Where do these Mobius transformations map the coordinate half-planes?

Question

They are

$$\frac{z-1}{z+1}, \frac{z+1}{z-1},\frac{z-i}{z+i},\frac{z+i}{z-i}.$$

All four look virtually identical, so I would like to know how to best distinguish between them.

For example, the first mapping, I notice that $1$ maps to $0$, $-1$ maps to $\infty$.

But since $1$ and $-1$ are symmetric points w.r.t. the imaginary axis, the Mobius transformations always maps symmetric points to symmetric points. So, the images, $0$ and $\infty$, being symmetric points, must mean that they are symmetric with respect to a circle. So the real axis gets mapped to some circle centered at the origin.

How do I tell how big this circle is? And since the real axis separates the upper and lower half planes, either the upper or lower plane gets mapped inside the circle, but how can I tell? I tried a sample point: I see that $i$ gets mapped to $i$, but is this telling me that the mapping is "fixing" the UHP? I doubt it, because it's mapping stuff to some open disk.

Edit: Also, with the first mapping, there doesn't seem to be a nice symmetry argument to exploit for the imaginary axis.

Thanks.

Answer 1

The first mapping, $$f : z \mapsto \frac{z - 1}{z + 1},$$ does not map the real axis to a circle centered at the origin: Indeed, we have $f(1) = 0$.

On the other hand, any imaginary number $iy$ is, by symmetry, equidistant from $-1$ and $1$, and so $$|f(iy)| = \left\vert\frac{iy - 1}{iy + 1}\right\vert = \frac{|iy - 1|}{|iy + 1|} =1,$$ that is, $f$ maps the imaginary axis to the unit circle (bijectively, if we include $f(\infty) = 1$). Using the above fact that $f(1) = 0$ (and continuity) gives that $f$ maps that right half-plane to the inside of the unit circle, and so the left half-plane to the outside.

As for the real axis, note that if $x$ is real, then so is $f(x) = \frac{x - 1}{x + 1}$ (unless $x = -1$, in which case we have $f(-1) = \infty$), so $f$ maps the real axis (including $\infty$) to itself.

The other three maps can be analyzed similarly, or we can use the fact that they are precisely the compositions of $f$ with the rotation maps $z \mapsto i^k z$, $k = 1, 2, 3$.

Answer 2

I'll do the first one for you, and you can do the rest. Consider the unit circle $z=e^{i\theta}$. Then $$\frac{z-1}{z+1}=\frac{e^{i\theta}-1}{e^{i\theta}+1}=\frac{1-e^{-i\theta}+e^{i\theta}-1}{1+e^{i\theta}+e^{-i\theta}+1}=\frac{2i\sin(\theta)}{2+2\cos(\theta)}=i\frac{\sin\theta}{1+\cos\theta}=i\tan\frac{\theta}{2}.$$ Thus, the unit circle gets mapped to the imaginary axis. The inside of the unit circle gets mapped to one side of the imaginary axis, while the outside gets mapped to the other. To tell which side gets mapped where, check $z=0$. This maps to $-1$. Thus, the inside of the unit circle gets mapped to the left side of the imaginary axis, while the outside of the nit circle gets mapped to the right. This method should work for all of your functions.

Answer 3

Every Mobius map can be broken down into a composition of scaling maps, translations and inversions. Just do a standard long division of polynomials, i.e., fo $\frac {az+b}{cz+d}$ , do along division $ (az+b)| (cz+d)$ and you will get this decomposition. Then you can apply these operations of inversion, scaling and translation, rotation in the right order to get your map. Maybe you could study more between games too!!