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 2   COMPLEX FUNCTIONS

\(
\(Needs["Graphics`ComplexMap`"];\)\ \)
Needs["Graphics`ImplicitPlot`"]; 
Needs["Graphics`Colors`"]; 
Needs["Algebra`InequalitySolve`"]; 
a = u = v = w = x = y = z = Z = 
                      0; 
Clear[a, ac, al, b, bc, bhc, bl, bverti, circlea, circleb, eq1, eq2, 
        eq3, eq4, eq5, eqn, eqns, f,  F, hc, hl, horiz, Iden, ineq, ineq1, 
        ineq2, ineq3, ineq4, ineq5, ineq6, ineq7, ineq8, ineq9, θ, r, 
        u, u0, u1, u2,  U, U0, v, vc, vl, verti, V, w, wplane, W, x,  X, y, 
        Y, z, Z, zplane];

Section 2.1	Functions of a Complex Variable

Section 2.2	Transformations and Linear Mappings

Section 2.3	The Mappings w = z^n and w = z^1/n

Section 2.4	Limits and Continuity

Section 2.5	Branches of Functions

Section 2.6 The Reciprocal Transformation  w = 1/z

The mapping  w = (1/z)  is called the reciprocal transformation and maps the z-plane one-to-one and onto the w-plane except for the point  z=0,  which has no image, and the point  w=0,  which has no preimage or inverse image.  Since  \(z\ \(\(z\)\(\ \)\)\&_\  = \(\(|\)\(z\)\( | \^2\)\)\),  we can express the reciprocal transformation as a composition:  \(w\  = \ \(\(\(Z\)\(\ \)\)\&_\ \ \ and\ \ Z\  = \ 
          z\/\(\(|\)\(z\)\( | \^2\)\)\)\).  

	The mapping  \(Z\  = \ z\/\(\(|\)\(z\)\( | \^2\)\)\)  is called the Inversion Mapping with respect to the unit circle  |z| = 1.  It has the property that a nonzero point is mapped onto the point  Z  such that  |Z || z| = 1  and  arg Z  =  arg z.  Hence it maps points inside the circle  |z| = 1  onto points outside the circle  |Z| = 1, and conversely.  Any point of unit modulus is mapped onto itself.  The geometric description of the reciprocal transformation is now evident from this composition.  It is an inversion followed by a reflection through the x-axis.  If we use the polar coordinate form  w = ρ (e^i)ϕ = 
          (1/r) (e^(-iθ))  where  z = r(e^i)θ,  then we see that the ray  r > 0,  θ = α  is mapped one-to-one and onto the ray  ρ > 0,  ϕ = -α.  Also, points that lie inside the circle  |z| = 1  are mapped onto points that lie outside the circle  |w| = 1, and vice versa.

It is convenient to extend the system of complex numbers by joining to it an "ideal" point denoted by  ∞  and called the point at infinity.  This new set is called the Extended Complex Plane.  The point  ∞  has the property that  (lim_(n  ->  ∞)) (Subscript[z, n]) = ∞  if and only if  \(lim\+\(n\  -> \ ∞\)\ \(\(|\)\(z\_n\)\(|\)\) = ∞\).

An ϵ-neighborhood of the point at infinity is the set {z :  |z|>(1/ϵ)}. The usual way to visualize the point at infinity is by using what we call the stereographic projection and is attributed to Riemann. Let Ω be a sphere of diameter 1 that is centered at the point (0, 0, (1/2)) in three-dimensional space where coordinates are denoted by the triple of real numbers (x, y, ξ). Here the complex number z=x+iy will be associated with the point  (x, y, 0).

The point 𝒩 = (0, 0, 1) on Ω is called the north pole of Ω. Let z be a complex number, and consider the line segment L in three-dimensional space that joins z to the north pole 𝒩 = (0, 0, 1). Then L intersects Ω in exactly one point ℒ. The correspondence z⟷ℒ is called the stereographic projection of the complex z plane onto the Riemann Sphere Ω, and is shown below.

Let us reconsider the mapping  w = (1/z).  Let us assign the images  w = ∞  and  w = 0  to the points  z = 0  and  z = ∞, respectively.  The reciprocal transformation can now be written as
	
	w=f[z]=(1/z)   when  z≠0,∞
w=f[z]=0     when  z=∞
w=f[z]=∞     when  z=0
	
It is easy to see that the transformation  w = f[z] = (1/z) is a one-to-one mapping of the extended complex  z-plane onto the extended complex  w-plane.

Extra Example for Page 85.  Investigate the limits of  w = f[z] = (1/z) as  z -> 0,and z->∞.

Solution for Extra Example Page 85.

f[z_] = (1/z); 
lim0  =  Limit[f[z], z -> 0]; 
lim1  =  
        Limit[f[z], z -> ∞]; 
Print["    f[z] = ", f[z]]; 
\(\(Print[\*""(lim_(z -> ∞)) f[z] = (lim_(z -> ∞)) "", f[z], " = ", 
        lim1];\)\ \)
\(\(Print[\*""(lim_(z -> 0)) f[z] = (lim_(z -> 0)) "", f[z], " = ", lim0];\)\ \)

Example 2.22,  Page 86.  Show that the image of the right half plane  A = {z : Re[z] > (1/2)} under the mapping  w = (1/z)   is the disk  (Subscript[D, 1]) (1) | w - 1 | <1  in the w-plane.

Solution 2.22.

Enter the function f[z] and find its inverse.

Remove[w, x, y, z, 
        Z]; 
Clear[eq1, eq2, f, ineq, u, v, zz]; 
TagSet[x, Im[x], 0]; 
TagSet[y, Im[y], 0]; 
TagSet[x, Re[x], x]; 
TagSet[y, Re[y], y]; 
f[z_] := (1/z); 
Print["      f[z] = ", f[z]]; 
Z = x + I y; 
Print["f[", Z, "] = ", f[Z]]; 
Print["f[", Z, "] = ", 
        ComplexExpand[f[Z], 
          TargetFunctions -> {Im, Re}]];

Solve for the inverse function and the real and imaginary parts.

F[x_, y_] = 
        ComplexExpand[f[x + I y], 
          TargetFunctions -> {Im, Re}]; 
solset = Solve[w == (1/z), z]; 
Z = 
        (Subscript[solset, [[1, 1, 2]]]); 
g[w_] = Z; 
G[u_, v_] = 
        ComplexExpand[g[u + I v], 
          TargetFunctions -> {Im, Re}]; 
Print["      w = ", f[z]]; 
Print["      w = ", f[x + I y]]; 
Print["      w = ", F[x, y]]; 
Print["u + \[ImaginaryI] v = ", 
        F[x, y]]; 
Print[""]; 
Print["u = ", ComplexExpand[Re[F[x, y]]]]; 
Print["v = ", ComplexExpand[Im[F[x, y]]]]; 
Print[" "]; 
Print[solset]; 
Print[" "]; 
Print["      w = ", g[z]]; 
Print["      w = ", g[x + I y]]; 
Print["      w = ", G[x, y]]; 
Print["u + \[ImaginaryI] v = ", 
        G[x, y]]; 
Print[""]; 
Print["u = ", ComplexExpand[Re[G[x, y]]]]; 
Print["v = ", 
        ComplexExpand[Im[G[x, y]]]];

Now find the image of the right half plane  Re[z] > (1/2).

Clear[f, Iden, ineq1, ineq2, ineq3, ineq4, ineq5, 
        wplane, z, zplane]; 
Print[ineq1  =  x > (1/2)]; 
Print[""]; 
Print[
        ineq2  =  
          ReplaceAll[
            ineq1 , {x -> (u/((u^2)  +  (v^2))), 
              y -> ((-v)/((u^2)  +  (v^2)))}]]; 
Print[
        ineq3  =  
          Distribute[ 2 ((u^2)  +  (v^2)) ineq2,  
            Greater]]; 
Print[
        ineq4  =  
          Distribute[ 1  -  2  u  +  ineq3,  
            Greater]]; 
Print[""]; 
Print[
        ineq5  =  
          MapAll[FullSimplify, ineq4]];

This last inequality  (((u - 1))^2) + (v^2) < 1  is the same as  |w - 1|<1  and is the disk of radius 1 centered at  w = 1  in the w-plane.

Clear[f, Iden]; 
Iden[z_] = z; 
f[z_] = (1/z); 
CartesianMap[
        Iden,  {0.5, 10, 0.5}, {-5, 5, 
          0.25}, 
PlotRange -> {{-0.1, 
              10}, {-5, 5}}, AspectRatio -> 1, 
        Ticks -> {Range[0, 10, 2], Range[-5, 5, 1]}, 
        AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Blue, Cyan}]; 
\(\(Print[\*""The half-plane Re[z] > \!\(1\/2\).""];\)\ \)
CartesianMap[
        f,  {0.5, 10, 0.5}, {-5, 5, 
          0.25}, 
PlotRange -> {{-0.1, 
              2.1}, {-1.1, 1.1}}, 
AspectRatio -> 
          1, Ticks -> {Range[0, 2, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["The image is the disk |w - 1| < 1,"]; 
Print["under the mapping  w = ", 
        f[z]];

We see that the image of the right half plane  Re[z] > (1/2) under the reciprocal transformation  w = (1/z)   is the disk  |w - 1|<1  in the w-plane.

Example 2.23,  Page 87.  For the transformation  w = (1/z),  find the image of the portion of the right half plane  Re[z] > (1/2)  that lies inside the circle (Subscript[C, 1]) = {z : |z - (1/2)| = 1}.

Solution 2.23.

First, find the image of the right half plane  Re[z] > (1/2).

Clear[f, Iden, ineq1, ineq2, ineq3, ineq4, ineq5, 
        ineq6, ineq7, ineq8, ineq9, wplane, z, zplane]; 
Print[ineq1  =  x > (1/2)]; 
Print[
        ineq2  = 
          ReplaceAll[
            ineq1 , {x -> (u/((u^2)  +  (v^2))), 
              y -> ((-v)/((u^2)  +  (v^2)))}]]; 
Print[
        ineq3  =  
          Distribute[ 2 ((u^2)  +  (v^2)) ineq2,  
            Greater]]; 
Print[
        ineq4  =  
          Distribute[ 1  -  2  u  +  ineq3,  
            Greater]]; 
Print[
        ineq5  =  
          MapAll[FullSimplify, ineq4]];

This last inequality  (((u - 1))^2) + (v^2) < 1  is the same as  |w - 1|<1  and is the disk of radius 1 centered at  w = 1  in the w-plane.

Now find the image of the interior of the circle|z - (1/2)|<1.

Print[
        ineq1  =  
          (x^2)  -  x  +  (1/4)  +  (y^2)  <  1]; 
Print[
        ineq2  =  
          ReplaceAll[
            ineq1 , {x -> (u/((u^2)  +  (v^2))), 
              y -> ((-v)/((u^2)  +  (v^2)))}]]; 
Print[ineq3  =  MapAll[Together, ineq2]]; 
Print[
        ineq4  =  
          Distribute[ 4 ((u^2)  +  (v^2)) ineq3,  
            Less]]; 
Print[ineq5  =  MapAll[Expand, ineq4]]; 
Print[
        ineq6  =  
          Distribute[ 4  u  - (u^2)  -  (v^2)  +  ineq5,  
            Less]]; 
Print[ineq7  =  Distribute[ (1/3) ineq6,  Less]]; 
Print[ineq8  =  MapAll[Expand, ineq7]]; 
Print[
        ineq9  =  
          Distribute[ ((((2/3)))^2)  + ineq8,  
            Less]];

This last inequality  (16/9) < (4/9) + ((4 u)/3) + (u^2) + (v^2)  is the same as  |w + (2/3)|>(4/3)  and is the exterior of the circle with center  w = ((-2)/3)  and radius  (4/3)  in the w-plane.

((((4/3)))^2) < ((((2/3) + u))^2) + 
        (v^2)
MapAll[
      Expand, ((((4/3)))^2) < ((((2/3) + u))^2) + 
          (v^2)]

Use Mathematica to graph the mapping.

Clear[f, Iden, W, z, Z]; 
Iden[z_] = z; 
f[z_] = (1/z); 
Z[z_] = Iden[z + (1/2)]; 
W[z_] = f[z + (1/2)]; 
\(\(zplane\  = \ 
        PolarMap[
          Z\ , {0, 1, 
            0.1}, {\(-π\)\/2, π\/2, π\/12}, 
PlotRange -> {{\(-0.1\), 2.1}, {\(-1.1\), 1.1}}, 
          AspectRatio -> 1, 
          Ticks -> {Range[\(-6\), 4, 1], Range[\(-2\), 8, 1]}, 
          AxesLabel -> {"x", "y"}, 
          PlotStyle -> {Red, Magenta}];\)\ \)
\(\(Print[\*""The portion of the half-plane Re[z] > (1/2),""];\)\ \)
\(\(Print[\*""that lies inside the circle |z-(1/2)| < 1.""];\)\ \)
\(\(wplane\  = \ 
        PolarMap[
          W\ , {0, 1, 
            0.1}, {\(-π\)\/2, π\/2, π\/12}, 
PlotRange -> {{\(-0.1\), 2.1}, {\(-1.1\), 1.1}}, 
          AspectRatio -> 1, 
          Ticks -> {Range[0, 2, 1], Range[\(-1\), 1, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {Red, Magenta}];\)\ \)
Print["The image is the part of the disk |w-1|<1,"]; 
\(\(Print[\*""that lies outside the circle |w+(2/3)| = (4/3),""];\)\ \)
Print["under the mapping  w = ", 
        f[z]];

We see that the image of the portion of the right half plane  Re[z] > (1/2)  that lies inside the circle|z - (1/2)|<1  under the reciprocal transformation  w = (1/z)  is the crescent shaped region in the w-plane which is the portion of the disk  |w - 1|<1  that lies outside the circle  |w + (2/3)| = (4/3).

To study images of "generalized circles,'' let us consider the equation  

			A ((x^2) + (y^2)) +  B x +  C y  +  D  =  0, 

where A, B, C, and D are real numbers. This equation represents either a circle or a line, depending on whether A≠0 or A=0, respectively. Transforming the equation to polar coordinates gives  

			A (r^2) +  r (B cos θ  +  C sin θ)  +  
          D  =  0.  

	Using the polar coordinate form of the reciprocal transformation, we find that the image of the curve in the above equation can be expressed by  

			A  +  ρ (B cos ϕ  -  C sin ϕ)  +  
          D (ρ^2)  =  0,

which represents either a circle or a line, depending on whether D≠0 or D=0, respectively. Therefore, we have shown that the reciprocal transformation w = (1/z) carries the class of lines and circles onto itself.

Example 2.24,  Page 88.  Consider the mapping  w = f (z) = (1/z).   
(a)  Find the images of the vertical lines  x = a.   (b)  Find the images the horizontal lines  y = b.

Solution 2.24.

(a)  First, find the images of the vertical lines  x = a.

Clear[a, ac, al, b, bc, bl, circlea, circleb, hc, 
        hl, horiz, x, y, u, v, vc, verti, vl, wplane, zplane]; 
Print[eq1  =  x  ==  a]; 
Print[
        eq2  =  
          ReplaceAll[
            eq1 , {x -> (u/((u^2)  +  (v^2))), 
              y -> ((-v)/((u^2)  +  (v^2)))}]]; 
Print[
        eq3  =  
          Distribute[ ((((u^2)  +  (v^2)))/a)  eq2,  
            Equal]]; 
Print[
        eq4  =  
          Distribute[ ((- u)/a)  +  eq3,  Equal]]; 
Print[
        eq5  =  
          Distribute[ (1/(((2  a))^2))  +  eq4,  
            Equal]];

The image of the vertical lines  x = a  is the circle|w - (1/(2  a))| = (1/(2 | a | )),  with center  (Subscript[w, 0]) = (1/(2  a))  and radius  (1/(2 | a | )).  

(b)  Second, find the images the horizontal lines  y = b.

Print[eq1  =  y  ==  b]; 
Print[
        eq2  =  
          ReplaceAll[
            eq1 , {x -> (u/((u^2)  +  (v^2))), 
              y -> ((-v)/((u^2)  +  (v^2)))}]]; 
Print[
        eq3  =  
          Distribute[ ((((u^2)  +  (v^2)))/b)  eq2,  
            Equal]]; 
Print[eq4  =  Distribute[ (v/b)  +  eq3,  Equal]]; 
Print[
        eq5  =  
          Distribute[(( 1)/(((2  b))^2))  +  eq4,  
            Equal]];

The image is the circle|w + (i/(2  b))| = (1/(2 | b | )),  with center  (Subscript[w, 0]) = ((- i)/(2  b))  and radius  (1/(2 | b | )).  

Now prepare some horizontal and vertical lines in Mathematica for graphing.

hl[b_]  =  
        Line[{{-2, b}, {2, b}}]; 
horiz  =  {hl[-1], hl[((-1)/2)], hl[(1/2)], 
          hl[1]}; 
bl  =  Graphics[horiz]; 
vl[a_]  =  Line[{{a, -2}, {a, 2}}]; 
verti  =  {vl[-1], vl[((-1)/2)], vl[(1/2)], 
          vl[1]}; 
al  =  Graphics[verti];

Now prepare the image circles in Mathematica for graphing.

hc[b_]  =  
        Circle[{0, ((-1)/(2  b))}, (1/Abs)[2  b]]; 
circleb  =  {hc[-1], hc[((-1)/2)], hc[(1/2)], 
          hc[1]}; 
bc  =  Graphics[circleb]; 
vc[a_]  =  Circle[{(1/(2  a)), 0}, (1/Abs)[2  a]]; 
circlea  =  {vc[-1], vc[((-1)/2)], vc[(1/2)], 
          vc[1]}; 
ac  =  Graphics[circlea];

Graph the horizontal and vertical lines and their images.

Show[al, bl, 
        PlotRange -> {{-2.2, 2.2}, {-2.2, 
              2.2}}, 
AspectRatio -> 1, 
        Axes -> 
          True, 
Ticks -> {Range[-4, 4, 1], 
            Range[-4, 4, 
              1]}, 
AxesLabel -> {"x", "y"}, 
        DefaultColor -> Blue]; 
Print["The vertical  lines  x = a,"]; 
Print["and horizontal lines y = b."]; 
Show[ac, bc, 
        PlotRange -> {{-2.2, 2.2}, {-2.2, 
              2.2}}, 
AspectRatio -> 1, 
        Axes -> 
          True, 
Ticks -> {Range[-4, 4, 1], 
            Range[-4, 4, 
              1]}, 
AxesLabel -> {"u", "v"}, 
        DefaultColor -> Blue]; 
\(\(Print[\*""The image circles |w-(1/(2  a))| = (1/(2 | a | )) and |w+(I/(2  b))| = (1/(2 | b | )),""];\)\ \)
\(\(Print[\*""under the mapping  w = \!\(1\/z\).""];\)\ \)

We see that the images of the vertical lines  x = a  and  horizontal lines  y = b  under the mapping  w = (1/z) are circles.

Library Research Experience for Undergraduates

Project I. Write a report on Mobius transformations.

1. Avital, Shmuel and Shlomo Libeskind ''An Algebraic and Geometric Approach to Two Step Iteration of Bilinear Functions (in The Teaching of Math.),'' Am. Math. M., Vol. 91, No. 1. (Jan., 1984), pp. 53-56, Jstor.   

2. Beardon, Alan F., ''Curvature, Circles, and Conformal Maps (in Notes),'' Am. Math. M., Vol. 94, No. 1. (Jan., 1987), pp. 48-53, Jstor.   

3. Booker, T. Hoy, (1989), ''Bilinear Basics,'' Math. Mag., V. 62, No. 4, pp. 262-267.

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

5. Brickman, Louis, ''The Symmetry Principle for Mobius Transformations (in Notes),'' Am. Math. M., Vol. 100, No. 8. (Oct., 1993), pp. 781-782., Jstor.   

6. Budden, F. J. (1969), ''Transformation Geometry in the Plane by Complex Number Methods,'' The Math. Gazette, V. 53, No. 383, pp. 19-31.

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

8. Eljoseph, Nathan, ''On the Iteration of Linear Fractional Transformations,'' Am. Math. M., Vol. 75, No. 4. (Apr., 1968), pp. 362-366, Jstor.   

9. Amir-Moez, (1967), ''Conformal Linear Transformations,'' Math. Mag., Vol. 40, pp. 268-270.

Section 2.6 Exercises for The Reciprocal Transformation  w = (1/z)
See textbook page 89.

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