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First year uni maths - geometry

2 questions that I got a bit stuck on.

1) 3 hyperbolic lines are given. All are z is an element of H (hyperbolic plane).

a: mod(z-1) =3
b: re(z) = 3
c: mod(z-3) = 2 root(3)

Find the vertices A, B, C.

A:= b intersection c
B:= a intersection c
C:= a intersection b

Note: I was unsure of how to calculate the intersections.


2) A mobius transformation is defined by:

lambda(z) = (10z-26)/(2z-2)

Find the unique fixed point of lambda in H(hyperbolic plane).
Show where it lies on the H line.
Reply 1
For #1 the fact that they're in the hyperbolic plane is irrelevant; just find the points of intersection as if you're just working in the upper half-plane with the Euclidean metric.

For #2, set z=λ(z)z=\lambda(z) and solve for zz. You might get more than one possible answer, but one of them won't lie in H.
Original post by nuodai
For #1 the fact that they're in the hyperbolic plane is irrelevant; just find the points of intersection as if you're just working in the upper half-plane with the Euclidean metric.

For #2, set z=λ(z)z=\lambda(z) and solve for zz. You might get more than one possible answer, but one of them won't lie in H.


I'm unsure as to how to find the intersections :s-smilie:
Reply 3
Original post by forever_and_always
I'm unsure as to how to find the intersections :s-smilie:


It's probably easier to work with Cartesian coordinates. For example, to find where a and b intersect note that z1=3|z-1|=3 is a circle with centre 1+0i and radius 3, so we can write this as (x1)2+y2=9(x-1)^2+y^2=9. The line Re(z)=3\text{Re}(z)=3 is precisely the line x=3x=3, and so these intersect whenever (31)2+y2=9(3-1)^2+y^2=9, which you can solve for yy. You get a positive and negative solution for y, so it should be clear which lies in the upper half-plane and which doesn't. You do a similar thing for the other ones.
Original post by nuodai
It's probably easier to work with Cartesian coordinates. For example, to find where a and b intersect note that z1=3|z-1|=3 is a circle with centre 1+0i and radius 3, so we can write this as (x1)2+y2=9(x-1)^2+y^2=9. The line Re(z)=3\text{Re}(z)=3 is precisely the line x=3x=3, and so these intersect whenever (31)2+y2=9(3-1)^2+y^2=9, which you can solve for yy. You get a positive and negative solution for y, so it should be clear which lies in the upper half-plane and which doesn't. You do a similar thing for the other ones.


The question says that I have to find the complex numbers which represent those vertices. With the method quoted, I end up with plus or minus root 5. But that doesn't give me a complex number :s-smilie:
Reply 5
Original post by forever_and_always
The question says that I have to find the complex numbers which represent those vertices. With the method quoted, I end up with plus or minus root 5. But that doesn't give me a complex number :s-smilie:


No, but you know that x=3x=3 and y=5y=\sqrt{5}, so if z=x+iyz=x+iy then what is zz? [All I did was flip between complex number notation and (x,y)-plane notation.]

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