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 9.4 Mapping Involving Trigonometric Functions

The trigonometric functions can be expressed with compositions that involve the exponential function followed by a bilinear function. We will be able to find images of certain regions by following the shapes of successive images in the composite mapping.

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 9.12, Page 380. The transformation is a one-to-one
conformal mapping of the vertical strip onto the disk .

> f:='f': z:='z':
f := z -> tan(z):
`w = f(z) ` = f(z);

Which is the same as the following function .

> g:='g': z:='z':
g := z -> I*(1-exp(I*2*z))/(1+exp(I*2*z)):
`g(z) ` = g(z);

The first function in this composition is . The image of the
vertical strip under the mapping is traced using a graph.

> f1:='f1': z:='z':
f1 := z -> exp(I*2*z):
`f1(z) ` = f1(z);
conformal(f1(z), z=-Pi/4-I*2..Pi/4+I*2,
title=`w = exp(i2z)`,
labels=[` u`,`v `], tickmarks=[3,5],
grid=[17,17], numxy=[17,17],
scaling=constrained,
view=[0..4,-4..4]);

The image of the vertical strip under the mapping is traced using a graph.

> f:='f': z:='z':
f := z -> tan(z):
`f(z) ` = f(z);
conformal(f(z), z=-Pi/4-I*2..Pi/4+I*2,
title=`w = tan(z)`,
labels=[` u`,`v `], tickmarks=[3,3],
grid=[17,17], numxy=[17,17],
scaling=constrained,
view=[-1..1,-1..1]);

Thus, the image of the vertical strip under is the disk .

Example 9.13, Page 381. The transformation is a one-to-one
conformal mapping of the vertical strip onto the w-plane slit along the
rays , and , . The image is traced using a graph.

> f:='f': z:='z':
f := z -> sin(z):
`f(z) ` = f(z);
conformal(f(z), z=-Pi/2-I*2.25..Pi/2+I*2.25,
title=`w = sin(z)`,
labels=[`u `,`v `], tickmarks=[7,7],
grid=[13,13], numxy=[13,13],
scaling=constrained,
view=[-3.4..3.4,-3.4..3.4]);

Thus, the image of the vertical strip is the -plane slit along the rays , and , .

Figure 9.18, Page 384. The transformation is a one-to-one conformal mapping of the
first quadrant , onto the semi-infinite strip , . The image is traced using a graph.

> f:='f': z:='z':
f := z -> arcsin(z):
`f(z) ` = f(z);
conformal(f(z), z=0.0001+I*0.0001..4+I*4,
title=`w = arcsin(z)`,
labels=[`u `,`v `], tickmarks=[4,6],
grid=[17,17], numxy=[17,17],
scaling=constrained,
view=[0..1.6,0..2.6]);

Thus, the image of the first quadrant , under is the semi-infinite strip , .

Example 9.14, Page 384. Find the principal value of .

> u:='u': v:='v': x:='x': y:='y':
u := proc(x,y)
arcsin((sqrt((x+1)^2+y^2) -
sqrt((x-1)^2+y^2))/2)
end:
v := proc(x,y)
sign(y)*arccosh((sqrt((x+1)^2+y^2) +
sqrt((x-1)^2+y^2))/2)
end:
`u(x,y) ` = u(x,y);
`v(x,y) ` = v(x,y); ` `;
`u(1,1) + I v(1,1) ` = u(1,1) + I*v(1,1);
`u(1,1) + I v(1,1) ` = evalc(u(1,1) + I*v(1,1));
`u(1,1) + I v(1,1) ` = evalf(u(1,1) + I*v(1,1)); ` `;
`arcsin(1+I) ` = arcsin(1+I);
`arcsin(1+I) ` = evalc(arcsin(1+I));
`arcsin(1+I) ` = evalf(arcsin(1+I));

End of Section 9.4.