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  Mapping Involving Elementary Functions

In Section 5.1 we saw that the function  w = f (z) = (e^z)  is a one-to-one mapping of the fundamental period strip  -π < y ≤ π  in the z-plane onto the w-plane with the point  w = 0  deleted.  Since  f' (z) ≠ 0,  the mapping  w = (e^z)  is a conformal mapping at each point  z  in the complex plane.  The family of horizontal lines  y = c  and   - π < c ≤ π,  and segments  x = a  and   - π < y ≤ π  form an orthogonal grid in the fundamental period strip.  Their images under the mapping  w = (e^z)  are the rays  ρ  > 0  and  ϕ = c  and the circles  |w| = (e^a),  respectively.  These images form an orthogonal curvilinear grid in the w-plane.

Example 9.8,  Page 372.  The transformation  \(w = \(f \((z)\) = \(e\^z\  - \ i\)\/\(e\^z\  + \ i\)\)\)  is a one-to-one conformal mapping of the horizontal strip  0 < y < π  onto the disk  |w|< 1.  Furthermore, the x-axis is mapped onto the lower semicircle bounding the disk, and the line  y = π  is mapped onto the upper semicircle.

Solution 9.8.

Enter the function  \(f \((z)\) = \(e\^z\  - \ i\)\/\(e\^z\  + \ i\)\).

Remove[w, z]; 
Clear[f, g, solset]; 
f[
          z_]  =   (((E^z) - I)/((E^z) + I)); 
Print["f[z] = ", f[z]];

To show  \(f \((z)\) = \(e\^z\  - \ i\)\/\(e\^z\  + \ i\)\)  is one-to-one conformal we need to find the inverse function

solset  =  
        Solve[w  ==  f[z], z]; 
g[w_]  =  
        (Subscript[solset, [[1, 1, 2]]]); 
Print["f[z] = ", f[z]]; 
Print[solset]; 
Print["g[w] = ", g[w]];  
Print[" "]; 
Print["f[g[w]] = ", f[g[w]], " = ", 
        Simplify[f[g[w]]]]; 
Print[" "]; 
Print["g[f[z]] = ", g[f[z]]]; 
Print["g[f[z]] = ", 
        Simplify[g[f[z]]], " = ", 
PowerExpand[
          Simplify[g[f[z]]]]];

The image is traced using a graph.

Clear[f, z]; 
f[
          z_]  =   (((E^z) - I)/((E^z) + I)); 
\(\(CartesianMap[
        Identity, {\(-6\), 6, 
          0.5}, {0, π, π\/12}, 
PlotRange -> {{\(-6.4\), 6.4}, {0, 3.2}}, 
        AspectRatio -> 
          1\/2, 
Ticks -> {Range[\(-6\), 6, 2], 
            Range[0, 3, 1]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Magenta, Pink}];\)\ \ \)
Print["The horizontal strip 0 < y < π."]; 
CartesianMap[f, {-6, 6, 0.5}, {0, π, (π/12)}, 
        PlotRange -> {{-1.1, 1.1}, {-1.1, 
              1.1}}, 
AspectRatio -> 1, 
        Ticks -> {Range[-1, 1, 1], Range[-1, 1, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The image is the disk |w|< 1,"]; 
Print["under the mapping  w = ", 
        f[z]];

We see that the transformation  \(w = \(f \((z)\) = \(e\^z\  - \ i\)\/\(e\^z\  + \ i\)\)\)  maps the horizontal strip  0 < y < π  onto the disk  |w|< 1.

Example 9.9,  Page 373.  The transformation  w = f (z) = Log (((1  +  z)/(1  -  z)))  is a one-to-one conformal mapping of the unit disk  |z|< 1  onto the horizontal strip  |v|< (π/2).  Furthermore, the upper semicircle of the disk is mapped onto the line  v = (π/2)  and the lower semicircle onto  v = ((- π)/2).

Solution 9.9.

Enter the function  f (z) = Log (((1  +  z)/(1  -  z))).

Remove[w, z]; 
Clear[f, g, solset, w, z]; 
f[z_]  =   Log[((1 + z)/(1 - z))]; 
Print["f[z] = ", f[z]];

To show  f (z) = Log (((1  +  z)/(1  -  z)))  is one-to-one conformal we need to find the inverse function.

solset  =  
        Solve[w  ==  f[z], z]; 
g[w_]  =  
        (Subscript[solset, [[1, 1, 2]]]); 
Print["f[z] = ", f[z]]; 
Print[solset]; 
Print["g[w] = ", g[w]]; 
Print[" "]; 
Print["g[f[z]] = ", g[f[z]], " = ", 
        Simplify[g[f[z]]]]; 
Print[" "]; 
Print["f[g[w]] = ", f[g[w]]]; 
Print["f[g[w]] = ", 
        Simplify[f[g[w]]], " = ", 
PowerExpand[
          Simplify[f[g[w]]]]];

The image is traced using a graph.

Clear[f, z]; 
f[z_]  =    Log[((1 + z)/(1 - z))]; 
PolarMap[
        Identity, {0.000005, 0.999995, 0.11111}, 
{0, 
          2  π, (π/12)}, 
        PlotRange -> {{-1.1, 1.1}, {-1.1, 1.1}}, 
        PlotRegion -> {{0.15, 0.85}, {0.15, 0.85}}, 
        AspectRatio -> 1, 
        Ticks -> {Range[-1, 1, 1], Range[-1, 1, 1]}, 
        AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Green, 
            DarkGreen}];  
Print["The unit disk |z| < 1."]; 
PolarMap[
        f, {0.000005, 0.999995, 0.11111}, 
{0, 
          2  π, (π/12)}, 
        PlotRange -> {{-3, 3}, {-1.6, 1.6}}, 
        AspectRatio -> (1.6/3), 
        Ticks -> {Range[-3, 3, 1], 
            Range[((-π)/2), (π/2), (π/2)]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Green, 
            DarkGreen}];   
\(\(Print[\*""The image is the horizontal strip |v| < (π/2),""];\)\ \)
Print["under the mapping  w = ", 
        f[z]];

We see that the transformation  w = f (z) = Log (((1  +  z)/(1  -  z)))  maps the unit disk  |z|< 1  onto the horizontal strip  |v|< (π/2).

Example 9.10,  Page 373.  The transformation  w = f (z) = ((((1  +  z))^2)/(((1  -  z))^2))  is a one-to-one conformal mapping of the portion of the unit disk  |z|< 1  that lies in the upper half-plane  Im[z] > 0  onto the upper half-plane  Im[w] > 0.  Furthermore, the upper semicircular portion of the boundary is mapped onto the line negative u-axis, and the segment  -1 < x < 1, y = 0  is mapped onto the positive u-axis.

Solution 9.10.

Enter the function  f (z) = ((((1  +  z))^2)/(((1  -  z))^2)).

Remove[w, z]; 
Clear[f, g, solset]; 
f[z_]  =   ((((1 + z))^2)/(((1 - z))^2)); 
Print["f[z] = ", f[z]];

To show  f (z) = ((((1  +  z))^2)/(((1  -  z))^2))  is one-to-one conformal we need to find the inverse function.  Since there is two branches, one of them is appropriate for this problem.

solset  =  
        Solve[w  ==  f[z], z]; 
g[w_]  =  
        (Subscript[solset, [[1, 1, 2]]]); 
Print["f[z] = ", f[z]]; 
Print[solset]; 
Print["g[w] = ", g[w]]; 
Print[" "]; 
Print["f[g[w]] = ", f[g[w]], " = ", 
        Simplify[f[g[w]]]]; 
Print[" "]; 
Print["g[f[z]] = ", g[f[z]]]; 
Print["g[f[z]] = ", 
        PowerExpand[g[f[z]]], " = ", 
Simplify[
          PowerExpand[g[f[z]]]]];

The image is traced using a graph.

Clear[f, z]; 
f[z_]  =   ((((1 + z))^2)/(((1 - z))^2)); 
\(\(PolarMap[
        Identity, {0.000005, 0.999995, 0.11111}, 
{0, π, π\/12}, 
        AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Cyan, Blue}];\)\ \)
Print["The portion of the unit disk |z| < 1,"]; 
Print["that lies in the upper half-plane Im[z] > 0."]; 
PolarMap[f, {0.000005, 0.999995, 0.11111}, {0, π, (π/12)}, 
        PlotRange -> {{-4.2, 4.2}, {0, 4.5}}, 
        AspectRatio -> (4.5/8), 
        Ticks -> {Range[-4, 4, 2], Range[0, 4, 2]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Cyan, Blue}]; 
Print["The image is the upper half-plane Im[w] > 0,"]; 
Print["under the mapping w = ", 
        f[z]];

We see that the transformation  w = f (z) = ((((1  +  z))^2)/(((1  -  z))^2))  maps the portion of the unit disk  |z|< 1  that lies in the upper half-plane  Im[z] > 0  onto the upper half-plane  Im[w] > 0.

Example 9.11,  Page 374.  Consider the function  w = f (z) = (Sqrt[(z^2)  -  1]),  which is the composition of the functions  Z = (z^2) - 1  and  w = (Sqrt[Z])  where the branch of the square root is  \(\@Z = \(\(\@R\) \((cos \((θ\/2)\)\  + \ i\ sin \((θ\/2)\))\)\ \ \ where\ \ θ = 
          arg \((Z)\)\ \ and\ \ 0 ≤ θ < 2  π\)\).  Then the transformation  w = f (z)  maps the upper half-plane  Im[z] > 0  one-to-one and onto the upper-half plane  Im[w] > 0  slit along the segment  u = 0, 0 < v ≤ 1.

Solution 9.11.

Enter the function  f (z) = (Sqrt[(z^2)  -  1]).

Remove[w, z]; 
Clear[f, g, solset]; 
f[z_]  =  (Sqrt[(z^2)  -  1]); 
Print["f[z] = ", f[z]];

To show  w = f(z) = (Sqrt[(z^2) - 1])  is one-to-one conformal we need to find the inverse function.

f[
          z_]  =  (Sqrt[(z^2)  -  1]); 
solset  =  Solve[w  ==  f[z], z]; 
g[w_]  =  
        (Subscript[solset, [[2, 1, 2]]]); 
Print["f[z] = ", f[z]]; 
Print[solset]; 
Print["g[w] = ", g[w]]; 
Print[" "]; 
Print["f[g[w]] = ", f[g[w]], " = ", 
        PowerExpand[f[g[w]]]]; 
Print[" "]; 
Print["g[f[z]] = ", g[f[z]], " = ", 
        PowerExpand[g[f[z]]]];

However, in Mathematica  the branch cut for square root is along the negative x-axis.  For graphing purposes, we need the version of the function f(z) with a branch cut along the positive x-axis  f (z) = i (Sqrt[ 1 - (z^2)]).

f[
          z_]  =   I (Sqrt[1  -  (z^2)]); 
Print["f[z] = ", (Sqrt[(z^2)  -  1])]; 
Print["The branch we will use for graphing"]; 
Print["f[z] = ", f[z]];

The image is traced using a graph.

Clear[f, z]; 
f[z_]  =   I (Sqrt[1  -  (z^2)]); 
CartesianMap[
        Identity, {-3, 3, 0.5}, {0.00005, 1.35005, 
          0.09}, 
PlotRange -> {{-3, 3}, {0, 
              1.5}}, AspectRatio -> 
          (1.35/6), 
Ticks -> {Range[-3, 3, 1], 
            Range[0, 1, 0.5]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The upper half-plane Im[z] > 0."]; 
CartesianMap[f, {-3, 3, 0.5}, {-0.00005, 1.35005, 0.09}, 
        PlotRange -> {{-3, 3}, {0, 2}}, AspectRatio -> (1/3), 
        Ticks -> {Range[-3, 3, 1], Range[0, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The image is the upper-half plane Im[w] > 0,"]; 
\(\(Print["slit along the segment u = 0, 0 < v ≤ 1,"];\)\ \)
Print["under the mapping  w = ", 
        f[z]];

We see that the transformation  w = f (z) = i (Sqrt[1  -  (z^2)])  maps the upper half-plane  Im[z] > 0  one-to-one and onto the upper-half plane  Im[w] > 0  slit along the segment  u = 0, 0 < v ≤ 1.

The Riemann Surface for  w = ((((z^2)  -  1))^(1/2)).

Using the other branch of the square root, we find that w = (Subscript[f, 2]) (z) = -(Subscript[f, 1]) (z), is a one-to-one conformal mapping from the domain set (Subscript[D, 2]) consisting of the z-plane slit along the segment -1 ≤ x ≤ 1, y = 0, onto the range set (Subscript[H, 2]) consisting of the w-plane slit along the segment u = 0, -1 ≤ v ≤ 1. The sets (Subscript[D, 1]) and (Subscript[H, 1]) for (Subscript[f, 1]) (z) and (Subscript[D, 2]) and (Subscript[H, 2]) form the Riemann Surface for the mapping, as shown below.

 
			The Riemann surfaces for the mapping  w = f (z) = ((((z^2)  -  1))^(1/2)).

Section 9.3  Exercises for Mapping Involving Elementary Functions
See textbook page 379.

Section 9.4	Mappings by Trigonometric Functions

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