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 2.6 The Reciprocal Transformation

The mapping is called the reciprocal transformation and maps the plane one-to-one and onto the plane except for the point , which has no image, and the point , which has no preimage or inverse image. Using exponential notation , we see that if = , then we have

= = = .

It is convenient to extend the system of complex numbers by joining to it an "ideal" point denoted by and called the point at infinity . This new set is called the extended complex plane . The reciprocal transformation maps the "extended complex -plane" one-to-one and onto the "extended complex -plane".

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 2.22, Page 86. Show that the image of the right half plane under the transformation is the disk .

> u:='u': v:='v': x:='x': X:='X': Y:='Y': Z:='Z':
assume(X, real); assume(Y, real);
Z := X + I*Y:
ineq0 := Re(Z) > 1/2:
ineq1 := subs(X='x',ineq0):
ineq2 := subs({x=u/(u^2 + v^2),y=-v/(u^2 + v^2)}, ineq1):
ineq3 := 2*(u^2 + v^2)*1/2 < 2*(u^2 + v^2)*u/(u^2+v^2):
ineq4 := (1-2*u < 1-2*u) + ineq3:
ineq1; ineq2; ineq3; ineq4;

Thus, the image is the disk .

> f:='f': z:='z':
f := z -> 1/z:
`f(z) ` = f(z);
conformal(f(z), z=0.5-5*I..10+5*I,
title=`Image of Re z > 1/2 under w = 1/z`,
grid=[30,31], numxy=[31,31],
scaling=constrained,
labels=[` u`,`v `],
tickmarks=[5,5],
view=[-0.1..2.1,-1.1..1.1]);

Example 2.23, Page 87. Find the image of the portion of the right half plane that lies inside the circle under the mapping .

> f:='f': F:='F': z:='z':
f := z -> 1/z:
`f(z) ` = f(z);
F := z -> subs(Z=z+1/2,f(Z)):
`F(z) ` = F(z);
conformal(F(Re(z)*exp(I*Im(z))), z=0.01..1+I*2*Pi,
title=`Image of |z-1/2|<1 under w = 1/z`,
grid=[25,25],numxy=[25,25],
scaling=constrained,
labels=[` u`,` v`],
tickmarks=[5,5],
view=[-3..1,-2..2]);

Using the result from Example 2.16 and the above information. We conclude that the image is the crescent shaped region in the w-plane which is the portion of the disk that lies outside the circle .

Example 2.24, Page 88.
(a) Find the images of the vertical lines under the mapping .

> a:='a': u:='u': v:='v': x:='x': y:='y':
eq1 := x = a:
eq2 := subs({x=u/(u^2 + v^2),y=-v/(u^2 + v^2)}, eq1):
eq3 := (u^2 + v^2)/a*eq2:
eq4 := (- u/a = - u/a) + eq3:
eq5 := (1/(2*a)^2 = 1/(2*a)^2) + eq4:
eq1; eq2; eq3; eq4; eq5;

Thus, the image is the circle , with center and radius .

(b) Find the images of the horizontal lines under the mapping .

> b:='b': u:='u': v:='v': x:='x': y:='y':
eq1 := y = b:
eq2 := subs({x=u/(u^2 + v^2),y=-v/(u^2 + v^2)}, eq1):
eq3 := (u^2 + v^2)/b*eq2:
eq4 := (v/b = v/b) + eq3:
eq5 := (1/(2*b)^2 = 1/(2*b)^2) + eq4:
eq1; eq2; eq3; eq4; eq5;

>

Thus, the image is the circle ,with center and radius .

> f:='f': z:='z':
f := z -> 1/z:
`f(z) ` = f(z);
e := 0.000001:
conformal(f(z), z=-2-2*I..e+2+e+2*I,
title=`Image of lines under w = 1/z`,
grid=[5,5], numxy=[17,17],
scaling=constrained,
labels=[` u`,`v `],
tickmarks=[5,5],
view=[-1.1..1.1,-1.1..1.1]);

End of Section 2.6.