followed by a magnification by the factor abs(`f '(`*z[0]*`)`) , followed by a rigid translation by the vector A-z[0] . Consequently, the mapping w = `S(z)` preserves the angles at the point z[0] . We now show that the mapping w = `f(z)` also preserves angles at z[0] .

 

 

Theorem 9.1 Let f be an analytic function in the domain D , and let z[0] be a point in D . If `f '(`*z[0]*`)` <> 0 , then f is conformal at z[0] .

 

Load Maple's "conformal mapping" procedure.
Make sure this is done only ONCE during a Maple session.

> with(plots):

 

Warning, the name changecoords has been redefined


Example 9.1, Page 357. Show that the mapping `w = f(z) = cos(z)` i s conformal at the points i , 1 , and pi+i .

Find the angle of rotation alpha = `arg(f '(z))` and the scale factor abs(`f '(z)`) at the points i , 1 , and pi+i .

> f:='f': z:='z': Z:='Z':
f := z -> cos(z):
df := diff(f(Z),Z):
Df := z -> subs(Z=z,df):
`f(z) ` = f(z);
`f '(z) ` = Df(z);` `;
Z0 := I:
` `; `At Z0 ` = Z0;
f1 := Df(Z0):
`f '(Z0) ` = f1;
`rotation ` = argument(f1),` scale ` = abs(f1);
Z0 := 1:
` `; `At Z0 ` = Z0;
f1 := Df(Z0):
`f '(Z0) ` = f1;
`rotation ` = argument(f1),` scale ` = abs(f1);
Z0 := Pi + I:
` `; `At Z0 ` = Z0;
f1 := Df(Z0):
`f '(Z0) ` = f1;
`rotation ` = argument(f1),` scale ` = abs(f1);

`f(z) ` = cos(z)

`f '(z) ` = -sin(z)

` `

` `

`At Z0 ` = I

`f '(Z0) ` = -I*sinh(1)

`rotation ` = -1/2*Pi, ` scale ` = sinh(1)

` `

`At Z0 ` = 1

`f '(Z0) ` = -sin(1)

`rotation ` = Pi, ` scale ` = sin(1)

` `

`At Z0 ` = Pi+I

`f '(Z0) ` = I*sinh(1)

`rotation ` = 1/2*Pi, ` scale ` = sinh(1)

 


 Example 9.2, Page 358. Graph the transformation w = z^2 .

> f:='f': z:='z':
f := z -> z^2:
`f(z) ` = f(z);

`f(z) ` = z^2

Find the image of the square 0 `` <= `` x `` <= `` 1, 0 `` <= `` y `` <= `` 1.

 

> f:='f': z:='z':
f := z -> z^2:
`f(z) ` = f(z);
conformal(f(z), z=0..1+I,
title=`w = z^2`,
labels=[`u`,`v `], tickmarks=[3,3],
grid=[9,9], numxy=[9,9],
scaling=constrained,
view=[-1..1,0..2]);

`f(z) ` = z^2

[Maple Plot]

 

End of Section 9.1.