2.6 The Mobius Transforms

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The reciprocal transformation is a special case of a general class of complex

functions called the Fractional linear or Mobius transforms. In the old days,

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 Mobius functions is:

w = f(z) =

az + b

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 Mobius function by the corresponding matrix:

az + b

cz + d _ _ a b

c d _

2.6. THE M OBIUS 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 Mobius functions we get

another Mobius function and the matrix multiplication gives the corresponding

coe_cients. This is easily veri_ed, and shows that providing ad 6= bc the

Mobius function

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 Mobius functions,

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 Mobius functions

since the point at1is handled straightforwardly.

Exercise 2.6.1 Verify that if ad 6= bc the Mobius function can be de_ned

for 1 in a sensible manner. What if ad = bc?

Exercise 2.6.2 Con_rm that any Mobius function takes circles to circles.

What happens when ad = bc?

A rather special case is when the image by a Mobius function of a circle is

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

w = f(z) =

z 􀀀1

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

z = f􀀀1w =

w + 1

􀀀w + 1

which tells us that for w = iv we have

z =

1 + iv

1 􀀀 iv

=

(1 􀀀 v2) + 2iv

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 Mobius function has to be determined by its value at three points: it

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 determine

the Mobius function which takes the three points where you now they

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 equations

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 Mobius function which takes the interior of the unit

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

f(z) =

i(1 􀀀 z)

z + 1

2.7. THE EXPONENTIAL FUNCTION 69

Exercise 2.6.3 Find a Mobius function which takes the interior of the disk

jz 􀀀 (1 + 2i)j < 3

to the half-plane with positive imaginary part.

Exercise 2.6.4 Draw the images of the rays from the origin under the function

w =

z

z 􀀀 1

= 1+

1

z 􀀀1

Exercise 2.6.5 Investigate the e_ect of composing some of the maps you

have met so far. Show that a Mobius function can be written as a suitable

composite of inversions and a_ne maps, and deduce directly that it has to

take circles to circles.

Exercise 2.6.6 Calculate the image of the lines having imaginary part constant

under the map

f(z) = (z2

􀀀1)1=2

Exercise 2.6.7 What is the Riemann surface for the above map?

The Mobius functions are of some interest because they are closed under

composition, and also for historical reasons. All books on Complex Analysis

mention them. I should have been excommunicated if I had left them out,

and I am already regarded as having heretical tendencies, so I have put them

in. You are strongly encouraged to do the above exercises so that (a) you

will be able to make an informed guess at some of the applications and (b) so

that when you meet them in the examination you will approach them with

con_dence and a clear conscience.

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