Inversive Geometry by Shirleen Stibbe

Intro to Inversive Gerometry

Shirleen Stibbe

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M203 Pure Mathematics Summerschool

Inversion in a circle, centre 0, radius r A' = (X, Y) = (kx, ky), k ≠ 0 t: A → A' OA.OA' = r2

r O

A = (x, y)

NB OA = |A| = x 2 + y 2 ,

|A'| = k|A|

Example: Invert A = (3, 4) in the unit circle (r = 1) r2 = 1 = |(3, 4)| x |(3k, 4k)| = So k =

1/25

and A' =

5 x 5k = 25k (3/25, 4/25)

  points outside ⇔ points inside   points on the circle fixed   t: A' →A , and

t-1 = t (self inverse)

  what happens to the centre???? Can't cope - yet. So get rid of it in the meantime. Punctured plane 2

Inversion in the unit circle C (centre 0, radius 1) A = (x, y)

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

k>0

OA ⋅ OAʹ′ = x 2 + y 2 × k x 2 + y 2 = k( x 2 + y 2 ) = r 2 = 1 so

1 2

x +y

and

⎛ x y ⎞ ⎟ ( X, Y ) = ⎜⎜ 2 , 2 2 2 ⎟ x + y ⎠ ⎝ x + y

Strategy: HB p62 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! 3

Inversion in C - the easy way   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 Images Equation

Curve

Inverts to

c≠0 d≠0

circle not 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

line through 0

line through 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

Examples: Invert

i) L, the line x = 1/2, and ii) C, the circle centre (1, 0), radius 2

in the unit circle C L Draw a picture

C -1

Do some algebra: 

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

Even easier way: 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 L C L'

C -1

-1/3

Now for the really cunning part Use complex numbers

Inversion in C is a doddle: z → 1/ z

⎛ x y ⎞ ⎟ z = x + iy ~ ( x, y ) → ⎜⎜ 2 , 2 2 2 ⎟ ⎝ x + y x + y ⎠ ~

x + iy x2 + y2

x + iy 1 1 = = ( x − iy )( x + iy ) x − iy z

Euclidean transformations in C z+c

Translation:

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

Reflection in x-axis: t(z) = z

b a

z z

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

az θ z

compositions of reflections Scaling: t(z) = kz, k ∈ R, k > 0 8

Rotate through angle θ = reflect in line p, then reflect in line q

t(z)

θ/2

Translate through vector c = reflect in line p, then reflect in line q

q t(z)

|c|

½|c| 9

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

translate through -c, scale by 1/r

z−c , r 1 t:z→ z

t1 : z → Invert in C:

Convert C to C:

scale by r, translate through +c Note: t1 = t2-1

t 2 : z → rz + c tC:

t 2−1

⎯⎯ ⎯→

z−c r

t C ( z) =

⎯⎯→

t 2  t  t 2−1( z)

r z−c

⎯⎯→

r2 +c z−c

r2 = +c z−c

Discovery! tC and t are conjugates 10

A late arrival at the Inversion Ball Introducing the very lovely

Point at Infinity

∞

  extend the plane: C ∪ {∞} = Ĉ   solves (nearly) all our problems   0 → ∞ and ∞ → 0 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

Generalised Circle: Ordinary circle if it doesn't contain ∞ Extended line if it contains ∞ 11

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, z ≠ 0,

z≠∞

If C is an extended line L ∪ {∞}:   t(∞ ) = ∞   t(z) = reflection in L,

z≠∞

Inversive Group Elements:

set of inversive transformations

Operations:

composition of functions

Properties:

a) map gen circles to gen circles b) preserve magnitude of angles 12

Möbius Functions

M:C →Ĉ

M(z) :

az + b az + b M( z ) = cz + d cz + d a, b, c, d ∈ C, ad – bc ≠ 0

M( z) =

direct - preserves orientation of angles composite of even number of inversions

M( z ) :

indirect - reverses orientation of angles composite of odd number of inversions

⎧∞ if c = 0 M(∞) = ⎨ ⎩a / c if c ≠ 0

Working form

M( −d / c ) = ∞

⎛ z − α ⎞ M( z) = K ⎜ ⎟ ⎝ z − β ⎠ K = a / c, α = −b / a, β = −d / c

r2 Example1: Express + c as a Möbius transformation z−c r2 r 2 + c( z − c ) cz + (r 2 − cc ) z−α +c = = = c z−c z−c z−c z−c where α = c − r 2 / c

M( z) =

Möbius and Matrices

Associated matrix:

⎛ a b ⎞ A = ⎜ ⎟, ⎝ c d ⎠

az + b cz + d

−1

⎛ d − b ⎞ = ⎜ ⎟ ⎝ − c a ⎠

Composition of transformations = matrix multiplication

Example2:

M( z) =

z −i z−2

i) calculate M(2), M(i) and M( ∞ ) ii) show that M-1M(z) = z

M(2) = ∞,

M(i) = 0

1− i / z = 1 |z|→∞ 1 − 2 / z

M(∞ ) = lim

ii)

−1

Matrix is

⎛1 − i ⎞ A = ⎜ ⎟ ⎝1 − 2 ⎠

⎛ − 2 i ⎞ = ⎜ ⎟ so ⎝ − 1 1⎠

ad − bc = −2 + i ≠ 0

0 ⎞ ⎛ − 2 i ⎞ ⎛1 − i ⎞ ⎛ − 2 + i A A = ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ − 2 + 1⎠ ⎝ − 1 1⎠ ⎝1 − 2 ⎠ ⎝ 0 −1

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:

∞

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.

Find a Möbius transformation M(z) that takes a, b and c to the real axis [ Easiest way: send them to 0, 1 and ∞ ]

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.

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.

If not, then it doesn't.

If M(∞) is real, then ∞ lies on C, so C is an extended line. Otherwise, C is a 'proper' circle. 16

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

z−a M( z) = K , z −b

M( 4 + 4i) = 0 M(1 − i) = ∞

⎛ z − ( 4 + 4i) ⎞ ⇒ M( z) = K⎜ ⎟ ⎝ z − (1 − i) ⎠

⎛ 5 − ( 4 + 4i) ⎞ ⎛ 1 − 4i ⎞ M(5) = K⎜ = K ⎟ ⎜ ⎟ = 1 ⎝ 4 + i ⎠ ⎝ 5 − (1 − i) ⎠

So K =

4+i i (1 − 4i) = = i 1 − 4i 1 − 4i

iz + 4 − 4i ⎛ z − ( 4 + 4i) ⎞ Therefore M( z) = i ⎜ = ⎟ z − ( 1 − i ) z − 1+ i ⎝ ⎠ 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.

M(3i) =

− 3 + 4 − 4i 1 − 4i = = −1 3i − 1 + i − 1 + 4i

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. 17