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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 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=[15,15], numxy=[50,50],  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

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.