proof of Möbius circle transformation theorem

proof of Möbius circle transformation theorem

Case 1:

f

⁢

(

z

)

=

a

⁢

z

+

b

.

Case 1a: The points on

|

z

-

C

|

=

R

can be written as

z

=

C

+

R

⁢

e

i

⁢

θ

. They are mapped to the points

w

=

a

⁢

C

+

b

+

a

⁢

R

⁢

e

i

⁢

θ

which all lie on the circle

|

w

-

(

a

⁢

C

+

b

)

|

=

|

a

|

⁢

R

.

Case 1b: The line

Re

⁢

(

e

i

⁢

θ

⁢

z

)

=

k

are mapped to the line

Re

⁢

(

e

i

⁢

θ

⁢

w

a

)

=

k

+

Re

⁢

(

b

a

)

.

Case 2:

f

⁢

(

z

)

=

1

z

.

Case 2a: Consider a circle

passing through

the origin. This can be written as

|

z

-

C

|

=

|

C

|

. This circle is mapped to the line

Re

⁢

(

C

⁢

w

)

=

1

2

which does not pass through the origin. To show this, write

z

=

C

+

|

C

|

⁢

e

i

⁢

θ

.

w

=

1

z

=

1

C

+

|

C

|

⁢

e

i

⁢

θ

.

Re

⁢

(

C

⁢

w

)

=

1

2

⁢

(

C

⁢

w

+

C

⁢

w

¯

)

=

1

2

⁢

(

C

C

+

|

C

|

⁢

e

i

⁢

θ

+

C

¯

C

¯

+

|

C

|

⁢

e

-

i

⁢

θ

)

=

1

2

⁢

(

C

C

+

|

C

|

⁢

e

i

⁢

θ

+

C

¯

C

¯

+

|

C

|

⁢

e

-

i

⁢

θ

⁢

e

i

⁢

θ

e

i

⁢

θ

⁢

C

/

|

C

|

C

/

|

C

|

)

=

1

2

⁢

(

C

C

+

|

C

|

⁢

e

i

⁢

θ

+

|

C

|

⁢

e

i

⁢

θ

|

C

|

⁢

e

i

⁢

θ

+

C

)

=

1

2

Case 2b: Consider the line which does not pass through the origin.
This can be written as

Re

⁢

(

a

⁢

z

)

=

1

for

a

≠

0

. Then

a

⁢

z

+

a

⁢

z

¯

=

2

which is mapped to

a

w

+

a

¯

w

¯

=

2

. This is simplified as

a

⁢

w

¯

+

a

¯

⁢

w

=

2

⁢

w

⁢

w

¯

which becomes

(

w

-

a

/

2

)

⁢

(

w

¯

-

a

¯

/

2

)

=

a

⁢

a

¯

/

4

or

|

w

-

a

2

|

=

|

a

|

2

which is a circle passing through the origin.

Case 2c: Consider a circle which does not pass through the origin. This can be written as

|

z

-

C

|

=

R

with

|

C

|

≠

R

. This circle is mapped to the circle

|

w

-

C

¯

|

C

|

2

-

R

2

|

=

R

|

|

C

|

2

-

R

2

|

which is another circle not passing through the origin. To show this, we will demonstrate that

C

¯

|

C

|

2

-

R

2

+

C

-

z

R

⁢

z

¯

z

⁢

R

|

C

|

2

-

R

2

=

1

z

Note:

|

C

-

z

R

⁢

z

¯

z

|

=

1

.

C

¯

|

C

|

2

-

R

2

+

C

-

z

R

⁢

z

¯

z

⁢

R

|

C

|

2

-

R

2

=

z

⁢

C

¯

-

z

⁢

z

¯

+

z

¯

⁢

C

z

⁢

(

|

C

|

2

-

R

2

)

=

C

⁢

C

¯

-

(

z

-

C

)

⁢

(

z

¯

-

C

¯

)

z

⁢

(

|

C

|

2

-

R

2

)

=

|

C

|

2

-

R

2

z

⁢

(

|

C

|

2

-

R

2

)

=

1

z

Case 2d: Consider a line passing through the origin. This can be written as

Re

⁢

(

e

i

⁢

θ

⁢

z

)

=

0

. This is mapped to the set

Re

⁢

(

e

i

⁢

θ

w

)

=

0

, which can be rewritten as

Re

⁢

(

e

i

⁢

θ

⁢

w

¯

)

=

0

or

Re

⁢

(

w

⁢

e

-

i

⁢

θ

)

=

0

which is another line passing through the origin.

Case 3: An arbitrary

Mobius transformation

can be written as

f

⁢

(

z

)

=

a

⁢

z

+

b

c

⁢

z

+

d

. If

c

=

0

, this falls into Case 1, so we will assume that

c

≠

0

. Let

f

1

⁢

(

z

)

=

c

⁢

z

+

d

f

2

⁢

(

z

)

=

1

z

f

3

⁢

(

z

)

=

b

⁢

c

-

a

⁢

d

c

⁢

z

+

a

c

Then

f

=

f

3

∘

f

2

∘

f

1

. By Case 1,

f

1

and

f

3

map circles to circles and by Case 2,

f

2

maps circles to circles.

Title

proof of Möbius circle transformation theorem

Canonical name

ProofOfMobiusCircleTransformationTheorem

Date of creation

2013-03-22 13:38:00

Last modified on

2013-03-22 13:38:00

Owner

brianbirgen (2180)

Last modified by

brianbirgen (2180)

Numerical id

5

Author

brianbirgen (2180)

Entry type

Proof

Classification

msc 30E20

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