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

We not take our first look at the geometric interpretation of a complex function.  If  D  is the domain of definition of the real-valued functions  u(x,y)  and  v(x,y),  then the system of equations   

			u = u (x, y)  and  v = v (x, y) 

describes a transformation or mapping from  D  in the xy-plane into the uv-plane.  Therefore, the function  

			w = f (z) = u (x, y) + i v (x, y)  

can be considered as a mapping or transformation from the set  D  in the z-plane onto the range  R  in the w-plane.  

	If  A  is a subset of the domain of definition  D, then the set  B = {w = f (z) : z ∈ A}  is called the image of the set  A, and  f  is said to map A onto B.  The image of a single point is a single point, and the image of the entire domain  D  is the range  R.  The mapping  w = f (z)  is said to be from A into S  if the image of  A  is contained in  S.  The inverse image of a point  w  is the set of all points  z  in  D  such that  w = f (z).  The inverse image of a point may be one points, several points, or none at all.  If the latter case occurs, then the point  w  is not in the range of  f.

	The function  f  is said to be one-to-one if it maps distinct points  (Subscript[z, 1]) ≠ (Subscript[z, 2])  onto distinct points  f ((Subscript[z, 1])) ≠ f ((Subscript[z, 2])).  If  w = f (z) maps the set  A  one-to-one and onto the set  B,  then for each  w  in  B  there exists exactly one point  z  in  A  such that  w = f (z). Then loosely speaking, we can solve the equation  w = f (z)  by solving for  z  as a function of  w.  That is, the inverse function  z = g (w) can be found, and the following equations hold: 
	
			g (f (z))=z   for all  z∈A
and
f (g (w))=w   for all  z∈B

Conversely, if w=f(z)  and z=g(w) are functions that map A into B and B into A, respectively, and the above equations hold, then f maps the set A one-to-one and onto the set B. The one-to-one property is easy to show, for if we have f ((Subscript[z, 1])) = f ((Subscript[z, 2])), then g (f ((Subscript[z, 1]))) = g (f ((Subscript[z, 2]))), and using the equation g(f(z))=z , we obtain (Subscript[z, 1]) = (Subscript[z, 2]). To show that f is onto, we must show that each point w in B is the image of some point in A. Let (Subscript[w, 0]) be any point in B. Then (Subscript[z, 0]) = g ((Subscript[w, 0])) lies in A and f ((Subscript[z, 0])) = (Subscript[w, 0]). In other words, (Subscript[z, 0]) is the point in A that has the desired property. Thus, we conclude that f is a one-to-one mapping from A onto B. Similar arguments show that g maps the set B one-to-one and onto the set A.
	
	Further, f is a one-to-one mapping from D onto T and if A is a subset of D, then f is a one-to-one mapping from A onto its image B. One can also show that if ζ = f (z) is a one-to-one mapping from A onto S, and w = g (ζ) is a one-to-one mapping from S onto B, then the composite mapping w = g (f (z)) is a one-to-one mapping from A onto B.

	We usually indicate the inverse of f by the symbol (f^(-1)). If the domains of f and (f^(-1)) are A and B respectively, 
	
			z = (f^(-1)) (f (z)) for all z  ∈  A,  
		and
			w = f ((f^(-1)) (w)) for all z  ∈  B.  
Also,
			w = f (z) iff (f^(-1)) (w) = z, and z = (f^(-1)) (w) iff f (z) = w.

Example 2.6,  Page 55.  If w = f (z) = i z for any complex number z, find (f^(-1)) (w).

Solution 2.6.

f[z_] = I z; 
solset = Solve[w == f[z], z]; 
g[w_] = 
        (Subscript[solset, [[1, 1, 2]]]); 
Print["w = f[z] = ", f[z] ]; 
Print[solset]; 
\(\(Print[\*""z = \!\(f\^\(-1\)\)[w] = "", 
        g[w]\ ];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(f\^\(-1\)\)[f[z]] = "", 
        g[f[z]]\ ];\)\ \)
Print[""]; 
\(\(Print[\*""f[\!\(f\^\(-1\)\)[w]] = "", 
        f[g[w]]\ ];\)\ \)

Remark 2.1. Once we have specified (f^(-1)) (w) = -i w for all complex numbers w, we see that there is nothing magical about using the symbol w. We could just as easily write (f^(-1)) (z) = -i z for all complex numbers z.

We now show how to find the image B of a specified set A under a given mapping u+iv=w=f(z) . The set A is usually described with an equation or inequality involving x and y. Using inverse functions we can construct a chain of equivalent statements leading to a description of the set B in terms of an equation or an inequality involving u and v.

Example 2.7,  Page 55.  Show that the function  f[z] = i z  maps the line  y = x + 1  onto the line  v = - u - 1.

Solution 2.7.

Enter the function  w = f[z] = i z.

Clear[x, y, z]; 
Clear[eq1, eq2, eq3, f, solset, u, v]; 
TagSet[x, Im[x], 0]; 
TagSet[y, Im[y], 0]; 
TagSet[x, Re[x], x]; 
TagSet[y, Re[y], y]; 
f[z_] := I z; 
Print["    f[", z, "] = ", 
        f[z]]; 
Print["f[", x  +  I y, "] = ", 
        f[x  +  I y]]; 
Print["f[", x  +  I y, "] = ", 
        Expand[f[x  +  I y]]];

Form the line  y = x + 1  and its image.

eq1 = 
        u == ComplexExpand[Re[f[x + I y]]]; 
eq2 = 
        v == ComplexExpand[Im[f[x + I y]]]; 
eq3 = y == x  +  1; 
sol1 = Solve[eq1, y]; 
sol2 = Solve[eq2, x]; 
Print["u + \[ImaginaryI] v = ", 
        Expand[f[x + I y]]]; 
Print[""]; 
Print["u = ", 
        ComplexExpand[
          Re[f[x + I y]]]]; 
Print["v = ", 
        ComplexExpand[
          Im[f[x + I y]]]]; 
Print[""]; 
Print[sol1]; 
Print[sol2]; 
Print[""]; 
eq4 = 
        Eliminate[{eq1, eq2, eq3}, {x, y}]; 
Print["In the z-plane."]; 
Print[eq3]; 
Print[""]; 
Print[
        ReplaceAll[
          eq3, {(Subscript[sol1, [[1, 1]]]), 
            (Subscript[sol2, [[1, 1]]])}]]; 
Print[""]; 
Print["In the w-plane."]; 
Print[eq4];

Plot the mapping  w = f[z] = i z.

ImplicitPlot[
        eq3, {x, -2, 2}, {y, -2, 
          2}, 
Ticks -> {Range[-1, 6, 1], 
            Range[-3, 4, 1]}, 
AspectRatio -> 1, 
        AxesLabel -> {"x", "y"}]; 
Print["The line in the z-plane,"]; 
Print[eq3]; 
ImplicitPlot[
        eq4, {u, -2, 2}, {v, -2, 
          2}, 
Ticks -> {Range[-1, 6, 1], 
            Range[-3, 4, 1]}, 
AspectRatio -> 1, 
        AxesLabel -> {"u", "v"}]; 
Print["The line in the w-plane,"]; 
Print[eq4]; 
Print["under the mapping w = f[z] = ", 
        f[z]];

We see that he transformation  w = f[z] = i z  maps the line  y = x + 1  onto the line  v = - u - 1.

We now turn our attention to the investigation of some elementary mappings.  Let  B = a + ib  denote a fixed complex number.  Then the transformation  

			w = T (z) = z + B = 
            x + a + i (y + b)  

is a one-to-one mapping of the z-plane onto the w-plane and is called a Translation.  This transformation can be visualized as a rigid translation whereby the point  z  is displaced through the vector  a + ib  to its new position  w = T (z).  The inverse mapping is given by  

			 z = (T^(-1)) (w) = w - B = 
            u - a + i (v - b)  

and shows that  T  is a one-to-one mapping from the z-plane onto the w-plane.

	Let  α  be a fixed real number.  Then the transformation  

			\(w = \(R \((z)\) = \(z\ e\^iα = \(r\ \(e\^iθ\) e\^iα = 
              r\ e\^\(i \((θ + α)\)\)\)\)\)\)  

is a one-to-one mapping of the z-plane onto the w-plane and is called a Rotation.  It can be visualized as a rigid rotation whereby the point  z  is rotated about the origin through an angle  α   to its new position  w = T (z).  If we use polar coordinates  w = ρ (e^i)ϕ  in the w-plane, then the inverse mapping is given by  

			\(z = \(\(R\^\(-1\)\) \((w)\) = \(w\ e\^\(-iα\) = \(ρ\ \(e\^iϕ\) e\^\(-iα\) = ρ\ e\^\(i \((ϕ - α)\)\)\)\)\)\).  

This shows that  R  is a one-to-one mapping from the z-plane onto the w-plane.

Example 2.8,  Page 57.  The ellipse centered at the origin with a horizontal major axis of 4 units and vertical minor axis of 2 units can be represented by the parametric equation  

		s[t] = 2  Cos[t] + i Sin[t], for 0 ≤ t ≤ 2  π.  

Suppose we wanted to rotate the ellipse by an angle of (π/6) radians as well as shift the center of the ellipse two units to the right and one unit up. Using complex arithmetic we can easily generate a parametric equation r[t] that does this:

			\(r[t] = 
        s[t]\ e\^\(i\ π/6\) + \((2 + i)\)\)  
			r[
          t] = (e^(((i π)/6))) (2 Cos[
                    t] + i Sin[
                    t]) + (2 + i)  
			r[
          t] = (Sqrt[3]) Cos[t]-(1/2) 
            Sin[t]+2  + i (Cos[t] + ((Sqrt[3])/2)  Sin[t] + 
            1)
			r[t] = ((Sqrt[3]) Cos[t] - (1/2) Sin[t] + 2),Cos[t] +((Sqrt[3])/2  Sin[t] + 1)

Solution 2.8.

Clear[r, s, t]; 
s[t_] = 
        2  Cos[t] + I Sin[t]; 
r[t_] = 
        s[t] (E^(I π/6)) + (2 + I); 
ParametricPlot[{Re[s[t]], Im[s[t]]}, {t, 0, 2  π}, 
        PlotStyle -> Magenta, 
        PlotRange -> {{-2.1, 2.1}, {-1.05, 1.05}}, 
        AspectRatio -> (1/2), 
        Ticks -> {Range[-2, 2, 1], 
            Range[-2, 2, 1]}]; 
Print["s[t] = ", 
        s[t], " for 0 ≤ t ≤ 2π."]; 
ParametricPlot[{Re[r[t]], Im[r[t]]}, {t, 0, 2  π}, 
        PlotStyle -> Magenta, 
        PlotRange -> {{0, 4}, {-0.5, 2.5}}, 
        AspectRatio -> (3/4), 
        Ticks -> {Range[0, 4, 1], 
            Range[0, 3, 1]}]; 
Print["r[t] = ", 
      r[t], " for 0 ≤ t ≤ 2π."]

We have one additional type of mapping to discuss in this section. Let  K > 0  be a fixed positive real number.  Then the transformation  

			w  =  S (z)  =  K z  =  
            K x + i K y  

is a one-to-one mapping of the z-plane onto the w-plane and is called a magnification.  If   K > 1,  it has the effect of stretching the distance between points by the factor  K.  If   K < 1,  then it reduces the distance between points by the factor  K.  The inverse mapping is given by  

			z  =  (S^(-1)) (w)  =  (1/K) 
              w  =  (1/K) u  +  i  (1/K) v  

and shows that  S  is a one-to-one mapping from the z-plane onto the w-plane.

	Let  A = K(e^i)α  and  B = a + ib, where  K > 0  is a fixed positive real number.  Then the transformation  

			w  =  W (z)  =  A z + B 

is a one-to-one mapping of the z-plane onto the w-plane and is called a linear transformation.  It can be considered as the composition of a rotation, a magnification, and a translation.  It has the effect of rotating the plane though an angle given by  α = Arg A,  followed by a magnification by the factor  K = |A|, followed by a translation by the vector  B = a + ib.  The inverse mapping is given by  z = (W^(-1)) (w) = (1/A) w - (B/A)  and shows that  W  is a one-to-one mapping from the z-plane onto the w-plane.

Example 2.9,  Page 58.  Show that the linear transformation  w = f[z] = i z + i  maps the right half plane  Re[z] > 1  onto the upper half plane  Im[w] > 2.

Solution 2.9.

Enter the function  w = f[z] = i z + i.

Clear[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_] := I z + I; 
Print["    f[", z, "] = ", 
        f[z]]; 
Print["f[", x  +  I y, "] = ", 
        f[x  +  I y]]; 
Print["f[", x  +  I y, "] = ", 
        Expand[f[x  +  I y]]];

Solve for the inverse function.

solset = 
        Solve[w == I z + I, z]; 
Z = 
        (Subscript[solset, [[1, 1, 2]]]); 
Print["      w = ", f[z]]; 
Print["      w = ", f[x + I y]]; 
Print["      w = ", 
        Expand[f[x + I y]]]; 
Print["u + \[ImaginaryI] v = ", 
        Expand[f[x + I y]]]; 
Print[""]; 
Print["u = ", 
        ComplexExpand[Re[f[x + I y]]]]; 
Print["v = ", 
        ComplexExpand[Im[f[x + I y]]]]; 
Print[" "]; 
Print[solset]; 
Print[" "]; 
Print["z = ", Z]; 
Print["z = ", ComplexExpand[Z]]; 
Z = 
        ReplaceAll[Z , 
          w -> u + I v]; 
Print[""]; 
Print["z = ", Z]; 
Print["z = ", Expand[Z]]; 
Print[""]; 
Print["x = ", ComplexExpand[Re[Z]]]; 
Print["y = ", 
        ComplexExpand[Im[Z]]];

To solve for  Re[z] > 1,   we use the following computations.

Print[Re[z] > 1]; 
Print[x > 1]; 
Print[
        ineq = ComplexExpand[Re[Z]] > 1]; 
solset = InequalitySolve[ineq, v]; 
Print[""]; 
Print[solset];

The solution is the upper half plane  Im[w] > 2.  
Now use Mathematica to graph the transformation.

\(
\(Iden[z_] = z;\)\ \)
f[z_] = I z + I; 
CartesianMap[
        Iden,  {1, 5, 0.5}, {-6, 4, 
          0.5}, 
PlotRange -> {{-1, 6}, {-3, 
              4}}, AspectRatio -> 
          1, 
Ticks -> {Range[-1, 6, 1], 
            Range[-3, 4, 1]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The half-plane  Re[z] > 1."]; 
CartesianMap[
        f,  {1, 5, 0.5}, {-6, 4, 
          0.5}, 
PlotRange -> {{-1, 6}, {-1, 
              6}}, AspectRatio -> 
          1, 
Ticks -> {Range[-1, 6, 1], 
            Range[-1, 6, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Magenta, Pink}];  
Print["The image half-plane Im[w] > 2,"]; 
Print["under the mapping  w = ", 
        f[z]];

We see that the linear transformation  w = f[z] = i z + i  maps the right half plane  Re[z] > 1  onto the upper half plane  Im[w] > 2.

It is easy to see that translations and rotations preserve angles.  Since magnifications rescale distance by a factor  K,  it follows that triangles are mapped onto similar triangles, and so angles are preserved.  Since a linear transformation can be considered as the composition of a rotation, a magnification, and a translation, it follows that linear transformations preserve angles.  Consequently, any geometric object is mapped onto an object that is similar to the original object: hence linear transformations can be called similarity mappings.

Note. The usage of the phrase "linear transformation" in a "complex analysis course" is different than that the usage in "linear algebra courses".

Example 2.10,  Page 60.  Show that the image of the open disk  (Subscript[D, 1])(-1-i)={z : |z + 
                1 + i|< 1}  under the transformation  w = (3 - 4  i) z  +  6 + 2  i  is the open disk  (Subscript[D, 5])(-1+3 i)={w : |w + 1 - 
                3 i|< 5}.

Solution 2.10.

Enter the function  w = f[z] = (3 - 4  i) z  +  6 + 
            2  i .

Clear[w, x, y, z, 
        Z]; 
Clear[f, ineq, solset, term]; 
f[z_] = (3 - 4  I) z + 6 + 
          2  I; 
Print["    f[", z, "] = ", 
        f[z]]; 
Print["f[", x  +  I y, "] = ", 
        f[x  +  I y]]; 
Print["f[", x  +  I y, "] = ", 
        ComplexExpand[
          f[x  +  I y]]];

Solve for the inverse function, i.e. solve for  z  in terms of  w.  Then find z + 1 + i.

solset  =  
        Solve[w  ==  f[z], z]; 
Z  =  
        Factor[(Subscript[solset, [[1, 1, 2]]])]; 
term  =  Together[Z + 1 + I]; 
Print[solset]; 
Print[""]; 
Print["        z = ", Z]; 
Print[""]; 
Print[z + 1 + I, " = ", 
        Z + 1 + I]; 
Print[z + 1 + I, " = ", 
        term];

To solve for |z + 1 + i|< 1,  we use the computations.

Print[
        Abs[z + 1 + I]  <  1]; 
Print[""]; 
Print[
        ineq  =  
          Abs[term[[1]]] Abs[term[[2]]]  <  
            1]; 
Print[""]; 
Print[Distribute[5 ineq,  Less]];

Which is the disk  (Subscript[D, 5])(-1+3 i)={w : |w + 1 - 
                3 i|< 5}  in the w-plane.  Use Mathematica to graph the transformation.

\(
\(Clear[Z, W];\)\ \)
Iden[z_]  =  z; 
f[z_]  =  (3  -  4  I) z  +  6  +  
          2  I; 
Z[z_]  =  Iden[z - 1 - I]; 
W[z_]  =  f[z - 1 - I]; 
\(\(PolarMap[
        Z\ , {0, 1, 0.2}, {0, 
          2  π, π\/12}, 
PlotRange -> {{\(-6\), 4}, {\(-2\), 8}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[\(-6\), 4, 1], 
            Range[\(-2\), 8, 1]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {Cyan, Blue}];\)\ \)
Print["The open disk |z+1+\[ImaginaryI]| < 1."]; 
\(\(PolarMap[
        W\ , {0, 1, 0.2}, {0, 
          2  π, π\/12}, 
PlotRange -> {{\(-6\), 4}, {\(-2\), 8}}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[\(-6\), 4, 1], 
            Range[\(-2\), 8, 1]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {Cyan, Blue}];\)\ \)
\(\(Print["The image is the open disk |w+1-3I| < 5,"];\)\ \)
Print["under the mapping w = ", 
        f[z]];

We see that the image of the open disk  (Subscript[D, 1])(-1-i)={z : |z + 
                1 + i|< 1}  under the transformation  w = (3 - 4  i) z  +  6 + 2  i  is the open disk  (Subscript[D, 5])(-1+3 i)={w : |w + 1 - 
                3 i|< 5}.

Example 2.11,  Page 60.  Show that the image of the right half plane  Re[z] > 1  under the linear transformation  w = (-1 + i) z  -  2 + 3  i  is the half plane  v > u + 7.

Solution 2.11.

Enter the function w = f[z].

Clear[w, x, y, z, 
        Z]; 
Clear[f, ineq1, ineq2, ineq3]; 
TagSet[x, Im[x], 0]; 
TagSet[y, Im[y], 0]; 
TagSet[x, Re[x], x]; 
TagSet[y, Re[y], y]; 
f[z_] := (-1 + I) z - 2 + 
          3  I; 
Print["    f[", z, "] = ", 
        f[z]]; 
Print["f[", x  +  I y, "] = ", 
        f[x  +  I y]]; 
Print["f[", x  +  I y, "] = ", 
        ComplexExpand[
          f[x  +  I y]]];

Solve for the inverse function, i.e. solve for  z  in terms of  w.

solset  =  
        Solve[w  ==  (-1  +  I) z  -  2  +  
              3  I,  z]; 
zz  =  
        (Subscript[solset, [[1, 1, 2]]]); 
Z  =  
        ReplaceAll[zz , w -> u  +  I v]; 
Print["w = ", f[z]]; 
Print["w = ", f[x  +  I y]]; 
Print["w = ", 
        Expand[f[x + I y]]]; 
Print[""]; 
Print["u = ", 
        ComplexExpand[Re[f[x + I y]]]]; 
Print["v = ", 
        ComplexExpand[Im[f[x + I y]]]]; 
Print[""]; 
Print[solset]; 
Print[""]; 
Print["z = ", zz]; 
Print["z = ", Z]; 
Print["z = ", Expand[Z]]; 
Print[""]; 
Print["x = ", ComplexExpand[Re[Z]]]; 
Print["y = ", 
        ComplexExpand[Im[Z]]];

To solve for Re[z] > 1 ,  we use the computations.

Print[
        Re[z]  >  1]; 
Print[""]; 
Print[x  >  1]; 
Print[ineq1  =  ComplexExpand[Re[Z]]  >  1]; 
Print[ineq2  =  MapAll[Together, ineq1]]; 
Print[
        ineq3  =  
          Distribute[2 ineq2,  Greater]]; 
Print[""]; 
Print[
        Distribute[ineq3  +  u  +  5,  
          Greater]];

Or the following commands will also find the solution.

Print[
        Re[z]  >  1]; 
Print[x  >  1]; 
Print[
        ineq  =  
          ComplexExpand[Re[Z]]  >  1]; 
solset = InequalitySolve[ineq, v]; 
Print[""]; 
Print[solset];

Which is the right half plane  v > u + 7  in the w-plane.  Use Mathematica to graph the transformation.

\(
\(Iden[z_]\  = \ z;\)\ \)
f[z_]  =  (-1  +  I) z  -  2  +  
          3  I; 
CartesianMap[
        Iden,  {1, 5, 0.5}, {-6, 7, 
          0.5}, 
PlotRange -> {{-1, 6}, {-1, 
              6}}, AspectRatio -> 
          1, 
Ticks -> {Range[-1, 6, 1], 
            Range[-1, 6, 1]}, AxesLabel -> {"x", "y"}, 
        PlotStyle -> {DarkGreen, Green}]; 
Print["The right half plane Re[z] > 1."]; 
CartesianMap[
        f,  {1, 5, 0.5}, {-6, 7, 
          0.5}, 
PlotRange -> {{-10, 2}, {-2, 
              10}}, AspectRatio -> 
          1, 
Ticks -> {Range[-10, 2, 2], 
            Range[2, 10, 2]}, AxesLabel -> {"u", "v"}, 
        PlotStyle -> {DarkGreen, Green}];  
Print["The image is the half plane v > u + 7,"]; 
Print["under the mapping w = ", 
        f[z]];

We see that the image of the right half plane  Re[z] > 1  under the linear transformation  w = (-1 + i) z  -  2 + 3  i  is the half plane  v > u + 7.

Section 2.2  Exercises for  Transformations and Linear Mappings
See textbook page 61.

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

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