COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
Internet:  http://www.jbpub.com/  
This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 9  CONFORMAL MAPPING

S = T = w = W = x = y = z = Z = 
                      0; 
Clear[cos, f, F, g, graphs, Iden, S, set, sin, T, tan, w, W, w1, w2, 
        w3, wdot, wdots, wplane, z, z1, z2, z3, zdot, zdots, zplane, Z, 
        Zplane]; 
Needs["Graphics`ComplexMap`"]; 
Needs["Graphics`Colors`"];

Section 9.1	Basic Properties of Conformal Mappings

Section 9.2	Bilinear Transformations

Section 9.3	Mappings Involving Elementary Functions

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.

Example 9.12,  Page 380.  The transformation  w = tan(z)  is a one-to-one conformal mapping of the vertical strip  |x|< (π/4)  onto the disk  |w|< 1.

Solution 9.12.

Enter the function  f(z) = tan(z).

Remove[w, z]; 
Clear[f, f1, f2, g]; 
f[z_]  =   Tan[z]; 
Print["f[z] = ", f[z]];

Using the trigonometric identities

cos[
          z_] = (((E^iz) + (E^(-iz)))/2); 
sin[
          z_] = (((E^iz) - (E^(-iz)))/(2 I));

We obtain the following form for  f(z) = tan(z).

tan[z_] = 
        sin[z]/cos[z]; 
Print["tan[z] = ", tan[z]]; 
Print["tan[z] = ", 
        Together[tan[z]]];

Which is the same as the following function  \(g \((z)\) = \(i\ \((1\  - \ e\^\(i\ 2\ z\))\)\)\/\(1\  + \ e\^\(i\ 2\ z\)\)\).

g[
          z_]  =   ((I (1 - (E^(I 2 z))))/(1 + (E^(I 2 z)))); 
Print["tan[z] = g[z] = ", g[z]];

The first function in this composition is  \(Z = \(\(f\_1\) \((z)\) = e\^\(i\ 2\ z\)\)\),  and the second function is a bilinear transformation \(w = \(\(f\_2\) \((Z)\) = \(i\ \((1\  - \ Z)\)\)\/\(1\  + \ Z\)\)\),  and then  \(w = \(f \((z)\) = \(\(f\_2\) \((\(f\_1\) \((z)\))\) = \(i\ \((1\  - \ e\^\(2\ i\ z\))\)\)\/\(1\  + \ e\^\(2\ i\ z\)\)\)\)\).

f1[
          z_]  =   (E^(I 2 z)); 
f2[
          Z_]  =   ((I (1 - Z))/(1 + Z)); 
\(\(Print[\*""Z = \!\(f\_1\)[z] = "", 
        f1[z]];\)\ \)
\(\(Print[\*""w = \!\(f\_2\)[Z] = "", 
        f2[Z]];\)\ \)
Print[""]; 
Print["The mapping is: z -> Z -> w."]; 
Print[""]; 
\(\(Print[\*""w = \!\(f\_2\)[\!\(f\_1\)[z]] = "", 
        f2[f1[z]]];\)\ \)

The image is traced using a graph.

Clear[f, z]; 
f[z_]  =   Tan[z]; 
CartesianMap[
        Identity, {((-π)/4), (π/4), (π/24)}, {-2, 2, 0.2}, 
        AspectRatio -> 2, 
        Ticks -> {Range[((-π)/4), (π/4), (π/8)], 
            Range[-2, 2, 1]}, 
        PlotRange -> {{-1.05, 1.05}, {-2.1, 2.1}}, 
        AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Magenta, Pink}];  
\(\(Print[\*""The vertical strip ((-π)/4) < x < (π/4).""];\)\ \)
CartesianMap[
        f1, {((-π)/4), (π/4), (π/24)}, {-2, 2, 0.2}, 
        AspectRatio -> (6.1/3.1), 
        Ticks -> {Range[0, 3, 1], 
            Range[-3, 3, 
              1]}, 
PlotRange -> {{-0.05, 
              3.5}, {-3.5, 3.5}}, AxesLabel -> {"X", "Y"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The intermediate half-plane Im[Z] > 0,"]; 
\(\(Print[\*""under the mapping Z = \!\(f\_1\)[z] = "", 
        f1[z]];\)\ \)
CartesianMap[
        f, {((-π)/4), (π/4), (π/24)}, {-2, 2, 0.2}, 
        AspectRatio -> 1, 
        Ticks -> {Range[-1, 1, 1], Range[-1, 1, 1]}, 
        PlotRange -> {{-1.05, 1.05}, {-1.05, 1.05}}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The final image is the unit disk |w| < 1,"]; 
\(\(Print[\*""under the mapping w = \!\(f\_2\)[Z] = "", 
        f2[Z]];\)\ \)
\(\(Print[\*""w = \!\(f\_2\)[\!\(f\_1\)[z]] = "", 
        f2[f1[z]], " = ", f[z]];\)\ \)

We see that the transformation  w = tan(z)  maps the vertical strip  |x|< (π/4)  onto the disk  |w|< 1.

Example 9.13,  Page 381.  The transformation  w  =  f(z) = sin(z)  is a one-to-one conformal mapping of the vertical  strip  |x|< (π/2)  onto the w-plane slit along the rays  u ≤ -1, v = 0  and  1 ≤ u, v = 0.

Solution 9.13.

Enter the function  f(z) = sin(z).

Remove[z]; 
Clear[f]; 
f[z_]  =  Sin[z]; 
Print["f[z] = ", f[z]];

The image is traced using a graph.

Clear[f, z]; 
f[z_]  =  Sin[z]; 
CartesianMap[
        Identity, {((-π)/2), (π/2), (π/12)}, {-2.5, 2.5, 
          0.25}, 
AspectRatio -> 1.3, 
        Ticks -> {Range[((-π)/2), (π/2), (π/4)], 
            Range[-2, 2, 1]}, 
        PlotRange -> {{-2.0, 2.0}, {-2.6, 2.6}}, 
        AxesLabel -> {"x", "y"}, 
        PlotStyle -> {DarkGreen, 
            Green}];  
\(\(Print[\*""The vertical strip |x| < \!\(π\/2\).""];\)\ \)
CartesianMap[
        f, {((-π)/2), (π/2), (π/12)}, {-2.25, 2.25, 0.25}, 
        AspectRatio -> 1, 
        Ticks -> {Range[-3, 3, 1], Range[-3, 3, 1]}, 
        PlotRange -> {{-3.05, 3.05}, {-3.05, 3.05}}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, 
            Green}];  
Print["The image the w-plane slit along the"]; 
Print["rays u ≤ -1, v = 0 and 1 ≤ u, v = \
0,"]; 
Print["under the mapping  w = ", 
        f[z]];

We see that the transformation  w  =  f(z) = sin(z)  maps the vertical  strip  |x|< (π/2)  onto the w-plane slit along the rays  u ≤ -1, v = 0  and  1 ≤ u, v = 0.

Extra Example, Page 384.  The transformation  w = f(z) = arcsin(z)  is a one-to-one conformal mapping of the first quadrant  x > 0, y > 0  onto the semi-infinite strip  |u|<(π/2), v > 0.

Solution for Extra Example, Page 384.

Enter the function  f(z) = arcsin(z).

Remove[w, z]; 
Clear[f]; 
f[z_]  =  ArcSin[z]; 
Print["f[z] = ", f[z]];

The image is traced using a graph.

Remove[z]; 
Clear[f]; 
f[z_]  =  ArcSin[z]; 
CartesianMap[
        Identity, {0, 4, 0.2}, {0, 4, 
          0.2}, 
Ticks -> {Range[0, 4, 1], 
            Range[0, 4, 1]}, 
PlotRange -> {{-0.1, 
              4.1}, {-0.1, 4.1}}, 
        AspectRatio -> 
          1, 
AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["The first quadrant 0 < x, 0 < y."]; 
CartesianMap[
        f, {0, 4, 0.2}, {0, 4, 
          0.2}, 
Ticks -> {Range[
              0, (π/2), (π/4)], 
            Range[0, 2, 
              0.5]}, 
PlotRange -> {{0, (π/2)}, {0, 
              2.5}}, AspectRatio -> (5/π), 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
\(\(Print[\*""The semi-infinite strip |u| < (π/2), v > 0,""];\)\ \)
Print["under the mapping  w = ", 
        f[z]];

We see that the transformation  w = f(z) = arcsin(z)  maps the first quadrant  x > 0, y > 0  onto the semi-infinite strip  |u|<(π/2), v > 0.

Example 9.14,  Page 314.  Find the principal value of  ArcSin (1 + i).  Use the formula  
\(ArcSin[z] = 
        ArcSin[\(\@\(\((x + 1)\)\^2 + y\^2\) - \@\(\((x - 1)\)\^2 + y\^2\)\)\/2] + i\ Sign[
              y]\ ArcCosh[\(\@\(\((x + 1)\)\^2 + y\^2\) + \@\(\((x - 1)\)\^2 + y\^2\)\)\/2]\).

Solution 9.14.

Remove[x, y]; 
Clear[u, v]; 
u[x_, y_]  =  
        ArcSin[(((Sqrt[(((x + 1))^2) + (y^2)]) - (Sqrt[(((x - 1))^2) + (y^2)]))/2)]; 
v[x_, y_]  =  
        Sign[y] ArcCosh[(((Sqrt[(((x + 1))^2) + (y^2)]) + (Sqrt[(((x - 1))^2) + (y^2)]))/2)]; 
Print["u[x,y] = ", u[x, y]]; 
Print["v[x,y] = ", v[x, y]]; 
Print[" "]; 
Print["ArcSin[x + \[ImaginaryI] y] = ", 
        u[x, y] + I v[x, y]]; 
Print[" "]; 
Print["ArcSin[1+\[ImaginaryI]] = ", 
        u[1, 1] + I v[1, 1]]; 
Print["ArcSin[1+\[ImaginaryI]] = ", 
        N[u[1, 1] + I v[1, 1]]]; 
Print[""]; 
Print["Compare with Mathematica's function"]; 
Print["ArcSin[1+\[ImaginaryI]] = ", 
        ArcSin[1 + I]]; 
Print["ArcSin[1+\[ImaginaryI]] = ", 
        ComplexExpand[
          ArcSin[1 + I]]]; 
Print["ArcSin[1+\[ImaginaryI]] = ", 
        N[ArcSin[1 + I]]];

Therefore,  ArcSin[
          1 + i] = i ArcCosh[ (((Sqrt[5])  +  1)/2)] + 
          ArcSin[(((Sqrt[5])  -  1)/2)].

Is there any reason to assume that there exists a conformal mapping for some specified domain  D  onto another domain  G?  The theorem concerning the existence of conformal mappings is attributed to Riemann and can be found in the book by Lars V. Ahlfors Complex Analysis (New York: McGraw-Hill Book Co.) Chapter 6, 1966.

Theorem 9.4,  Page 385.  (Riemann Mapping Theorem)  If  D  is any simply connected domain in the plane (other than the entire plane itself), then there exists a one-to-one conformal mapping  w = f (z)  that maps  D  onto the unit disk  |w|<1.

Library Research Experience for Undergraduates

Project I. Write a report on Riemann mapping theorem.

1. Bak, Joseph, and Donald J. Newman, (1982), Complex Analysis, N.Y., Springer-Verlag.

2. Blackum, Harry A., (1987), ''Some Examples of Riemann's Surfaces,'' J. of Undergraduate Math., Vol. 19, No. 1, pp. 1-13.

3. Greenleaf, Frederick P., (1972), Introduction to Complex Variables, Philadelphia, Saunders.

4. Marsden, Jerrold E. and Michael J. Hoffman, (1987), Basic Complex Analysis, 2nd Ed., San Francisco, W. H. Freeman.

5. Walsh, J. L., ''History of the Riemann Mapping Theorem,'' Am. Math. M., Vol. 80, No. 3. (Mar., 1973), pp. 270-276., Jstor.

Section 9.4  Exercises for Mapping Involving Trigonometric Functions
See textbook page 308.

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