Math, asked by soumank3472, 11 months ago

Show that symmetry does not depend complex analysis

Answers

Answered by Anonymous
0

Answer:

Step-by-step explanation:

Let C

C

be a circle or a line belonging to C¯¯¯¯

C

¯

and let z2,z3,z4

z

2

,

z

3

,

z

4

. Two points z

z

and z∗

z

are said to be symmetric with respecto to C

C

if (z,z2,z3,z4)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=(z∗,z2,z3,z4)

(

z

,

z

2

,

z

3

,

z

4

)

¯

=

(

z

,

z

2

,

z

3

,

z

4

)

.

(i)

(

i

)

Prove that the previous definition doesn't depend on the chosen points z2,z3,z4∈C

z

2

,

z

3

,

z

4

C

but of C

C

.

(ii)

(

i

i

)

Prove that for each z∈C¯¯¯¯

z

C

¯

there is a unique symmetric point z∗

z

with respect to C

C

. The function that assigns to each z

z

its correspondent z∗

z

with respect to C

C

is called symmetry with respect to C

C

. Show that for each Möbius transformation T

T

which maps R¯¯¯¯

R

¯

to C

C

, the function

T∘T−1¯¯¯¯¯¯¯¯¯:C¯¯¯¯→C¯¯¯¯

T

T

1

¯

:

C

¯

C

¯

is the symmetry with respect to C

C

.

My attempt

(i)

(

i

)

Let C

C

be a circle centered at c

c

of radius R

R

Using invariance of cross ratio under Möbius transformations, and using that zi−c=R

z

i

c

=

R

for i=2,3,4

i

=

2

,

3

,

4

and zz¯¯¯=|z|2

z

z

¯

=

|

z

|

2

we get

(z,z2,z3,z4)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=(z−c,z2−c,z3−c,z4−c)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=(z¯¯¯−c¯¯,z2−c¯¯¯¯¯¯¯¯¯¯¯¯¯,z3−c¯¯¯¯¯¯¯¯¯¯¯¯¯,z4−c¯¯¯¯¯¯¯¯¯¯¯¯¯)=(z¯¯¯−c¯¯,R2z2−c,R2z3−c,R2z4−c)=(R2z¯¯¯−c¯¯,z2−c,z3−c,z4−c)=(R2z¯¯¯−c¯¯+c,z2,z3,z4)

(

z

,

z

2

,

z

3

,

z

4

)

¯

=

(

z

c

,

z

2

c

,

z

3

c

,

z

4

c

)

¯

=

(

z

¯

c

¯

,

z

2

c

¯

,

z

3

c

¯

,

z

4

c

¯

)

=

(

z

¯

c

¯

,

R

2

z

2

c

,

R

2

z

3

c

,

R

2

z

4

c

)

=

(

R

2

z

¯

c

¯

,

z

2

c

,

z

3

c

,

z

4

c

)

=

(

R

2

z

¯

c

¯

+

c

,

z

2

,

z

3

,

z

4

)

So if C

C

is a circle, from this equation one deduces the dependence only on C

C

.

(ii)

(

i

i

)

If C

C

is a circle, from the formula z∗=R2z¯¯¯−c¯¯+c

z

=

R

2

z

¯

c

¯

+

c

it follows the uniqueness of z∗

z

.

I need help to show (i)

(

i

)

and uniqueness of C

C

if C

C

is a line. I also don't know what to do to show that T∘T−1¯¯¯¯¯¯¯¯¯:C¯¯¯¯→C¯¯¯¯

T

T

1

¯

:

C

¯

C

¯

is the symmetry with respect to

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