Module

for

The Reciprocal Transformation w=1/z

2.5  The Reciprocal Transformation

The mapping    is called the reciprocal transformation and maps the z-plane one-to-one and onto the w-plane except for the point  z=0,  which has no image, and the point  w=0,  which has no preimage or inverse image.  Use the exponential notation   in the w-plane.  If  ,  we have

.

The geometric description of the reciprocal transformation is now evident.  It is an inversion (that is, the modulus of () is the reciprocal of the modulus of z) followed by a reflection through the x axis.  The ray  ,  is mapped one-to-one and onto the ray  .  Points that lie inside the unit circle     are mapped onto points that lie outside the unit circle and vice versa.  The situation is illustrated in Figure 2.21.

Figure 2.21  The reciprocal transformation  .

We can 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.  You will see shortly that the point    has the property, loosely speaking, that    iff  .

An -neighborhood of the point at infinity is the set  .  The usual way to visualize the point at infinity is by using what we call the stereographic projection, which is attributed to Riemann.  Let    be a sphere of diameter 1 that is centered at the point    in three-dimensional space where coordinates are specified by the triple of real numbers .   Here the complex number    will be associated with the point  .

The point   on    is called the north pole of  .  If we let z be a complex number and consider the line segment L in three-dimensional space that joins z to the north pole  ,  then L intersects  in exactly one point .  The correspondence    is called the stereographic projection of the complex z plane onto the Riemann sphere  .

A point    of unit modulus will correspond with  .  If z has modulus greater than 1, then    will lie in the upper hemisphere where for points    we have  .  If z has modulus less than 1, then    will lie in the lower hemisphere where for points    we have  .  The complex number   corresponds with the south pole,  .

Now you can see that indeed    iff    iff  .  Hence    corresponds with the "ideal" point at infinity.  The situation is shown in Figure 2.22.

Figure 2.22  The complex plane and the Riemann sphere  .

Let us reconsider the mapping  .  Let us assign the images    and    to the points    and  , respectively.  We now write the reciprocal transformation as

Note that the transformation    is a one-to-one mapping of the extended complex z plane onto the extended complex w plane.  Further, f is a continuous mapping from the extended z plane onto the extended w plane.  We leave the details to you.

Extra Example.  Investigate the limits of   as  .

Explore Solution for Extra Example.

Example 2.22.  Show that the image of the right half plane    under the mapping    is the closed disk    in the w-plane.

Solution.  Proceeding as we did in Example 2.7, we get the inverse mapping of    as  .  Then

which describes the disk  .  As the reciprocal transformation is one-to-one, preimages of the points in the disk    will lie in the right half-plane  .  Figure 2.23 illustrates this result.

Figure 2.23  The image of    under the mapping  .

Explore Solution 2.22.

Remark.  Alas, there is a fly in the ointment here.  As our notation indicates,    and    are not equivalent.  The former implies the latter, but not conversely.  That is, makes sense when  ,  whereas does not.  Yet Figure 2.23 seems to indicate that f maps    onto the entire disk   ,  including the point .  Actually, it does not, because has no preimage in the complex plane.  The way out of this dilemma is to use the complex point at infinity.  It is that quantity that gets mapped to the point , for as we have already indicated, the preimage of 0 under the "extended" mapping    is indeed .

Example 2.23.  For the transformation  ,  find the image of the portion of the right half plane    that lies inside the closed disk  .

Solution.  Using the result of Example 2.22, we need only find the image of the closed disk    and intersect it with the closed disk  .  To begin, we note that

.

Because ,  we have, as before,

which is an inequality that determines the set of points in the w plane that lie on and outside the circle  .  Note that we do not have to deal with the point at infinity this time, as the last inequality is not satisfied when .  When we intersect this set with  ,  we get the crescent-shaped region shown in Figure 2.24.

Figure 2.24  The image of the half disk under is a crescent-shaped region.

Explore Solution 2.23.

To study images of "generalized circles," we consider the equation

,

where A, B, C, and D are real numbers.  This equation represents either a circle or a line, depending on whether , respectively.  Transforming the equation to polar coordinates gives

.

Using the polar coordinate form of the reciprocal transformation, we can express the image of the curve in the preceding equation as

,

which represents either a circle or a line, depending on whether , respectively.  Therefore, we have shown that the reciprocal transformation carries the class of lines and circles onto itself.

Example 2.24.  Consider the mapping  .
(a)  Find the images of the vertical lines  x = a.   (b)  Find the images the horizontal lines  y = b.

Solution.  Taking into account the point at infinity, we see that the image of the line x=0 is the line u=0;  that is, the y axis is mapped onto the v axis.
Similarly, the x axis is mapped onto the u axis.  Again, the inverse mapping is  ,  so if  ,  the vertical line    is mapped onto the set of (u,v) points satisfying  .  For (u,v)~=(0,0), this outcome is equivalent to

,

which is the equation of a circle in the w plane with center    and radius  .  The point at infinity is mapped to (u,v)=(0,0).

Similarly, the horizontal line    is mapped onto the circle

which has center    and radius  .  Figure 2.25 illustrates the images of several lines.

Figure 2.25  The images of horizontal and vertical lines under the reciprocal transformation.

Explore Solution 2.24.

You are now ready to study Section 10.2 Bilinear Transformations - Mobius Transformations.

Riemann Sphere

Mobius - Bilinear Transformation

The Next Module is

Differentiable and Analytic Functions

(c) 2012 John H. Mathews, Russell W. Howell