COMPLEX ANALYSIS: Maple Worksheets, 2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
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CHAPTER 2 COMPLEX FUNCTIONS

Section 2.2 Transformations and Linear Mappings

We not take our first look at the geometric interpretation of a complex function. If is the domain of definition of the real-valued functions and , then the system of equations and describes a transformation or mapping from in the xy-plane into the uv-plane. Therefore, the function can be considered as a mapping or transformation from the set in the z-plane onto the range in the w-plane.

If is a subset of the domain of definition , then the set is called the image of the set , and 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 is said to be from A into S if the image of is contained in . The inverse image of a point w is the set of all points in such that . 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 is said to be one-to-one if it maps distinct points onto distinct points . If maps the set one-to-one and onto the set , then for each w in there exists exactly one point in such that . Then loosely speaking, we can solve the equation by solving for as a function of . That is, the inverse function can be found, and the following equations hold:

for all

and

for all

We now turn our attention to the investigation of some elementary mappings. Let denote a fixed complex number. Then the transformation 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 is displaced through the vector to its new position .

The inverse mapping is given by and shows that is a one-to-one mapping from the z-plane onto the w-plane.

Load Maple's "eliminate" and "conformal mapping" procedures.
Make sure this is done only ONCE during a Maple session.

> readlib(eliminate):
with(plots):

Warning, the name changecoords has been redefined

Example 2.6, Page 55. Show that the function maps the line onto the line .

> f:='f': x:='x': X:='X': y:='y': Y:='Y': z:='z': Z:='Z':
assume(X, real); assume(Y, real);
Z := X + I*Y:
f := z -> I*z:
`f(z) ` = f(z);
`Find the image of the line y = x + 1`;
eqns := {u = Re(f(Z)), v = Im(f(Z)), y = x + 1}:
eqns2 := (subs(X=x,Y=y,eqns)): eqns2;
`Eliminate x and y from these equations.`;
eliminate(eqns2,{x,y});

Thus we see that the solution is or .

Example 2.9, Page 58. Show that the linear transformation maps the right half plane onto the upper half plane .

> f:='f': u:='u': v:='v': w:='w': x:='x': y:='y': z:='z':
f := z -> I*z + I:
`w ` = f(z);
`u + I v ` = f(x + I*y);
`u + I v ` = evalc(f(x + I*y)); ` `;
`Solve for z in terms of w.`;
solset := expand(solve(W = f(z), z)):
g := w -> subs(W=w,solset):
`z ` = g(w);
`x + I y ` = g(u + I*v);
`x + I y ` = evalc(g(u + I*v));
`We will use the substitutions:`;
eqns := {x=v-1, y=-u}: eqns; ` `;
`Now find the image of the right half plane.`;
ineq := Re(z) > 1: ineq;
ineq := x > 1: ineq;
ineq := subs(eqns,ineq): ineq;
ineq := ineq + (1<1): ineq;

This solution is the upper half plane .

> f:='f': z:='z':
f := z -> I*z + I:
`f(z) ` = f(z);
conformal(f(z), z=1-6*I..5+4*I,
title=`w = I*z + I`,
grid=[9,11],numxy=[9,11],
scaling=constrained,
labels=[`u `,`v `],
view=[-4.25..6.25,-0.25..6.25]);

Example 2.10, Page 60. Show that the image of the open disk under the transformation is the open disk .

> f:='f': u:='u': v:='v': w:='w': x:='x': y:='y': z:='z':
f := z -> (3 - 4*I)*z + 6 + 2*I:
`w ` = f(z);
`u + I v ` = f(x + I*y);
`u + I v ` = evalc(f(x + I*y)); ` `;
`Solve for z in terms of w.`;
solset := expand(solve(W = f(z), z)):
g := w -> subs(W=w,solset):
`z ` = g(w);
`x + I y ` = g(u + I*v);
`x + I y ` = evalc(g(u + I*v));
`We will use the substitutions:`;
eqns := {x=3*u/25-2/5-4*v/25, y=3*v/25-6/5+4*u/25}:
eqns; ` `;
`Now find the image of the disk.`;
ineq := abs(z+1+I)^2 < 1: ineq;
ineq := (x+1)^2 + (y+1)^2 < 1: ineq;
ineq := subs(eqns,ineq): ineq;
ineq := map(expand,ineq): ineq;
ineq := ineq - (2/5<2/5): ineq;
ineq := 25*ineq: ineq;

Which is the disk in the w-plane.

> f:='f': F:='F': z:='z':
f := z -> (3 - 4*I)*z + 6 + 2*I:
`f(z) ` = f(z);
F := z -> subs(Z=z-1-I,f(Z)):
conformal(F(Re(z)*exp(I*Im(z))), z=0.01..1+I*2*Pi,
title=`w = (3-4i)*z + 6 + 2i`,
grid=[13,13], numxy=[13,13],
scaling=constrained,
labels=[`u `,`v `],
view=[-6..4,-2..8]);

Example 2.11, Page 60. Show that the image of the right half plane under the linear transformation is the half plane .

> f:='f': u:='u': v:='v': w:='w': x:='x': y:='y': z:='z':
f := z -> (-1 + I)*z - 2 + 3*I:
`w ` = f(z);
`u + I v ` = f(x + I*y);
`u + I v ` = evalc(f(x + I*y)); ` `;
`Solve for z in terms of w.`;
solset := expand(solve(W = f(z), z)):
g := w -> subs(W=w,solset):
`z ` = g(w);
`x + I y ` = g(u + I*v);
`x + I y ` = evalc(g(u + I*v));
`We will use the substitutions:`;
eqns := {x=-u/2-5/2+v/2, y=-v/2+1/2-u/2}:
eqns; ` `;
`Now find the image of the right half plane.`;
ineq := Re(z) > 1: ineq;
ineq := x > 1: ineq;
ineq := subs(eqns,ineq): ineq;
ineq := ineq + (5/2<5/2): ineq;
ineq := 2*ineq: ineq;
ineq := ineq + (u<u): ineq;

Which is the half plane in the w-plane.

> f:='f': z:='z':
f := z -> (-1 + I)*z - 2 + 3*I:
`f(z) ` = f(z);
conformal(f(z), z=1-6*I..5+7*I,
title=`w = (-1+I)*z - 2 + 3*I`,
grid=[9,14], numxy=[9,14],
scaling=constrained,
labels=[`u `,` v`],
view=[-14.25..3.25,-3.25..14.25]);

End of Section 2.2.