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Another way of representing any circle is in the form
AAzz Bz Bz CC 0
for complex numbers A B C.
If A 0 this is a straight line if C 0 it passes through the origin.
For this form also it is easy to con rm that inversion takes circles to circles
where a straight line is just a rather extremal case of a circle. Malcolm Hood
told me this one.
These representations are sneaky and probably cheating but it is telling
you something important namely some representations for things will make
some problems dead easy and others make it horribly di cult. Thinking
about this early on can save you a lot of grief.
Exercise 2.5.5 Can you see why a parametric representation of the circle
of the form z a r cos i b r sin could be a serious blunder in trying
to show that inversions take circles to circles
Remark
64 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS
The moral we draw from this little excursion is that being true and faithful
to a human being is possibly a ne and splendid thing being faithful and
true to a principle or ideology might be a ne and splendid thing or it might
be a sign of a sentimental nature gone wild. But being faithful and true to
a brand of beans or a choice of representation of an object is to confuse the
nger pointing at the moon with the moon itself and a sure sign of total
fatheadedness. The poor devil who believes deeply that the only true and
proper way to represent a circle in the plane is by writing down
x a 2 y b 2 r2
is to be pitied as someone who has confused the language with the thing being
talked about and is t only for politics. The more ways you have of talking
and thinking about things the easier it is to draw conclusions and the harder
it is to be led astray. It is also a lot more fun.
The converse is also true the inversion of a straight line is a circle through
the origin.
To see this let ax by c 0 be the equation of a straight line. Turn this
into polars to get
ar cos br sin c 0
Now put r 1 s to get the inversion
a s cos b s sin c 0
and rearrange to get
s2 as c cos bs c sin 0
a 2c
which is a circle passing through the origin with centre at .
b 2c
It is easy to see that the points at in nity on each end of the line get sent
to the origin.
This suggests that we could simplify the description by working not in the
plane but in the space we would get by adjoining a point at in nity .
We do this by putting a sphere of radius 1 2 sitting on the origin of R3 and
identify the z 0 plane with C. Now to map from the sphere to the plane
2 3
4 5
take a line from the north pole of the sphere which is at the point 0
1
1
2.5. THE FUNCTION F Z 65
Z
Q
P
Q
P
Figure 2.21 The Riemann Sphere
and draw it so it cuts the sphere in P and the plane at P . Now this sets up
a one one correspondence between the points of the sphere other than the
north pole and the points of the plane. The unit circle in the plane is sent
to the equator of the sphere.
Now we put the point at in nity of the plane in at the north pole of the
sphere.
An inversion of the plane now gives an inversion of the sphere which sends
the South pole the origin to the North pole all we do is to project down
so that the point Q goes directly to the point Q vertically below it and
vice versa. In other words we re ect in the plane of the equator.
Exercise 2.5.6 Verify that this rule ensures that a point in the plane is sent
to its inversion when we go from the point up to the sphere then re ect in
the plane of the equator then go back to the plane.
Exercise 2.5.7 Suppose we have a disk which contains the origin on its
boundary. What would you expect the inversion of the disk to look like
Suppose we have a disk which contains the origin in its interior. What would
you expect the inversion to look like
66 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS
Sketches of the general situation should take you only a few minutes to work
out it is probably easiest to visualise it on the Riemann Sphere.
Exercise 2.5.8 What would you expect to get qualitatively if you invert a
triangle shaped region of C Does it make a di erence if the triangle contains
the origin
Draw some pictures of some triangles and what you think their inversions
would look like.
Note that if you do an inversion and then invert the result you get back to
where you started. In other words the inversion is its own inverse map. Since
the same is true of conjugation the map f z 1 z also has this property.
Exercise 2.5.9 What happens if you invert a half plane made by taking all
the points on one side of a line through the origin What if the half plane is
the set of points on one side of a line not through the origin
I haven t said anything much about the conjugation because it is really very
trivial just re ect everything in the X axis.
2.6 The M
obius Transforms
The reciprocal transformation is a special case of a general class of complex
functions called the Fractional linear or M transforms. In the old days
obius
they also were called bilinear but this word now means something else and
is no longer used by the even marginally fashionable.
The general form of the M functions is
obius
az b
w f z
cz d
where a b c d are complex numbers. If c 0 d 1 we have the a ne maps
and if a 0 b 1 c 1 d 0 we have the reciprocal map. It is tempting
to represent each M function by the corresponding matrix
obius
az b
a b
c d
cz d
2.6. THE MOBIUS TRANSFORMS 67
which makes the identity matrix correspond nicely to the identity map w z.
One reason it is tempting is that if we compose two M functions we get
obius
another M function and the matrix multiplication gives the correspond
obius
ing coe cients. This is easily veri ed and shows that providing ad bc the
6
M function
obius
az b
cz d
has an inverse and indeed it tells us what it is.
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