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Hyperbolic Geometry

I am having a bit of a problem with the question below. I have done question (i) and found the answer to be 3+2i. I am having a very hard time trying to understand what is required of question (ii). Any help would be greatly appreciated.

Define a Mobius transformation, M, of the hyperbolic upper half plane by
M(z) = (5z-13)/(z-1)

(i) Find the unique fixed point, P, of M in the hyperbolic plane and show that it lies on the h-line
'n' given by mod(z-6)=sqrt13
(ii) Determine the image, M(n), of the h-line n.
(iii) Find the angle at P between the positive ray n+ and its image, M^(n+). (Positive
here means emanating from P in the direction of increasing real part, x.)

Reply 1

lalyala

Original post by nuodai

n

is given by the set of points

\left \{ z \in H\, :\, \left| z-6 \right| = \sqrt{13} \right \}

, and the transformation

M

sends

z \mapsto M(z)

. Let

w=M(z)

and find

z

in terms of

w

. Then you can substitute into the expression

\left| z-6 \right| = \sqrt{13}

to find the image.

Does that mean I find the inverse of the Mobius transformation? I got z=(w-13)/(w-5)
Where do I substitute the expression?

Reply 2

lalyala

Original post by nuodai

Sorry I just realised I made a mistake. Woops. You don't need to find

z

in terms of

w

.

The transformation maps

z \mapsto w=M(z)

, and so the line

|z-6|=\sqrt{13}

maps to the line

|w-6|=\sqrt{13}

. You just need to find this in terms of

z

Sorry, now I'm a bit confused. How do I do that exactly?

Reply 3

nuodai

Original post by lalyala

Sorry, now I'm a bit confused. How do I do that exactly?

I don't know what's wrong with me today... forget everything I said in my last two posts -- I was right the first time, but I'll go from the top.

A Möbius transformation is just a function which maps complex numbers to complex number. Here your Möbius transformation is

M

. The image of a point

z

is the point

M(z)

, and the image of a set of points is the set of images of all the points. That is,

M(S) = \{ M(z)\, :\, z \in S \}

. Here, your set is

n

, which is the hyperbolic line given by the set of complex numbers

z

for which

|z-6|=\sqrt{13}

. This means that

M(n) = \{ M(z)\, :\, |z-6| = \sqrt{13} \}

.

Now, if you write

w=M(z)

then

z=M^{-1}(w)

, and so substituting this into the above we get

M(n) = \{ M(M^{-1}(w))\, :\, |M^{-1}(w) - 6| = \sqrt{13} \}

= \{ w\, :\, |M^{-1}(w) - 6| = \sqrt{13} \}

So you need to substitute

z=M^{-1}(w)

into

|z-6|=\sqrt{13}

Reply 4

lalyala

Original post by nuodai

M

. The image of a point

z

is the point

M(z)

, and the image of a set of points is the set of images of all the points. That is,

M(S) = \{ M(z)\, :\, z \in S \}

. Here, your set is

n

, which is the hyperbolic line given by the set of complex numbers

z

for which

|z-6|=\sqrt{13}

. This means that

M(n) = \{ M(z)\, :\, |z-6| = \sqrt{13} \}

.

Now, if you write

w=M(z)

then

z=M^{-1}(w)

, and so substituting this into the above we get

M(n) = \{ M(M^{-1}(w))\, :\, |M^{-1}(w) - 6| = \sqrt{13} \}

= \{ w\, :\, |M^{-1}(w) - 6| = \sqrt{13} \}

So you need to substitute

z=M^{-1}(w)

into

|z-6|=\sqrt{13}

|(w-13)/(w-5) - 6| = sqrt(13)
Am I on the right track here?

Reply 5

nuodai

Original post by lalyala

|(w-13)/(w-5) - 6| = sqrt(13)
Am I on the right track here?

Yup. Now you can rearrange to simplify the expression (and interpret it geometrically if necessary... which would make sense if this is a geometry question).

Reply 6

lalyala

Original post by nuodai

Yup. Now you can rearrange to simplify the expression (and interpret it geometrically if necessary... which would make sense if this is a geometry question).

Am I correct to substitute w=x+iy and rearrange until its a complex number? Or would that just complicate things further? Sorry need a bit more guidance. Really appreciate your help.

Reply 7

nuodai

Original post by lalyala

Am I correct to substitute w=x+iy and rearrange until its a complex number? Or would that just complicate things further? Sorry need a bit more guidance. Really appreciate your help.

It really depends what's easiest for you. You can rearrange to put it in the form

\left| w-\alpha \right| = c|w - \beta|

, which defines either a line or a circle depending on the values of

\alpha, \beta, c

... but if you've not come across this form before, then write

w=x+iy

and find a Cartesian equation for the new hyperbolic line (i.e. line or circle). You can then convert it back to complex numbers using the fact that a circle with centre

a \in \mathbb{C}

and radius

r>0

is given by

|z-a| = r

Reply 8

lalyala

Original post by nuodai

It really depends what's easiest for you. You can rearrange to put it in the form

\left| w-\alpha \right| = c|w - \beta|

, which defines either a line or a circle depending on the values of

\alpha, \beta, c

... but if you've not come across this form before, then write

w=x+iy

and find a Cartesian equation for the new hyperbolic line (i.e. line or circle). You can then convert it back to complex numbers using the fact that a circle with centre

a \in \mathbb{C}

and radius

r>0

is given by

|z-a| = r

Okay, I think I'll do the latter as it seems like something the syllabus covered in comparison to the first option.
Is the answer supposed to be in the form of |z-a|=r or x+iy or... something else?

Reply 9

nuodai

Original post by lalyala

Okay, I think I'll do the latter as it seems like something the syllabus covered in comparison to the first option.
Is the answer supposed to be in the form of |z-a|=r or x+iy or... something else?

I don't suppose it matters what form you leave it in. Since this is a geometry course, I'd recommend describing it geometrically (i.e. "the hyperbolic line corresponding to the Euclidean circle with centre a and radius r").