Show that M(z) maps the unit circle to itself.

[raphael3d]

Homework Statement

consider the family of complex mappings:

z -> M_a(z) = (z-a)/(áz-1)

(a constant)
(á is complex conjugate of a)

Show that M_a(z) maps the unit circle to itself.

Homework Equations

the solution should look something like this i guess:
M_a(e^i*alpha) = e^i*alpha

The Attempt at a Solution

first i represented the unit circle with e^i*alpha, and inserted it into the mapping: M_a(e^i*alpha), but then i got stuck.

 

[tiny-tim]

hi raphael3d!

hint: zz' = 1

 

[raphael3d]

the length is 1 = |z| = |z|² = zz'
but how should that fact help me?

am i on the right track with
Ma(ei*alpha) = ei*alpha ?

thank you

 

[tiny-tim]

Forget coordinates.

Stick with zz' = 1 (for both the input and the result).​

 

[raphael3d]

=> M(zz')

you mean a reduction to 1-a/a'-1 ? (cant even visualize this)

or something like |M(z)| = 1

 

[Dick]

tiny-tim is suggesting that you try and show |M(z)|=1.

 

[raphael3d]

so here i am again, stuck at |z-a|/|a'z-1| = 1

=> |z-a|=|a'z-1|

i can't see where z'z comes into the game.

feel quit stupid for apparently not seeing the obvious here...

 

[tiny-tim]

rewrite …

|z-a|=|a'z-1|

 

[raphael3d]

=>|z-a|=|a'z-zz'|
=>|z-a|=|z(a'-z')|
=> stuck.

 

[tiny-tim]

square each side!

 

[raphael3d]

is there any other or more elegant way than writing for instance |z-a|²=(x1-x2)²+(y1-y2)² ?

 

[Dick]

is there any other or more elegant way than writing for instance |z-a|²=(x1-x2)²+(y1-y2)² ?

Interpret 'square each side' to mean calculate M(z)*M(z)'.

 

[raphael3d]

you mean M(z)*M(z') of i am not mistaken?

assuming M(z)=z
=> M(z)*M(z') = z*z' = |z|^2=1

 

[Dick]

you mean M(z)*M(z') of i am not mistaken?

assuming M(z)=z
=> M(z)*M(z') = z*z' = |z|^2=1

I meant what I said. If b is a complex number, bb'=|b|^2. Put b=M(z).

 

[raphael3d]

meaning M(z)*M(z)' = |M(z)|²,

but how can i evaluate |(z-a)/(a'z-1)|² ?

thank you for your patience.

 

[Dick]

meaning M(z)*M(z)' = |M(z)|²,

but how can i evaluate |(z-a)/(a'z-1)|² ?

thank you for your patience.

Multiply M(z) by its complex conjugate. Expand the numerator and denominator. What do you get?

 

[Delta2]

=>|z-a|=|a'z-zz'|
=>|z-a|=|z(a'-z')|
=> stuck.

|z-a|=|z||z'-a'|=|(z-a)'| which apparently is true because |-y|=|y|=|y'| for every y (y=z-a in this case). Also used that |z|=1 and
|z1z2|=|z1||z2| for z1=z, z2=a'-z'.

I think you also have to prove that for every y such that |y|=1 there is a z such that |z|=1 and M(z)=y.

 

[raphael3d]

but what does the complex conjugate of (z-a)/(a'z-1) looks like?

i have a mental blackout as it seems..

 

[Dick]

but what does the complex conjugate of (z-a)/(a'z-1) looks like?

i have a mental blackout as it seems..

I guess you do. The conjugate of (z-a) is (z'-a'). The conjugate of (a'z-1) is (az'-1). The conjugate of the quotient is the quotient of the conjugates. Do you see why?

 

[raphael3d]

well after multiplication of M(z)*M(z)', i get 1. i guess that's it.thank you dick.