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

In Section 2.3 we defined the principal square root function and investigated some of its properties.  We left some unanswered questions concerning the choices of square roots.  We now look into this problem because it is similar to situations involving other elementary functions.

	In our definition of a function in Section 2.1 we specified that each value of the independent variable in the domain is mapped onto one and only one value of the dependent variable.  As a result, one often talks about a single-valued function, which emphasizes the only one part of the definition and allows us to distinguish such functions from multiple-valued functions which we now introduce. 

	Let  w = f (z)  denote a function whose domain is the set  D  and whose range is the set  R.  If  w  is a value in the range, then there is an associated inverse function  z = g (w)  that assigns to each value  w  the value (or values)  of  z  in  D  for which the equation  f (z) = w  holds true.  But unless  f  takes on the value  w  at most once in  D, then the inverse function  g  is necessarily many valued, and se say that  g  is a Multivalued Function.  For example, the inverse of the function  w = f (z) = (z^2)  is the square root function  z = (f^(-1)) (w) = (w^(1/2)).  We see that for each value  z  other than  z=0,  the two points  z  and  -z  are mapped onto the same point    w = f (z);   hence  g  is in general a two-valued function.

The study of limits, continuity, and derivatives loses all meaning if an arbitrary or ambiguous assignment of function values is made. For this reason we did not allow multivalued functions to be considered when we defined these concepts. When working with inverse functions, it is necessary to carefully specify one of the many possible inverse values when constructing an inverse function. The idea is the same as determining implicit functions in calculus. If the values of a function f are determined by an equation that they satisfy rather than by an explicit formula, then we say that the function is defined implicitly or that f is an implicit function. In the theory of complex variables we study a similar concept.

Let  w = f (z)  be a multiple-valued function. A branch of f is any single-valued function (Subscript[f, 0]) that is continuous in some domain (except, perhaps, on the boundary), and at each point z in the domain, assigns one of the values of f(z). Associated with the branch of a function is the Branch Cut.

	We now investigate the branches of the Square Root function.

Example 2.20,  Page 79.  Find the image of the plane cut along the negative x-axis under the mappings 

		w = (Subscript[f, 1]) (z) =   (z^(1/
                  2)) = \(\(r\^\(1/2\)\) \((\ 
          cos \((θ\/2)\) + i\ sin \((θ\/2)\)\ )\)\),  
	and
		w = (Subscript[f, 2]) (z) = - (z^(1/2))= \(\(r\^\(1/
              2\)\) \((cos \((\(θ + 2  π\)\/2)\) + i\ sin \((\(θ + 2  π\)\/2)\))\)\).

Solution 2.20.

Use Mathematica to graph the mapping  w = (Subscript[f, 1]) (z) = (z^(1/2))= \(\(r\^\(1/
              2\)\) \((cos \((θ\/2)\) + i\ sin \((θ\/2)\))\)\) where -π < θ < π.

Remove[z]; 
Clear[f, Iden]; 
Iden[z_]  =  z; 
f[z_]  =  (z^(1/2)); 
zplane  = 
        PolarMap[
          Iden, {0, 4, 0.5}, {-3.14,  3.14,  
            (3.14/12)}, 
PlotRange -> {{-4, 
                4.4}, {-4, 4.4}}, AspectRatio -> 1, 
          Ticks -> {Range[-4, 4, 2], Range[-4, 4, 2]}, 
          AxesLabel -> {"x", "y"}, 
          PlotStyle -> {DarkGreen, 
              Green}]; 
\(\(Print["The rays r > 0, θ = α, and circles r = c."];\)\ \)
wplane  =  
        PolarMap[f , {0, 4, 0.5}, {-3.14,  3.14,  (3.14/12)}, 
          PlotRange -> {{-0.2, 2.2}, {-2.2, 2.2}}, 
          AspectRatio -> 2, 
          Ticks -> {Range[0, 2, 1], Range[-2, 2, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {DarkGreen, 
              Green}]; 
\(\(Print[\*""The image rays r>0, θ=(α/2), and circles r=(Sqrt[c]),""];\)\ \)
\(\(Print[\*""under the mapping  w = \!\(f\_1\)(z) = "", 
        f[z], "."];\)\ \)
\(\(Print[\*""w = (Subscript[f, 1])(z) = (r^(1/2))(cos((θ/2))+I sin((θ/2))) where -π<θ<π""];\)\ \)

We can compare the above polar mapping with the cartesian coordinate mapping.

CartesianMap[
        Iden,  {-4, 4}, {-4, 
          4}, 
PlotRange -> {{-4, 4.4}, {-4, 
              4.4}}, AspectRatio -> 
          1, 
Ticks -> {Range[-4, 4, 2], 
            Range[-4, 4, 2]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["Vertical lines x=a, and horizontal lines y=b."]; 
CartesianMap[
        f,  {-4, 4}, {-4, 
          4}, 
PlotRange -> {{-0.2, 
              2.2}, {-2.2, 2.2}}, AspectRatio -> 2, 
        Ticks -> {Range[0, 2, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
\(\(Print[\*""The image hyperbolas (u^2)-(v^2) = a and 2uv = b,""];\)\ \)
\(\(Print[\*""under the mapping  w = \!\(f\_1\)(z) = "", 
        f[z], "."];\)\ \)
\(\(Print[\*""w = (Subscript[f, 1])(z) = (r^(1/2))(cos((θ/2))+I sin((θ/2))) where -π<θ<π""];\)\ \)

Use Mathematica to graph the mapping  w=(Subscript[f, 2]) (z)=- (z^(1/2))=(r^(1/
                        2))\((cos \((\(θ + 2  π\)\/2)\) + i\ sin \((\(θ + 2  π\)\/2)\))\)  where  -π<θ<π .

Remove[z]; 
Clear[f, Iden]; 
Iden[z_]  =  z; 
f[z_] = - (z^(1/2)); 
zplane  =  
        PolarMap[
          Iden, {0, 4, 0.5}, {-3.14,  3.14,  
            (3.14/12)}, 
PlotRange -> {{-4, 
                4.4}, {-4, 4.4}}, AspectRatio -> 1, 
          Ticks -> {Range[-4, 4, 2], Range[-4, 4, 2]}, 
          AxesLabel -> {"x", "y"}, 
          PlotStyle -> {Magenta, 
              Pink}];  
\(\(Print["The rays r>0, θ=α, and circles r=c."];\)\ \)
wplane  =  
        PolarMap[f , {0, 4, 0.5}, {-3.14,  3.14,  (3.14/12)}, 
          PlotRange -> {{-2.2, 0.2}, {-2.2, 2.2}}, 
          AspectRatio -> 2, 
          Ticks -> {Range[-2, 0, 1], Range[-2, 2, 1]}, 
          AxesLabel -> {"u", "v"}, 
          PlotStyle -> {Magenta, Pink}];  
\(\(Print[\*""The image rays r>0, θ=\!\(\"α+4π\"\/2\), and circles r=(Sqrt[c]),""];\)\ \)
\(\(Print[\*""under the mapping  w = \!\(f\_2\)(z) = "", 
        f[z], "."];\)\ \)
\(\(Print[\*""w = (Subscript[f, 2])(z) = (r^(1/2))(cos(((θ + 4  π)/2))+I sin(((θ + 4  π)/2))) where -π<θ<π""];\)\ \)

We can compare the above polar mapping with the cartesian coordinate mapping.

CartesianMap[
        Iden,  {-4, 4}, {-4, 
          4}, 
PlotRange -> {{-4, 4.4}, {-4, 
              4.4}}, AspectRatio -> 
          1, 
Ticks -> {Range[-4, 4, 2], 
            Range[-4, 4, 2]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Blue, Cyan}]; 
Print["Vertical lines x=a, and horizontal lines y=b."]; 
CartesianMap[
        f,  {-4, 4}, {-4, 
          4}, 
PlotRange -> {{-2.2, 
              0.2}, {-2.2, 2.2}}, AspectRatio -> 2, 
        Ticks -> {Range[0, 2, 1], Range[-2, 2, 1]}, 
        AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Blue, Cyan}]; 
\(\(Print[\*""The image hyperbolas (u^2)-(v^2) = a and 2uv = b,""];\)\ \)
\(\(Print[\*""under the mapping  w = \!\(f\_2\)(z) = "", 
        f[z], "."];\)\ \)
\(\(Print[\*""w = (Subscript[f, 2])(z) = (r^(1/2))(cos(((θ + 4  π)/2))+I sin(((θ + 4  π)/2))) where -π<θ<π""];\)\ \)

We have illustrated the two branches of the square root  w=(Subscript[f, 1]) (z)=  (z^(1/2))   and  w=(Subscript[f, 2]) (z) = - (z^(1/2)).

Example 2.21,  Page 80.  Show that the function  (Subscript[f, 1]) (z) is discontinuous along the negative real axis, where 

		w = (Subscript[f, 1]) (z) = (z^(1/2))= \(\(r\^\(1/
              2\)\) \((cos \((θ\/2)\) + i\ sin \((θ\/2)\))\)\).

Solution 2.21.

Enter the function  w = f[r, θ] = (z^(1/2))= \(\(r\^\(1/
              2\)\) \((cos \((θ\/2)\) + i\ sin \((θ\/2)\))\)\)  and find the limits as z approaches a point on the negative x-axis.

Clear[f, r, θ]; 
f[
          r_, θ_]  =  (r^(1/
                2)) (Cos[(θ/2)] + I Sin[(θ/2)]); 
lim1  = 
        PowerExpand[
          Limit[f[r, θ], θ -> π]]; 
lim2  = 
        PowerExpand[
          Limit[f[r, θ], θ -> -π]]; 
\(\(Print[\*""Let z approach the point z = r (E^I)π on negative x-axis.""];\)\ \)
\(\(Print[\*""  z = r (E^I)π""];\)\ \)
Print["f[z] = ", f[r, θ]]; 
Print[""]; 
Print["Approach from the upper half-plane"]; 
\(\(Print[\*""(lim_(θ -> π))   r (E^I)π  = "", 
        r\ E\^\(I\ π\)];\)\ \)
\(\(Print[\*""(lim_(θ -> π)) f[r (E^I)π] = "", 
        lim2];\)\ \)
Print[""]; 
Print["Approach from the lower half-plane"]; 
\(\(Print[\*""(lim_(θ -> -π))   r (E^I)π  = "", 
        r\ E\^\(\(-I\)\ π\)];\)\ \)
\(\(Print[\*""(lim_(θ -> -π)) f[r (E^I)π] = "", 
        lim2];\)\ \)

Since the two limits are different,  (Subscript[f, 1]) (z) = (z^(1/2))= \(\(r\^\(1/
              2\)\) \((cos \((θ\/2)\) + i\ sin \((θ\/2)\))\)\)  is discontinuous at all points  z = -r + 0  i,   r > 0  along the negative real axis.

Remark.  Likewise (Subscript[f, 1]) (z)is discontinuous along the negative real axis, where w = (Subscript[f, 2]) (z) = - (z^(1/2))= \(\(r\^\(1/
              2\)\) \((cos \((\(θ + 2  π\)\/2)\) + i\ sin \((\(θ + 2  π\)\/2)\))\)\).

Section 2.5  Exercises for  Branches of Functions
See textbook page 83.

Section 2.6	The Reciprocal Transformation w  = 1/z

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