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Mobius Transformations
By Chris Park on Tuesday, February 12, 2002 - 08:41 am:

For a, b, c, d which are elements of real numbers, show that :

f(x) = (ax + b) / (cx + d) is one-one when ad-bc does not equal 0 and find f-1 in this case.

And also, what happens when ad-bc = 0.
Does f-1 exist?

Thanks

By David Loeffler on Tuesday, February 12, 2002 - 09:15 am:

Chris,

This is actually a very well known problem, as functions of this form are very important in complex analysis; more or less all their important properties still hold when you let the variables and coefficients be complex. They are known as Moebius transformations.

Anyway, as for f being one-to-one, it's quite easy to prove directly: suppose f(x) = y. Then ax+b = y(cx+d), and (cy-a)x = b-dy; so x = (b-dy)/(cy-a). This is uniquely determined; there can only be one such x. So f is one-to-one (the usual technical term is that f is injective).

There is a flaw in my argument: I assumed cy-a was not zero. If this is the case there is no x for which f(x) = y = a/c. So f-1 is undefined at one point. However, subject to the condition that ad-bc is not zero, there will be only one such point.

If ad-bc = 0 the function is simply a constant.

David

By Chris Park on Tuesday, February 12, 2002 - 12:38 pm:

I'm sorry David
Some of what you said did make sense, but I got lost sometimes, could you try and explain again, juz more step by step, it takes me a while to absorb new things, thanks~ and others can u help me too please :)

By David Loeffler on Tuesday, February 12, 2002 - 01:14 pm:

Well, let's go through this carefully. I'll take a slightly different approach this time - in my first post I answered two of your questions at once, so I'll do one at a time here.

We want to show that f(x) is a one-to-one function. This means that there can't be two different values of x which give the same value of f(x).

Let's suppose that this is not true, i.e. that there are distinct real numbers u and v such that f(u) = f(v).

We then have to have (au+b)/(cu+d) = (av+b)/(cv+d) - that's just the definition of f.

I'm going to multiply this out: it's equivalent to saying that

(au+b)(cv+d) = (av+b)(cu+d)

or, expanding all the brackets,

acuv + adu + bcv + bd = acuv + adv + bcu + bd

Lots of these terms cancel so we get

adu + bcv = adv + bcu

Rearranging this a bit, it's the same as saying that

(ad-bc)(u-v) = 0

Now, we assumed at the beginning that u and v were different. So we can only have this equal to zero if ad = bc.

Hence unless ad - bc = 0, f is a one-to-one function.

Let me know if you get this, and if so I'll post the next bit (finding the inverse).

David

By Chris Park on Wednesday, February 13, 2002 - 11:13 am:

But David, with f(u) = f(v), doesn't that mean u and v have to be equal??
And can u answer the next part as well, I sort of understand what you're doing.

By David Loeffler on Wednesday, February 13, 2002 - 01:34 pm:

Chris, that's exactly what we're trying to prove. Not all functions have the property that f(u) = f(v) can only hold if u=v. Functions having this property are called one-to-one, and we're trying to prove that f defined as you described above is one-to-one.

Just as an example, consider the function f(x) = x2. Then f(-1) = f(1) = 1; so f isn't one-to-one.

Also, consider the function f1(x) = (2x+4)/(x+2). This is of the form you describe, but has ad - bc = (2)(2) - (4)(1) = 0.

Now, you've probably noticed that f1(x) is in fact always equal to 2; so it isn't one-to-one, since f(1)=f(0) for example.

As for the next part, we've proved that if ad-bc is not zero, we can't have f(u) = f(v) for distinct u, v. Hence for any y, there can only be one x such that f(x) = y. So, given y, let's find that unique x.

We must have

f(x) = y

(ax+b)/(cx+d) = y

ax+b = y(cx+d)

ax + b = cxy + dy

ax - cxy = dy - b

(a-cy)x = dy-b

x = (dy-b)/(a-cy)

So the inverse function is the function g(y) = (dy-b)/(a-cy).

David

By Chris Park on Thursday, February 14, 2002 - 10:50 pm:

Ok
lets seee

For the first part of the question, well i understand you now, it makes sense!
But for the 2nd part, i dont...can u try and explain again, sorry :(
The question was asking what would happen when ad-bc = 0 and does f-1 exist?
is that asking if f-1 exists when ad-bc=0 or ad-bc does not equal 0? or doesnt it matter?

By Philip Ellison on Friday, February 15, 2002 - 11:51 am:

Okay, do you understand the proof for the function being one-to-one? Basically, what David is showing is that (by rearranging the equation), as (ad-bc)(u-v)=0 it means that one of the two brackets must be equal to 0 (hopefully this makes sense). When ad-bc is not equal to 0, this means that u-v must be equal to 0, i.e. u=v. This means that for each value of f(x), there can only be one value of x (as we have just shown that any possibilities would have to be equal).

For the next part of the question, for ad-bc=0, the function ceases to be one-to-one and hence you cannot find the inverse function (as, given the value of f(x) there are infinitely many possibilities for x). This can be seen by considering again the equation (ad-bc)(u-v)=0. As, when ad-bc=0, the first bracket is equal to 0, the second bracket can take any value you like. i.e. you can assign whatever values you like to u and v and still satisfy the equation. From this you should be able to see that f-1 does not exist when ad-bc=0, but does otherwise.

Once we have established that the function is one-to-one, David has shown how to rearrange the equation to find its inverse. Basically, what this is doing is making x the subject of the equation. This gives us the method needed to get from y (i.e. f(x)) to our original value of x.

I haven't introduced anything new, but hopefully I've managed to explain some of David's steps in a little bit more detail (though not nearly as eloquently :P).

Where did you get this problem from, as your teachers should be able to help you out with it if you are really having problems (it's in one of the first couple of Pure modules of the A-level course)?