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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.
The sneaky argument for the inversion also goes through for M func
obius
tions i.e. they take circles on the Riemann Sphere to other circles. It is clear
that the Riemann Sphere is the natural place to discuss the M functions
obius
since the point at 1 is handled straightforwardly.
Exercise 2.6.1 Verify that if ad bc the M function can be de ned
6 obius
for 1 in a sensible manner. What if ad bc
Exercise 2.6.2 Con rm that any M function takes circles to circles.
obius
What happens when ad bc
A rather special case is when the image by a M function of a circle is
obius
a straight line. It follows that the image of the interior of the disk bounded
by the circle is a half plane.
Example 2.6.1 Find the image of the interior of the unit disk by the map
z 1
w f z
z 1
Solution
We see immediately that z 1 goes to in nity and so the bounding circle
must be sent to a straight line and the interior to a half plane.
A quick check shows that the real axis stays real and that 1 0 0
1 0 5 3 and the intersection of the real axis with the unit disk is
sent to the negative real axis. It is easy to verify that i i i i.
68 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS
The inverse can be written down at sight using the matrix representation
and is
w 1
1
z f w
w 1
which tells us that for w iv we have
1 iv 1 v2 2iv
z
1 iv 1 v2
which point lies on the unit circle. In other words the inverse takes the
imaginary axis to the unit circle so the image by f of the unit circle is the
imaginary axis. And since 0 1 we conclude that the image by f of the
interior of the unit disk has to be the half plane having negative real part.
Any M function has to be determined by its value at three points it
obius
looks at rst sight as though 4 will be required but one could scale top and
bottom by any complex number and still have the same function. This must
be true since if we have z1 w1 z2 w2 and z3 w3 we have three linear
equations in a b c d and we can put a 1 without loss of generality.
It follows that if you are given three points and their images you can deter
mine the M function which takes the three points where you now they
obius
need to go. There is a sneaky way of doing this which you will nd in the
books but the method is not actually shorter than solving the linear equa
tions in general so I shall not burden your memory with it. It is possible
however to use some intelligence in selecting the points
Example 2.6.2 Find a M function which takes the interior of the unit
obius
disk to the half plane with positive imaginary part.
Solution
We have to have the unit circle going to the real axis so we might as well
send 1 to 0. We can also send 1 to 1. Finally if we send 0 to i we have
our three points.
The 1 1 condition means that we have cz d c z 1 and 0 i
means we have az b az i while 1 0 forces a i. So a suitable
function is
i 1 z
f z
z 1
2.7. THE EXPONENTIAL FUNCTION 69
Exercise 2.6.3 Find a M function which takes the interior of the disk
obius
j z 1 2i j 3
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