# Thread: Transformation of D^+ to P^+

1. ## Transformation of D^+ to P^+

I want to take the upper half of the unit disc, $D^+$ and send it to the upper half of the plane $P^+$. My instructor hinted that if I can specify three distinct points that go to 1, 0, and the point at infinity under the transformation that I can use the cross-ratio to determine a bilinear transformation that will "almost work".

I'm totally lost in this course right now, so I wasn't even sure what the hint meant, but I tried *something* (even if ignorantly).

Looking at the $D^+$, I was hoping that if I sent $i \mapsto \infty$ and left 1 and 0 fixed, that the boundary would "unfurl" to the real axis and the interior of $D^+$ would map to $P^+$ as a consequence. In terms of the cross-ratio, this translates to $z \mapsto \frac{\frac{w-0}{w-i}}{\frac{1-0}{1-i}} = \frac{-iw}{w-i}$.

But I don't even know how to tell if this works. And what might my instructor mean by "almost work"? Thanks ahead of time for any advice!!

2. Originally Posted by cribby
I want to take the upper half of the unit disc, $D^+$ and send it to the upper half of the plane $P^+$. My instructor hinted that if I can specify three distinct points that go to 1, 0, and the point at infinity under the transformation that I can use the cross-ratio to determine a bilinear transformation that will "almost work".

I'm totally lost in this course right now, so I wasn't even sure what the hint meant, but I tried *something* (even if ignorantly).

Looking at the $D^+$, I was hoping that if I sent $i \mapsto \infty$ and left 1 and 0 fixed, that the boundary would "unfurl" to the real axis and the interior of $D^+$ would map to $P^+$ as a consequence. In terms of the cross-ratio, this translates to $z \mapsto \frac{\frac{w-0}{w-i}}{\frac{1-0}{1-i}} = \frac{-iw}{w-i}$.

But I don't even know how to tell if this works. And what might my instructor mean by "almost work"? Thanks ahead of time for any advice!!
The transformation $w = \frac{1+z}{1-z}$ takes the unit disc to the right-hand half-plane. It takes the diameter [–1,1] to the positive real axis, and from that it follows easily that it takes $D^+$ to the positive quadrant. The transformation $w\mapsto w^2$ takes the positive quadrant to $P^+$. So putting the two transformations together, you should find that the map $z\mapsto \frac{(1+z)^2}{(1-z)^2}$ takes $D^+$ to $P^+$.

3. Originally Posted by Opalg
The transformation $w = \frac{1+z}{1-z}$ takes the unit disc to the right-hand half-plane. It takes the diameter [–1,1] to the positive real axis, and from that it follows easily that it takes $D^+$ to the positive quadrant. The transformation $w\mapsto w^2$ takes the positive quadrant to $P^+$. So putting the two transformations together, you should find that the map $z\mapsto \frac{(1+z)^2}{(1-z)^2}$ takes $D^+$ to $P^+$.
That's awesome, and I thank you very much. I still think it would be very informative for me to understand how to get this map "from scratch", you know? Although my original attempt was wrong, could the approach be salvaged somewhat?

4. Originally Posted by cribby
That's awesome, and I thank you very much. I still think it would be very informative for me to understand how to get this map "from scratch", you know?
I don't know that there is a clear-cut way to find maps of this sort, except by breaking the transformation up into simple steps and then forming their composition.

One thing to remember is that bilinear transformations preserve angles. The set $D^+$ has a couple of right-angles corners in its boundary, whereas $P^+$ does not. So you can be sure that a bilinear transformation will not be sufficient on its own to transform $D^+$ into $P^+$. The best you can do with a bilinear transformation is to take the ends of the diameter of the unit circle to the points 0 and $\infty$. Then the operation of squaring has the effect of doubling angles, so it will convert the right angle at 0 into a straight line.

That may sound a bit vague, but it is the way that I used to find the map from $D^+$ to $P^+$.