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Section 2.2 Transformations and Linear Mappings

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C02-2.mws

COMPLEX ANALYSIS: Maple Worksheets,  2001
(c) John H. Mathews          Russell W. Howell
mathews@fullerton.edu     howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc.,      40  Tall  Pine  Drive,      Sudbury,  MA  01776
Tele.  (800) 832-0034;      FAX:  (508)  443-8000,      E-mail:  mkt@jbpub.com,      http://www.jbpub.com/


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 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 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*epsilon*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 z[1] <> z[2] onto distinct points  f(z[1]) <> f(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*epsilon*A  

and

     f(g(w)) = w  for all  w*epsilon*B  

 We now turn our attention to the investigation of some elementary mappings.  Let  B = a+i*b  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+i*b  to its new position  w = T(z) .  

The inverse mapping is given by  `w = `*T^`-1`*`(z) = w - B ` = u-a+i*(v-b) T^`-1` and shows that  T  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  f(z) = i*z maps the line  y = x+1  onto the line  v = -u-1 .

> 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});

`f(z) ` = I*z

`Find the image of the line y = x + 1`

{y = x+1, v = x, u = -y}

`Eliminate x and y from these equations.`

[{x = v, y = v+1}, {u+v+1}]

Thus we see that the solution is  u+v+1 = 0  or  v = -u-1 .


Example 2.9, Page 58.   Show that the linear transformation  w = i*z+i maps the right half plane  1 < Re(z)  onto the upper half plane  2 < Im(w) .

> 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;

`w ` = I*z+I

`u + I v ` = I*(x+I*y)+I

`u + I v ` = -y+I*(x+1)

` `

`Solve for z in terms of w.`

`z ` = -I*w-1

`x + I y ` = -I*(u+I*v)-1

`x + I y ` = -I*u+v-1

`We will use the substitutions:`

{x = v-1, y = -u}

` `

`Now find the image of the right half plane.`

1 < Re(z)

1 < x

0 < v-2

0 < v-2

This solution is the upper half plane  2 < Im(w) .

> 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]);

`f(z) ` = I*z+I

[Plot]

Example 2.10, Page 60.   Show that the image of the open disk  abs(z+1+i) < 1  under the transformation  w = (3-4*i)*z+6+2*i  is the open disk  abs(w+1-3*i) < 5 .

> 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;

`w ` = (3-4*I)*z+(6+2*I)

`u + I v ` = (3-4*I)*(x+I*y)+(6+2*I)

`u + I v ` = 3*x+4*y+6+I*(-4*x+3*y+2)

` `

`Solve for z in terms of w.`

`z ` = 3/25*w+4/25*I*w+(-2/5-6/5*I)

`x + I y ` = 3/25*u+3/25*I*v+4/25*I*(u+I*v)+(-2/5-6/5*I)

`x + I y ` = 3/25*u-4/25*v-2/5+I*(3/25*v+4/25*u-6/5)

`We will use the substitutions:`

{x = 3/25*u-4/25*v-2/5, y = 3/25*v+4/25*u-6/5}

` `

`Now find the image of the disk.`

abs(z+(1+I))^2 < 1

(x+1)^2+(y+1)^2 < 1

(3/25*u-4/25*v+3/5)^2+(3/25*v+4/25*u-1/5)^2 < 1

1/25*u^2+2/25*u+1/25*v^2-6/25*v < 3/5

1/25*u^2+2/25*u+1/25*v^2-6/25*v < 3/5

u^2+2*u+v^2-6*v < 15

Which is the disk  abs(w+1-3*i) < 5  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=[15,15], numxy=[50,50],
 scaling=constrained,
 labels=[`u   `,`v   `],
 view=[-6..4,-2..8]);

`f(z) ` = (3-4*I)*z+(6+2*I)

[Plot]

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

> 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;

`w ` = (-1+I)*z+(-2+3*I)

`u + I v ` = (-1+I)*(x+I*y)+(-2+3*I)

`u + I v ` = -x-y-2+I*(x-y+3)

` `

`Solve for z in terms of w.`

`z ` = -1/2*w-1/2*I*w+(-5/2+1/2*I)

`x + I y ` = -1/2*u-1/2*I*v-1/2*I*(u+I*v)+(-5/2+1/2*I)

`x + I y ` = -1/2*u+1/2*v-5/2+I*(-1/2*v-1/2*u+1/2)

`We will use the substitutions:`

{y = -1/2*v-1/2*u+1/2, x = -1/2*u+1/2*v-5/2}

` `

`Now find the image of the right half plane.`

1 < Re(z)

1 < x

0 < -1/2*u+1/2*v-7/2

0 < -1/2*u+1/2*v-7/2

0 < -u+v-7

u < v-7

Which is the  half plane  7+u < v  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]);

`f(z) ` = (-1+I)*z+(-2+3*I)

[Plot]

>

End of Section 2.2.