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Intro to Inversive Gerometry Shirleen Stibbe

M203 Pure Mathematics Summerschool

NB OA = |A| = , |A'| = k|A| 22 yx +

Inversion in a circle, centre 0, radius r

Can't cope - yet. So get rid of it in the meantime.

Punctured plane

  points outside points inside

  points on the circle fixed

  t: A' A , and t-1 = t (self inverse)

  what happens to the centre????

→

⇔

Example: Invert A = (3, 4) in the unit circle (r = 1)

r2 = 1 = |(3, 4)| x |(3k, 4k)| = 5 x 5k = 25k

So k = 1/25 and A' = (3/25, 4/25)

t: A A' OA.OA' = r2

→r

O A = (x, y)

A' = (X, Y) = (kx, ky),

0k ≠

2

Inversion in the unit circle C (centre 0, radius 1)

1r)yx(k yxk yxAOOA 2222222 ==+=+×+=ʹ′⋅

A = (x, y) A' = (X, Y) = (kx, ky), k > 0

To find the image of:   a circle: (x - a)2 + (y - b)2 - r = 0   a line: ax + by = 0

under inversion in the unit circle:

1) Replace x by x / (x2 + y2) y by y / (x2 + y2) in the equation

2) Simplify

No! Get a life!

Strategy: HB p62

22 yx1k+

= ⎟⎟⎠

⎞⎜⎜⎝

⎛

++= 2222 yx

y,yx

x)Y,X(and so

3

line through 0

line through 0

c = 0 d = 0

line not through 0

circle through 0

c = 0 d 0

circle through 0

line not through 0

c 0 d = 0

circle not through 0

circle not through 0

c 0 d 0

Inverts to Curve Equation

≠

≠

≠

≠

Inversion in C - the easy way

Images

  Write the equation of the curve C in the form: d(x2 + y2) + ax + by + c = 0 (*)

  C is a circle if d ≠ 0, a line if d = 0

  C passes through the origin if and only if c = 0

  To invert C in the unit circle: interchange c and d in (*)

  That is: t(C) is the curve with equation c(x2 + y2) + ax + by + d = 0

!

4

Oops - we're not doing very well

  Can't cope with the origin

  Doesn't preserve circles

  Doesn't preserve lines

  Doesn't preserve size

Some good news - at last

Angle theorem:

Inversion preserves size, reverses direction

of angles between curves - just as reflection does

5

Examples: Invert i) L, the line x = 1/2, and

ii) C, the circle centre (1, 0), radius 2

in the unit circle C

0 -1 1 2 3

L

C

C

Draw a picture

  Equation for L: 0(x2 + y2) - x + 1/2 = 0

  Equation for t(L): 1/2(x2 + y2) -x = 0 Simplifies to: (x-1)2 + y2 = 1

  Equation for C: 1(x2 + y2) -2x - 3 = 0

  Equation for t(C): -3(x2 + y2) -2x +1 = 0 Simplifies to (x+1/3)2 + y2 = 4/9

Do some algebra:

6

Even easier way:

-1 1 2 3

L

C

C

L'

C' -1/3

L: line not through 0 → circle through 0 symmetric about x axis (right angles preserved) points on the circle fixed (1/2, 0) (nearest to 0) → (2, 0) (furthest from 0) Image: circle centre (1, 0), radius 1

C: circle not through 0 → circle not through 0 symmetric about x axis (right angles preserved) (-1, 0) fixed (3, 0) (furthest from 0) → (1/3, 0) (nearest to 0) Image: circle centre (-1/3, 0), radius 2/3

7

Now for the really cunning part

Use complex numbers

⎟⎟⎠

⎞⎜⎜⎝

⎛

++→+= 2222 yx

y,yx

x)y,x(~iyxz

z1

iyx1

)iyx)(iyx(iyx

yxiyx~ 22 =

−=

+−+

=+

+

compositions of reflections

Translation: t(z) = z + c, c = a + ib

Reflection in x-axis: t(z) =

Rotation about 0: t(z) = az a = Cosθ + iSinθ

z

z/1z→Inversion in C is a doddle:

Scaling: t(z) = kz, k R, k > 0 ∈

z z+c

a b

z

z

z

az θ

Euclidean transformations in C

8

Rotate through angle θ

= reflect in line p, then reflect in line q

Translate through vector c

= reflect in line p, then reflect in line q

q

|c|

½|c|

p

z

t(z)

q

θ/2

p

z t(z)

9

Inversion in C, centre c = a +ib, radius r

Discovery! tC and t are conjugates

ccz

r)z(ttt)z(t2

122C +

−== −

ccz

rczr

rczz

2ttt 2

12 +

−⎯→⎯

−⎯→⎯

−⎯⎯→⎯−

tC:

Convert C to C: translate through -c, scale by 1/r

Invert in C:

Convert C to C: scale by r, translate through +c

z1z:t →

Note: t1 = t2-1 crzz:t2 +→

We know how to invert in the unit circle C, so …

  Abuse the plane to convert C to C

  Do the inversion in C

  Conceal the evidence; convert C back to C

,rczz:t1

−→

10

Introducing the very lovely

  extend the plane: C Ĉ

  solves (nearly) all our problems

  under inversion

  define a line as a (generalised) circle with infinite radius

  generalised circles generalised circles

  we're back in business

  inversion (nearly) back in line with reflection

A late arrival at the Inversion Ball

Generalised Circle:

Ordinary circle if it doesn't contain ∞ Extended line if it contains ∞

0 and 0 →∞∞→

→

=∞∪ }{

∞Point at Infinity

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Inversive Group

Elements: set of inversive transformations

Operations: composition of functions

Properties: a) map gen circles to gen circles

b) preserve magnitude of angles

Inversion in Ĉ

C a generalised circle in Ĉ

t = inversion in C

If C is a 'proper' circle, centre 0, radius r:

  t(0) =

  t( ) = 0

  t(z) = normal inversion,

If C is an extended line L ∪ { }:

  t( ) =

  t(z) = reflection in L,

∞

∞

∞ ∞

∞

∞≠≠ z,0z

∞≠z

12

Möbius Functions

M(z) : direct - preserves orientation of angles

composite of even number of inversions

indirect - reverses orientation of angles

composite of odd number of inversions

:)z(M

∞=−⎩⎨⎧

≠

=∞=∞ )c/d(M

0c if c/a0c if

)(M

Workingform

⎟⎠

⎞⎜⎝

⎛β−α−

=zzK)z(M

c/d ,a/b ,c/aK −=β−=α=

Example1: Express as a Möbius transformation ccz

r2+

−

dzcbza)z(M

dczbaz)z(M

++

=++

=M : C Ĉ →

czzc

cz)ccr(zc

cz)cz(cr c

czr 222

−α−

=−−+

=−−+

=+−

c/rc 2−=αwhere

a, b, c, d ∈ C, ad – bc ≠ 0

13

Example2:

i) calculate M(2), M(i) and M( )

ii) show that M-1M(z) = z

i)

2ziz)z(M

−−

=

∞

Associated matrix:

Composition of transformations = matrix multiplication

⎟⎠

⎞⎜⎝

⎛

−

−=⎟

⎠

⎞⎜⎝

⎛= −

acbd

A,dcba

A 1

ii)  Matrix is

dczbaz)z(M

++

=Möbius and Matrices

0i2bcad21i1

A ≠+−=−⎟⎠

⎞⎜⎝

⎛

−

−=

⎟⎠

⎞⎜⎝

⎛

+−

+−=⎟

⎠

⎞⎜⎝

⎛

−

−⎟⎠

⎞⎜⎝

⎛

−

−=⎟

⎠

⎞⎜⎝

⎛

−

−= −−

1200i2

21i1

11i2

AAso 11i2

A 11

1 z/21z/i1lim )(M

0)i(M ,)2(M

|z|=

−−

=∞

=∞=

∞→

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Fundamental Theorem

Given any 3 points in C, there exists a Möbius transformation that maps these to any other three given points.

Note:

Two points do not fix a generalised circle:

But three do:

∞ ∞

15

Putting Möbius to work

To determine whether four points, a, b, c, d, all lie on the same generalised circle:

1 The first 3 points (a, b and c) determine a generalised circle, C.

2 Find a Möbius transformation M(z) that takes a, b and c to the real axis

[ Easiest way: send them to 0, 1 and ∞ ]

3 The M you found maps the C, the whole C and

nothing but the C to the real axis, and M-1 maps the real axis back to the generalised circle, C.

4 Apply M to the 4th point, d.

If M(d) is real, say M(d) = x ∈ R, then M-1(x) = d lies on the generalised circle C.

5  If not, then it doesn't.

6  If M(∞) is real, then ∞ lies on C, so C is an extended line. Otherwise, C is a 'proper' circle.

16

17

Example

1 Find a Möbius transformation that maps 1 - i to 4 + 4i to 0, and 5 to 1.

∞

⎟⎠

⎞⎜⎝

⎛−−+−

=⇒∞=−

=+

−−

=)i1(z)i44(zK)z(M

)i1(M0)i44(M

,bzazK )z(M

1i4i41K

)i1(5)i44(5K)5(M =⎟

⎠⎞

⎜⎝⎛

+−

=⎟⎠

⎞⎜⎝

⎛−−+−

=

i i41)i41(i

i41i4 K So =

−−

=−+

=

i1zi44iz

)i1(z)i44(zi)z(MTherefore

+−−+

=⎟⎠

⎞⎜⎝

⎛−−+−

=

2 Do the points 1 - i, 4 + 4i, 5 and 3i all lie on the same generalised circle, and if so, what kind of circle is it?

1 - i, 4 + 4i, 5 lie on a generalised circle, C, which is mapped by M to the real axis.

1 i41

i41 i1i3

i443 )i3(M −=+−−

=+−−+−

=

which is real, so 3i must lie on C

M(∞) = K = i, which is not on the real axis, so ∞ is not on C

So C is a 'proper' circle and not an extended line.