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

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CHAPTER 9 CONFORMAL MAPPING

Section 9.2 Bilinear Transformations

Another important class of elementary mappings was studied by Augustus Ferdinand Mobius (1790-1868). These mappings are conveniently expressed as the quotient of two linear expressions and are commonly known as linear fractional or bilinear transformations. They arise naturally in mapping problems involving the function . In this section we will show how they are used to map a disk one-to-one and onto a half plane.

Let and denote four complex constants with the restriction that . Then the function

,

is called a bilinear transformation or Mobius transformation or linear fractional transformation .

If the expression for is multiplied through by the quantity , then the resulting expression has the bilinear form

.

We can collect terms involving and write . For values of the inverse transformation is given by

= = .

This permits us to uniquely determine a bilinear transformation if three distinct image values , , and are specified. To determine such a mapping, it is convenient to use an implicit formula involving and .

Theorem 9.3 (The Implicit Formula)

There exists a unique bilinear transformation that maps three distinct points , , and onto three distinct points , , and , respectively. An implicit formula for the mapping is given by the equation

= .

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

> with(plots):

Warning, the name changecoords has been redefined

Example 9.3, Page 364. Show that maps
the unit disk one-to-one and onto the upper half plane .

> f:='f': s:='s': w:='w': z:='z':
f := z -> I*(1-z)/(1+z):
`w = s(z) ` = f(z);

To show is one-to-one, find the inverse function.

> `z ` = solve(w=f(z),z);

To show is onto we will consider a graph.
The graph is not conclusive, but it appears to confirm our suspision.

> f:='f': z:='z':
f := z -> I*(1-z)/(1+z):
`s(z) ` = f(z);
conformal(f(Re(z)*exp(I*Im(z))), z=0.01..1+I*2*Pi,
title=`w = i(1-z)/(1+z)`,
labels=[`u`,`v`], tickmarks=[7,4],
grid=[25,25], numxy=[25,25],
scaling=constrained,
view=[-3..3,0..3]);

Example 9.4, Page 366. Construct the bilinear transformatin that maps
, , onto the points , , , respectively.

> w:='w': w1:='w1': w2:='w2': w3:='w3':
z:='z': z1:='z1': z2:='z2': z3:='z3':
s:='s': S:='S': w:='w': W:='W': z:='z': Z:='Z':
formula := (z-z1)*(z2-z3)/(z-z3)/(z2-z1) =
(w-w1)*(w2-w3)/(w-w3)/(w2-w1):
formula;
`Now make the substitutions:`;
z1 := -I: z2 := 1: z3 := I:
w1 := -1: w2 := 0: w3 := 1:
z[1] = z1, z[2] = z2, z[3] = z3;
w[1] = w1, w[2] = w2, w[3] = w3;
formula;
formula := subs({z=Z,w=W},formula):
sol := solve(formula, W):
S := z -> subs(Z=z, sol):
`w = s(z) ` = S(z);

Check our work and look at the images of , , and .

> `The points in the z-plane are:`;
z[1] = z1, z[2] = z2, z[3] = z3;
` `; `w = s(z) ` = S(z);
`The images in the w-plane are:`;
s(z[1]) = S(z1), s(z[2]) = S(z2), s(z[3]) = S(z3);
`The given points were:`;
w[1] = w1, w[2] = w2, w[3] = w3;

Example 9.5, Page 366. Find the bilinear transformatin that maps the points
, , onto the points , , , respectively.

> w:='w': w1:='w1': w2:='w2': w3:='w3':
z:='z': z1:='z1': z2:='z2': z3:='z3':
s:='s': S:='S': W:='W': z:='z': Z:='Z':
formula := (z-z1)*(z2-z3)/(z-z3)/(z2-z1) =
(w-w1)*(w2-w3)/(w-w3)/(w2-w1):
formula;
`Now make the substitutions:`;
z1 := -2: z2 := -1-I: z3 := 0:
w1 := -1: w2 := 0: w3 := 1:
z[1] = z1, z[2] = z2, z[3] = z3;
w[1] = w1, w[2] = w2, w[3] = w3;
formula;
formula := subs({z=Z,w=W},formula):
sol := solve(formula, W):
S := z -> subs(Z=z, sol):
`w = s(z) ` = S(z);

This is equivalent to the formula , and can be verified by the computation:

> `s(z) ` = S(z);
`w ` = ((1-I)*z + 2)/((1+I)*z + 2);
`s(z) - w ` = simplify(((1-I)*z + 2)/((1+I)*z + 2)-S(z));

Check our work and look at the images of , , and .

> `The points in the z-plane are:`;
z[1] = z1, z[2] = z2, z[3] = z3;
` `; `w = s(z) ` = S(z);
`The images in the w-plane are:`;
s(z[1]) = S(z1), s(z[2]) = S(z2), s(z[3]) = S(z3);
`The given points were:`;
w[1] = w1, w[2] = w2, w[3] = w3;

Example 9.6, Page 367. Show that maps the
disk one-to-one and onto the upper half plane .

> s:='s': S:='S': w:='w': z:='z':
S := z -> ((1-I)*z + 2)/((1+I)*z + 2):
`w = s(z) ` = S(z);

To show is one-to-one, we need to find the inverse function.

> `z ` = solve(w=S(z),z);

To show is onto, use the method of oriented points on the boundary curve.

> w:='w': w1:='w1': w2:='w2': w3:='w3':
z:='z': z1:='z1': z2:='z2': z3:='z3':
z1 := -2: z2 := -1-I: z3 := 0:
w1 := -1: w2 := 0: w3 := 1:
`The points in the z-plane are:`;
z[1] = z1, z[2] = z2, z[3] = z3;
` `; `w = s(z) ` = S(z);
`The images in the w-plane are:`;
s(z[1]) = S(z1), s(z[2]) = S(z2), s(z[3]) = S(z3);
`The given points were:`;
w[1] = w1, w[2] = w2, w[3] = w3;

To find the image of the disk under we use the
change of variable and find the image of under .

> g:='g': S:='S': z:='z': Z:='Z':
S := z -> ((1-I)*z + 2)/((1+I)*z + 2):
g := z -> subs(Z=z,S(Z-1)):
`s(z) ` = S(z);
`g(z) ` = g(z);
conformal(g(Re(z)*exp(I*Im(z))), z=0.01..1+I*2*Pi,
title=`w = ((1-i)z + 2)/((1+i)z + 2)`,
labels=[`u`,`v`], tickmarks=[7,4],
grid=[25,25], numxy=[25,25],
scaling=constrained,
view=[-3..3,0..3]);

Therefore, the image of the disk is the upper half plane .

Example 9.7, Page 368. Find the bilinear transformatin that maps the crescent-shaped
region that lies inside the disk and outside the circle onto a horizontal strip.

Solution. For convenience we choose the points
, , which map onto , , , respectively.
In this case we remove the terms involving in the implicit formula, because
this implies that = = .

> w:='w': w1:='w1': w2:='w2': w3:='w3':
z:='z': z1:='z1': z2:='z2': z3:='z3':
s:='s': S:='S': W:='W': z:='z': Z:='Z':
formula :=
(w-w1)/(w2-w1) = (z-z1)*(z2-z3)/(z-z3)/(z2-z1) :
formula;
`Now make the substitutions:`;
z1 := 4: z2 := 2+2*I: z3 := 0:
w1 := 0: w2 := 1: w3 := infinity:
z[1] = z1, z[2] = z2, z[3] = z3;
w[1] = w1, w[2] = w2, w[3] = w3;
formula;
formula := subs({z=Z,w=W},formula):
sol := expand(solve(formula, W)):
S := z -> subs(Z=z, sol):
`w = s(z) ` = S(z);

Check our work and look at the images of , , and .

> `The points in the z-plane are:`;
z[1] = z1, z[2] = z2, z[3] = z3;
` `; `w = s(z) ` = S(z);
`The images in the w-plane are:`;
s(z[1]) = S(z1), s(z[2]) = S(z2), s(z[3]) = infinity;
`The given points were:`;
w[1] = w1, w[2] = w2, w[3] = w3;

And we need to look at the points:
, , and , , .

> z4 := 1-I: z5 := 2: z6 := 1+I:
w4 := -2+I: w5 := I: w6 := 2+I:
`The points in the z-plane are:`;
z[4] = z4, z[5] = z5, z[6] = z6;
` `; `w = s(z) ` = S(z);
`The images in the w-plane are:`;
s(z[4]) = S(z4), s(z[5]) = S(z5), s(z[6]) = S(z6);
`The given points were:`;
w[4] = w4, w[5] = w5, w[6] = w6;

To illustrate the mapping, consider the inverse image of the horizontal strip
under the inverse transformation .

> sol := solve(formula,Z):
`z = T(w) ` = subs(W=w,sol);

To illustrate the mapping we draw a graph.

> T:='T': w:='w':
T := w -> 4*I/(I+w):
`T(w) ` = T(w);
conformal(T(w), w=-6+I*0.01..6+I*0.99,
title=`z = 4i/(i+w)`,
labels=[` x`,`y `], tickmarks=[5,5],
grid=[25,25], numxy=[25,25],
scaling=constrained,
view=[0..4,-2..2]);

End of Section 9.2.