5.1. The linear transformation and the inversion.
In this section we investigate the transformation which provides very convenient methods of nding a one-to-one mapping of one domain into another. Let us start with the a linear transformation
w = φ(z) := Az + B, (1)
where A and B are xed complex numbers, A ≠ 0.
We write (1) as
As we see this transformation is a composition of a rotation about the origin through the angle Arg (a)
a magni cation
and a translation
w = w3 = w2 + B.
Each of these transformations are one-to-one mappings of the complex plane onto itself and gap geometric ob jects onto congruent ob jects. In particular, the the ranges of lines and of circles are line and circles, respectively.
Now we consider the inversion dened by
w : = 1/z. (2)
It is easy to see that the inversion is a one-to-one mapping of the extended complex plane onto itself (0 → ∞ and vice versa ∞ → 0.)
We are going to show that the image of a line is either a line or a circle. Indeed, let rst l passes through the origin. The image of a point peiθ is Letting p to tend from the negative in nity to the positive one, we see that the image is another line through the origin with an angle of inclination
Let now L be given by the equation
L : Ax + By ≠ C; with C = 0; and |A| + |B| > 0. (3)
On writing w = u + iv; we find
and so
(4)
Making these substitutions into (3), we get
or, equivalently,
(5)
This is apparently a equation of a circle.
5.2 We are prepared to go de ne a transformation in the general sense.
Definition: The transformation is called a transformation.
If c = 0; then the Mobius transformation is linear. If c ≠ 0; a = 0 then the transformation is a inversion. Consider the case ac ≠ 0: Then w can be written as
(6)
which is as a matter of fact a decomposition of a linear transformation and and inverse function. We also notice that
Hence, w is conformal at every point z ≠ -d/c:
Having in mind this observation and the previous deliberations, we can summarize
Theorem 5.1. Let f be a Mobius transformation. Then
f can be expressed as a composition of magni cations, rotations, translations and inversions.
f maps the extended complex plane onto itself.
f maps the class of circled and lines to itself.
f is conformal at every point except its pole
5.3. The group of transformations. Let
(7)
be a mobius transformation. The inverse function z = f-1(w) (that is: the identity) can be computed directly:
We see that the inverse is again a transformation. Furthermore, we easily check that the composition of two transformations (e.g. if
then
is again a transformation of MÖbius. On the other hand, f º I (z) = I º f (z) = f (z): So, the set of all Mobius transformations is a group with respect to the composition.
Now we shall prove
Theorem 5.2. A transformation is uniquely determined by three points zi ; i = 1, 2, 3, zi ≠ zj ; i, j = 1; 2; 3.
Proof: We first introduce the term of a double point. We say that z0 is double point of f (z); if
f (z ) = z . (8)
It It is obvious that a transformation has not more that two double points unless it coincides with the identity. Indeed, if (8) is ful lled for three distinct points, then the quadratic equation
az + b = cz2 + dz
will have three distinct zeros, which implies a = d; b = c = 0: Notice that a linear transformation has only one double point.
Let now zi, i = 1, 2, 3 and wi; i = 1, 2, 3 be given, zi ≠ zj ; wi ≠ wj , i.j. = 1; 2; 3: We are looking for a transformation w = f (z) such that
f (zi) = wi. (9)
Consider the cross-ratio (z , z1 , z2 , z3) of the points z , zi , i = 1, 2, 3, that is
The function T (z); z ∈ C; zi - fixed maps C in a one-to-one way onto itself. Notice that the order in which the points are listed is crucial in this notation.
Then the desired transformation (9) is given by the composition
(z, z1, z2, z3) = (w, w1, w2, w3).
It remains only to equate w.
The next step is to show that w = f (z) is the only transformation with property (9).
Indeed, suppose to the contrary that there is another transformation which satis es (9). Then the transformation has three distinct double points which is impossible.
Example 5.1: Find a transformation T (z) that maps the real line R onto the unit circle C0(1):
Solution: We select three arbitrary real points, say -1,0,1 and three arbitrary points on C0(1), say -i,1, i. The transformation w = T (z); determined by the cross products, namely
(z, -1, 0, 1) = (w, -i, 1, i)
maps the real line R onto the circle C passing through the points i; 1; i: As we know from the elementary geometry,
Example 5.2: Let
Find the image of R, I and C0(1) under the mapping w = T (z).
Solution: We rst observe that
R → R.
Further, we see that
T (-1) = 0,
T (1) = 1
T (0) = -1,
T (∞) = 1,
T (i) = -i.
The image of I is a line or a circle passing through the points -1, 1, i, that is the unit circle. The image of C0(1) is a line or a circle that passes through 0, ∞ and is orthogonal to the images of R and of I; in our case orthogonal to R and to C0(1): This turns to be the imaginary axis.
5.4. The left-hand-rule. : Consider the unit circle C0(1): The points -1, -i, 1 determine the direction -1 → -i → 1 → 1 in traversing C0 (1). The interior of the circle, the unit disk D0 (1) lies to the left of this orientation. We use to say that the disk is the left region with respect to the orientation 1 → -i → 1. Analogously, the upper half plane is the left region with respect to the direction -1, 0, ∞. Since Mobius transformations transformations are conformal mappings, it can be shown that a transformation that takes the distinct points z1, z2, z3 to the respective points w1, w2, w3 must map the left region of the circle (or line) oriented by z1, z2, z3 onto the left region of the circle (or line) oriented by w1, w2, w3. Using the conformality, we summarize
Theorem 5.3. : Let G be a domain in and assume that G is left oriented with respect to the direction given by the points z1 → z2 → z3, zi be a transformationthat maps Then T maps G onto that domain which is left oriented to γwith respect to the direction w1 → w2 → w3.
Example 5.3: Let w = T (z) be the transformationfrom Example 2. Find the images of the upper half plane, of the left half plane and of the unit disk.
Solution: As we have seen,
The upper plane the left domain with respect to the direction -1 → 0 → ∞, hence the range domain will be left oriented with respect to 0, -1,1 (the images of -1, 0, ∞), e.g., the half plane below the real axis. Similarly, we nd that the left half plane is mapped in the unit disk, whereas the unit disk - in the left half plane.
5.5. The symmetry and the transformation.
Definition: Given the circle Ca(r) of radius r and centered at z = a; we say that the points are symmetric with respect to Ca(r) if
One can easily prove that in this denition each straight line or circle passing through z and z* intersects tCa(r) orthogonally. In the case of a line we have the usual symmetry with respect to it.
Theorem 5.3. Given a circle C and a transformationw = T (z); assume that z, z* are symmetric with respect to C. Then their images T (z), T (z*) are again symmetric with respect to the image T (C ) of C.
Example 5.4: Find all Mobius transformation that map the disk C0 (r) onto itself.
Solution; We x in an arbitrary way a point a ∈ D0(r): Its symmetric point with respect to the unit circle is
(10)
Any transformation of the form
(11)
with K being some constant maps the points a and a* at 0 and in nity, respectively. Because of the symmetry, the image of C0(r) will be a circle C centered at the zero. Now we are going to choose the constant K in such a way that C coincides with C0(r). Take a point z0 := reiφ and calculate S (z0). We have, by (10),
or
Putting
we arrive at
for some real
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