Hyperbolic Geometry Watch

lalyala
Badges: 0
Rep:
?
#1
Report Thread starter 7 years ago
#1
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.)
0
reply
lalyala
Badges: 0
Rep:
?
#2
Report Thread starter 7 years ago
#2
(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?
0
reply
lalyala
Badges: 0
Rep:
?
#3
Report Thread starter 7 years ago
#3
(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?
0
reply
nuodai
Badges: 14
#4
Report 7 years ago
#4
(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}.
0
reply
lalyala
Badges: 0
Rep:
?
#5
Report Thread starter 7 years ago
#5
(Original post by nuodai)
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}.
|(w-13)/(w-5) - 6| = sqrt(13)
Am I on the right track here?
0
reply
nuodai
Badges: 14
#6
Report 7 years ago
#6
(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).
0
reply
lalyala
Badges: 0
Rep:
?
#7
Report Thread starter 7 years ago
#7
(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.
0
reply
nuodai
Badges: 14
#8
Report 7 years ago
#8
(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.
0
reply
lalyala
Badges: 0
Rep:
?
#9
Report Thread starter 7 years ago
#9
(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?
0
reply
nuodai
Badges: 14
#10
Report 7 years ago
#10
(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").
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19
  • University of East Anglia
    Could you inspire the next generation? Find out more about becoming a Primary teacher with UEA… Postgraduate
    Sat, 27 Apr '19
  • Anglia Ruskin University
    Health, Education, Medicine and Social Care; Arts, Humanities and Social Sciences; Business and Law; Science and Engineering Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (551)
37.84%
No - but I will (114)
7.83%
No - I don't want to (102)
7.01%
No - I can't vote (<18, not in UK, etc) (689)
47.32%

Watched Threads

View All