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 1  COMPLEX NUMBERS

Needs["Graphics`Arrow`"]; 
Needs["Graphics`Colors`"]; 
a = b = c = e = P = r = t = θ = v = V = w = x = y = z = Z = 0; 
Clear[a, b, bdd, c, disk, dot, e, eqn, ext, int, k, list, P, r, r1, 
        r2, set, solset, t, θ, θ1, θ2, v, V, values, w, 
        x, x0, x1, x2, x3, xx, y, y0, y1, y2, y3, yy, z, z1, z2, z3, z4, 
        Z];

Section 1.1	The Origin of Complex Numbers

Section 1.2	The Algebra of Complex Numbers

Section 1.3	The Geometry of Complex Numbers

Section 1.4	The Geometry of Complex Numbers, Continued

Section 1.5	The Algebra of Complex Numbers, Revisited

Section 1.6  The Topology of Complex Numbers

In this section we investigate some basic ideas concerning sets of points in the plane.  The first concept is that of a curve.  Intuitively, we think of a curve as a piece of string placed on a flat surface in some type of meandering pattern.  More Formally, we define a curve to be the range of a continuous complex-valued function  z(t)  defined on the interval  [a,b].  That is, a curve  C  is the range of a function given by  z (t) = (x (t), y (t)) = 
          x (t) + i y (t)  for  a ≤ t ≤ b,  where both  x(t)  and  y(t)  are continuous real valued functions.  If both x(t) and y(t) are differentiable, we say that the curve is smooth.  A curve for which x(t) and  y(t) are differentiable except for a finite number of points is called piecewise smooth.

Definition  (Curve), Page 30.  We specify a curve C as

			C :  z (t) = 
        x (t) + i y (t)   for   a ≤ 
          t ≤ b,  

and say that z(t) is a parametrization for the curve C. Notice that with this parametrization, we are specifying a direction the curve C, and we say that C is a curve that goes from the initial point z (a) = (x (a), y (a)) = 
          x (a) + i y (a) to the terminal point z (b) = (x (b), y (b)) = 
          x (b) + i y (b). If we had another function whose range was the same set of points as z(t) but whose initial and final points were reversed, we would indicate the curve this function defines by -C.

Example 1.22, Page 40.  Find parametrizations for C and -C, where C is the straight line segment beginning at (Subscript[z, 0]) = (Subscript[x, 0]) + i (Subscript[y, 0]) and ending at (Subscript[z, 1]) = (Subscript[x, 1]) + i (Subscript[y, 1]).

Solution 1.22.

Enter  (Subscript[z, 0])  and  (Subscript[z, 1])  and construct the line segment.

Clear[t, x, y, z]; 
(Subscript[z, 0]) = (Subscript[x, 0]) + I (Subscript[y, 0]); 
(Subscript[z, 1]) = (Subscript[x, 1]) + I (Subscript[y, 1]); 
z[t_]  =  
        (Subscript[z, 0]) + ((Subscript[z, 1]) - (Subscript[z, 0])) t; 
\(\(Print[\*""\!\(z\_0\) = "", 
        z\_0];\)\ \)
\(\(Print[\*""\!\(z\_1\) = "", 
        z\_1];\)\ \)
Print[""]; 
\(\(Print[\*""z[t] = (Subscript[z, 0]) + ((Subscript[z, 1])-(Subscript[z, 0]))t""];\)\ \)
Print["z[t] = ", z[t]]; 
z[t_]  =  
        (Subscript[x, 0]) + ((Subscript[x, 1]) - (Subscript[x, 0])) 
            t + I ((Subscript[y, 0]) + ((Subscript[y, 1]) - (Subscript[y, 0])) 
                  t); 
Print[""]; 
Print["The equation of, C, a line segment"]; 
\(\(Print[\*""starting at (Subscript[z, 0]) and ending at (Subscript[z, 1]).""];\)\ \)
Print["z[t] = ", z[t]]; 
Print["       for  0 ≤ t ≤ 1"]; 
Print["z[0] = ", z[0]]; 
Print["z[1] = ", z[1]];

Clear[t, x, y, z]; 
(Subscript[z, 0]) = (Subscript[x, 0]) + I (Subscript[y, 0]); 
(Subscript[z, 1]) = (Subscript[x, 1]) + I (Subscript[y, 1]); 
z[t_]  =  
        (Subscript[z, 1]) + ((Subscript[z, 0]) - (Subscript[z, 1])) t; 
\(\(Print[\*""\!\(z\_0\) = "", 
        z\_0];\)\ \)
\(\(Print[\*""\!\(z\_1\) = "", 
        z\_1];\)\ \)
Print[""]; 
\(\(Print[\*""z[t] = \!\(z\_1\)+(\!\(z\_0\)-\!\(z\_1\))t""];\)\ \)
Print["z[t] = ", z[t]]; 
z[t_]  =  
        (Subscript[x, 1]) + ((Subscript[x, 0]) - (Subscript[x, 1])) 
            t + I ((Subscript[y, 1]) + ((Subscript[y, 0]) - (Subscript[y, 1])) 
                  t); 
Print[""]; 
Print["The equation of, -C, a line segment"]; 
\(\(Print[\*""starting at (Subscript[z, 1]) and ending at (Subscript[z, 0]).""];\)\ \)
Print["z[t] = ", z[t]]; 
Print["       for  0 ≤ t ≤ 1"]; 
Print["z[0] = ", z[0]]; 
Print["z[1] = ", z[1]];

Example for,  Page 40.  Find the equation of the line segment with the initial point  (Subscript[z, 0]) = -3 + 2  i  and the terminal point  (Subscript[z, 1]) =  1  +  i.

Solution for Page 40.

Enter the initial point  (Subscript[z, 0]) = -3 + 2  i  and the terminal point  (Subscript[z, 1]) =  1  +  i  and construct the line segment.

Clear[t, x, y, z]; 
(Subscript[z, 0]) = -3 + 2  I; 
(Subscript[z, 1]) =  1  +  I; 
(Subscript[x, 0]) = Re[(Subscript[z, 0])];   (Subscript[y, 0]) = Im[(Subscript[z, 0])];   (Subscript[x, 1]) = Re[(Subscript[z, 1])];   (Subscript[y, 1]) = 
      Im[(Subscript[z, 1])]; 
z[t_]  =  
        (Subscript[x, 0]) + ((Subscript[x, 1]) - (Subscript[x, 0])) 
            t + I ((Subscript[y, 0]) + ((Subscript[y, 1]) - (Subscript[y, 0])) 
                  t); 
dot0 = 
        Graphics[{PointSize[0.03], Green, 
            Point[{Re[(Subscript[z, 0])], Im[(Subscript[z, 0])]}]}]; 
dot1 = 
        Graphics[{PointSize[0.03], Red, 
            Point[{Re[(Subscript[z, 1])], Im[(Subscript[z, 1])]}]}]; 
\(\(gr = 
        ParametricPlot[{Re[z[t]], Im[z[t]]}, {t, 0, 
            1}, 
PlotRange -> {{\(-3\), 1}, {0, 
                2}}, 
Ticks -> {Range[\(-3\), 1, 1], 
              Range[0, 2, 1]}, 
AspectRatio -> 1\/2, 
          PlotStyle -> Magenta, 
          AxesLabel -> {"x", "y"}, 
DisplayFunction -> 
            Identity];\)\ \)
Show[gr, dot0, dot1, 
        DisplayFunction -> $DisplayFunction]; 
Print["Equation of a line segment:"]; 
Print["z[t] = ", 
        z[t], "  for  0 ≤ t ≤ 1"]; 
Print["Initial  point    z[0] = ",  z[0]]; 
Print["Terminal point    z[1] = ",  
        z[1]];

Example 1.23, Page 41.  Show that the circle C with center (Subscript[z, 0]) = (Subscript[x, 0]) + i (Subscript[y, 0]) and radius (Subscript[r, 0]) can be parameterized to form a simple closed curve.

Solution 1.23.

Enter  (Subscript[z, 0])  and  (Subscript[r, 0])  and construct the standard parameterization for the circle.
Use the complex exponential form z[t] = (Subscript[z, 0]) + (Subscript[r, 0]) (e^(i t)).

Clear[t, x, y, 
        z];  
(Subscript[z, 0]) = 2 + I;  
(Subscript[r, 0]) = 2;  
z[t_]  =  
        (Subscript[z, 0]) + (Subscript[r, 0]) (E^(I t));  
x[t_] = ComplexExpand[Re[z[t]]];  
y[t_] = ComplexExpand[Im[z[t]]];  
ParametricPlot[{x[t], y[t]}, 
{t, 0, 2  π}, 
        PlotRange -> {{0.0, 5.0}, {-1.5, 
              3.5}}, 
Ticks -> {Range[0.0, 5.0, 1.0], 
            Range[-1.0, 3.0, 
              1.0]}, 
AspectRatio -> 1, 
        Prolog -> {Thickness[0.01], Green}, 
        AxesLabel -> {"x", "y"}];  
Print["The equation of, C, a circle"];  
\(\(Print[\*""with center (Subscript[z, 0]) and radius (Subscript[r, 0]).""];\)\ \ \)
Print["z[t] = ", z[t]];  
\(\(Print["       for  0 ≤ t ≤ 2π"];\)\ \ \)
Print["x = ", x[t]];  
Print["y = ",  
        y[t], " for 0 ≤ t ≤ 2π"];

Example for Page 41.  The curve  x[t] = Sin[2  t] Cos[t],   y[t] = Sin[2  t] Sin[t]  for  0 ≤ t ≤ 2  π.  
Remark. The curve looks like a "four leafed rose".

Solution for Page 41.

Enter the curve and plot it.

Clear[t, x, y]; 
x[t_] = Sin[2  t] Cos[t]; 
y[t_] = Sin[2  t] Sin[t]; 
ParametricPlot[{x[t], y[t]}, 
{t, 0, 2  π}, 
        PlotRange -> {{-0.8, 0.8}, {-0.8, 
              0.8}}, 
Ticks -> {Range[-0.8, 0.8, 
              0.8], Range[-0.8, 0.8, 
              0.8]}, 
AspectRatio -> 1, 
        Prolog -> {Thickness[0.01], Green}, 
        AxesLabel -> {"x", "y"}]; 
Print["x = ", x[t]]; 
Print["y = ",  
        y[t], "  for  0 ≤ t ≤ 2π"];

Example 1.24,  Page 42.  The solution sets of the inequalities |z|<1, |z - i|<2, and |z + 1 + 2  i|<3 are neighborhoods of the points 0,i,and -1-2  i, with radius 1,2, and 3, respectively.  They can also be expressed as (Subscript[D, 1]) (0), (Subscript[D, 2]) (i), and (Subscript[D, 3]) (-1 - 2  i).

Definitions (Interior Point, Exterior Point, Boundary Point), Page 42.  The point  (Subscript[z, 0])  is said to be an interior point of the set  S  provided that there exists an ϵ-neighborhood of  (Subscript[z, 0])  that contains only points of  S;   (Subscript[z, 0])  is called an exterior point of  S  if there exists an ϵ-neighborhood of  (Subscript[z, 0])  that contains no points of  S.  If  (Subscript[z, 0])  is neither an interior point nor an exterior point of  S, then it is called a boundary point of  S  and has the property that each  ϵ-neighborhood  of  (Subscript[z, 0])  contains both points in  S  and points not in  S.

Example 1.25,  Page 43.  Let  S = {z : |z|<1}.  (a) Find the interior of S.  (b) Find the exterior of S.  (c) Find boundary of  S.

Solution 1.25 (a).  Find the interior of S.

Let  (Subscript[z, 0]) be a point of  S.  Then  |(Subscript[z, 0])|< 1  so that we can choose  ϵ = 1- | (Subscript[z, 0]) | >0.  If  z  lies in the disk  |z - (Subscript[z, 0])|< ϵ,  then  

\(\(\(|\)\(z\)\(|\)\) = \(\(|\)\(z\_0 + z - 
            z\_0\)\(|\)\(\(≤\)\(\ \)\(\(|\)\(z\_0\)\(|\)\(+\(\(|\)\(z - z\_0\)\(|\)\(\(<\)\(\ \)\(\(|\)\(z\_0\)\(|\)\(\ \)\(\(+\ ϵ\)\  < \ 1\)\)\)\)\)\)\)\)\).  

Hence the ϵ-neighborhood of  (Subscript[z, 0])  is contained in  S,  and  (Subscript[z, 0])  is an interior point of  S.  It follows that the interior of  S  is the open unit disk.

dot = 
        Graphics[{PointSize[0.03], 
            Point[{0.3, 0.7}]}]; 
r = 1 - (Sqrt[(0.3^2) + (0.7^2)]); 
disk  =  
        Graphics[{Yellow, 
            Disk[{0.3, 0.7}, r]}]; 
bdd  =  Graphics[{Circle[{0, 0}, 1]}]; 
\(\(Show[bdd, disk, dot, \ 
        PlotRange -> {{\(-1.5\), 1.5}, {\(-1.5\), 1.5}}, 
        Ticks -> {Range[\(-1.5\), 1.5, 0.5], 
            Range[\(-1.5\), 1.5, 
              0.5]}, 
\ \ \ \ \ \ \ \ \ \ AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}];\)\ \)
\(\(Print[\*""The ϵ-neighborhood of (Subscript[z, 0]) is contained in S.""];\)\ \)

int  =  
        Graphics[{Yellow, Disk[{0, 0}, 1]}]; 
\(\(Show[int, \ PlotRange -> {{\(-1.5\), 1.5}, {\(-1.5\), 1.5}}, 
        Ticks -> {Range[\(-1.5\), 1.5, 1.5], 
            Range[\(-1.5\), 1.5, 
              1.5]}, 
\ \ \ \ \ \ \ \ \ \ AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}];\)\ \)
Print["The interior of the set S is the unit disk {z: |z| < \
1}."];

Solution 1.25 (b).  Find the exterior of S.

If  |(Subscript[z, 0])|> 1 then a similar argument will produce an ϵ-neighborhood of  (Subscript[z, 0])  that lies entirely in the region  |z|> 1,  hence (Subscript[z, 0])  is an exterior point of  S.  It follows that the exterior of  S  is the open region  |z|> 1.

dot = 
        Graphics[{PointSize[0.03], 
            Point[{0.4, 1.1}]}]; 
r = (Sqrt[(0.4^2) + (1.1^2)]) - 1; 
disk  =  
        Graphics[{Cyan, Disk[{0.4, 1.1}, r]}]; 
bdd  =  Graphics[{Circle[{0, 0}, 1]}]; 
\(\(Show[bdd, disk, dot, \ 
        PlotRange -> {{\(-1.5\), 1.5}, {\(-1.5\), 1.5}}, 
        Ticks -> {Range[\(-1.5\), 1.5, 0.5], 
            Range[\(-1.5\), 1.5, 
              0.5]}, 
\ \ \ \ \ \ \ \ \ \ AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}];\)\ \)
\(\(Print[\*""The ϵ-neighborhood of (Subscript[z, 0]) is contained in the region |z| > 1.""];\)\ \)

\(
\(ext\  = \ 
        Graphics[{{Cyan, 
              Disk[{0, 0}, 3]}, 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {White, 
              Disk[{0, 0}, 1]}}];\)\ \)
\(\(Show[ext, \ 
        PlotRange -> {{\(-1.5\), 1.5}, {\(-1.5\), 
              1.5}}, 
\ \ \ \ \ \ \ \ \ \ AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}];\)\ \)
Print["The exterior of the set S is the open region {z: |z| > \
1}."];

Solution 1.25 (c).  Find the boundary of S.

If  (Subscript[z, 0]) = (e^(i (Subscript[θ, 0]))) is any point on the unit circle, then any ϵ-neighborhood of  (Subscript[z, 0])  will contain the point  \(\((1 - ϵ\/2)\) e\^\(i\ θ\_0\)\), which belongs to  S,  and  \(\((1 + ϵ\/2)\) e\^\(i\ θ\_0\)\)  which does not belong to  S.  It follows that the boundary of  S  is the unit circle  |z| =  1.

dot = 
        Graphics[{PointSize[0.03], 
            Point[{0.4, (Sqrt[1 - (0.4^2)])}]}]; 
r = 0.2; 
disk  =  
        Graphics[{YellowBrown, 
            Disk[{0.4, (Sqrt[1 - (0.4^2)])}, 
              r]}]; 
bdd  =  Graphics[{Circle[{0, 0}, 1]}]; 
\(\(Show[bdd, disk, dot, \ 
        PlotRange -> {{\(-1.5\), 1.5}, {\(-1.5\), 1.5}}, 
        Ticks -> {Range[\(-1.5\), 1.5, 0.5], 
            Range[\(-1.5\), 1.5, 
              0.5]}, 
\ \ \ \ \ \ \ \ \ \ AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}];\)\ \)
\(\(Print[\*""Every ϵ-neighborhood of (Subscript[z, 0]) contains points of S""];\)\ \)
Print["and points that do not belong to S."];

bdd  =  
        Graphics[{Brown, Thickness[0.01], Circle[{0, 0}, 1]}]; 
\(\(Show[bdd, \ PlotRange -> {{\(-1.5\), 1.5}, {\(-1.5\), 1.5}}, 
        Ticks -> {Range[\(-1.5\), 1.5, 0.5], 
            Range[\(-1.5\), 1.5, 
              0.5]}, 
\ \ \ \ \ \ \ \ \ \ AspectRatio -> 1, 
        Axes -> True, 
        AxesLabel -> {"x", "y"}];\)\ \)
Print["The boundary of the set S is the unit circle {z: |z| = \
1}."];

Theorem 1.6 (Jordan Curve Theorem), Page 46. The compliment of any simple closed curve  C  can be partitioned into two mutually exclusive domains  I  and  E  in such a way that  I  is bounded,  E  is unbounded, and  C  is the boundary for both  I  and  E.   In addition  I⋃E⋃C  is the entire complex plane.  

Definition (Interior, Exterior), Page 46.  The domain  I  is called the interior of  C,  and  the domain  E  is called the exterior of  C.

The Jordan curve theorem is a classic example of a result in mathematics that seems obvious but is very hard to demonstrate. Its proof is beyond the scope of this book. Jordan's original argument, in fact, was inadequate, and it was not until 1905 that a correct version was finally given by the American topologist Oswald Veblen. The difficulty lies in describing the interior and exterior of a simple closed curve analytically, and in showing that they are connected sets. For example, in which domain (interior or exterior) do the two points depicted in Figure 1.28 lie? If they are in the same domain, how, specifically, can they be connected with a curve? If you appreciated the subtleties involved in showing the right half plane is connected, you can begin to appreciate the obstacles that Veblen had to navigate.

Library Research Experience for Undergraduates

Project I. Write a paper on teaching complex analysis and discuss some of the new approaches.

1. Bannon, Thomas J., (1991), ''Fractals and Transformations,'' Math. Teach., V. 81, No. 3, pp. 178-185.

2. Barton, Ray, (1990), ''Chaos and Fractals,'' Math. Teach., V. 83, No. 7, pp. 524-529.

3. Boyd, James N., (1985), ''A Property of Inversion in Polar Coordinates, The Math. Teach., V. 78, No. 1, pp. 60-61.

4. Braden, Bart, (1991), ''The Vector Field Approach in Complex Analysis,'' Visualization in Teaching and Learning Math., Providence, R.I., Math. Assoc. of Amer., pp. 191-196.

5. Bruch, John C. and Roger C. Wood (1972), ''Teaching Complex Variables with an Interactive Computer System,'' IEEE Trans. on Ed., E-15, No. 1, pp. 73-80.

6. Cohen, Martin P., (1983), ''Inversion in a Circle: A Different Kind of Transformation,'' The Math. Teach., V. 86, No. 8, pp. 620-623.

7. Frantz, Marny and Sylvia Lazarnick, (1991), ''The Mandelbrot Set in the Classroom,'' The Math. Teach., V. 84, No. 3, pp. 173-177.

8. Goffinet, Daniel, (1991), ''Number Systems With a Complex Base: a Fractal Tool for Teaching Topology,'', Am. Math. M., V. 98, No. 3, pp. 249-255.

9. Harding, R. D., (1974), ''Computer Aided Teaching in Applied Math.,'' Int. J. of Math. Ed. in Sci. and Tech., V. 5, No. 4, pp. 447-455.

10. Kern, Jane F. and Cherry C. Mauk, (1990), ''Exploring Fractals--a Problem-solving Adventure Using Math. and Logo,'' Math. Teach., V. 83, No. 3, pp. 179-185.

11. Kim, David and Robert Travers, (1982), ''Those elusive imaginary zeros,'' The Math. Teach., V. 75, pp. 62-64.

12. Kimberling, Clark, (1987), ''Power of Complex Numbers,'' The Math. Teach., V. 80, No. 1, pp. 63-67.

13. Kleiner, Israel,(1988), ''Thinking the Unthinkable: The Story of Complex Numbers (with a Moral),'' The Math. Teach., V. 81, No. 7, p583-92.

14. Lambert, Gary E., (1979), ''A 'Complex' Proof for a Geometric Construction of a Regular Pentagon,'' The Math. Teach., V. 72, No. 1, pp 65-66.

15. Lounesto, Pertti and Risto Mikkola and Vesa Vierros, (1989), ''Geometric Algebra Software for Teaching Complex Numbers, Vectors and Spinors,'' J. Comp. in Math. and Sci. Teach., V. 9, No. 2, pp. 93-105.

16. Miller, Valerie A. and G. Scott Owen, (1991), ''Using Fractal Images in the Visualization of Iterative Techniques from Numerical Analysis, Visualization in Teaching and Learning Math., Providence, R.I., Math. Assoc. of Amer., pp. 197-206.

17. Musser, Gary L., (1978), ''Line Reflections in the Complex Plane-A Billiards Player's Delight,'' The Math. Teach., V. 71, No. 1, pp. 60-64.

18. Shilgalis, Thomas W., (1982), ''Geometric Transformations on a Microcomputer,'' The Math. Teach., V. 75, No. 1, pp. 16-19.

19. Travers, Robert and David Kim, (1982), ''Those Elusive Imaginary Zeros,'' The Math. Teach., V. 75, No. 1, pp. 62-64.

20. Wood, Roger C. and John C. Brunch (1972), ''Teaching Complex Variables with an Interactive Computer System,'' IEEE Trans. on Ed., V. E-15, No. 1, pp. 73-80.

Section 1.6  Exercises for The Topology of Complex Numbers.
See textbook page 47.

GO TO  THE  Table of Contents