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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.

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  • $\begingroup$ The first mapping does not map the real line to a circle centered at the origin: The image of $1$ under that map is the origin itself. $\endgroup$ Jul 1 '15 at 2:47
  • $\begingroup$ It does, however, map the imaginary axis to the unit circle, as each imaginary $z$ is equidistant from $1$ and $-1$, and hence $\left\vert\frac{z - 1}{z + 1}\right\vert = 1$. $\endgroup$ Jul 1 '15 at 2:51
  • $\begingroup$ Hmm...can I think of it this way, @Travis? I realize my mistake, I think. So, since 1 and -1 are symmetric w.r.t. the imaginary axis, and their images must also be symmetric w.r.t. the image of the imaginary axis (not the real axis), which I don't know yet. But 1 and -1 map to 0 and infinity, which are symmetric w.r.t. the image, which implies that the imaginary axis must get mapped to a circle. Is this valid? If so, how do I know how big the circle is? Your equidistant argument I have never heard of...thanks, $\endgroup$
    – User001
    Jul 1 '15 at 3:00
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    $\begingroup$ I've expanded my comments into a proper answer below, and I think it answers all these questions, but feel free to comment there if not. $\endgroup$ Jul 1 '15 at 3:16
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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 $z \mapsto i^k z$, $k = 1, 2, 3$.

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  • $\begingroup$ Hi @Travis, if this mapping fixes the real line, must it also fix the upper and lower half planes, too? I'm thinking of the connectedness argument - a continuous function maps connected sets to connected sets. Or, does the fact that f is not continuous at -1 make this connectedness argument invalid? Also, your getting the norm = 1, because of iy being equidistant from -1 and 1, but isn't iy really equidistant from 1+iy and -1+iy? I might not be understanding the meaning of equidistant, but I feel like we should just be extending iy out in a straight line, along the real axis.... $\endgroup$
    – User001
    Jul 1 '15 at 3:26
  • $\begingroup$ actually, I think got the equidistant argument, if I just use the modulus definition. and I'm pretty sure the connectedness argument is not valid. but feel free to comment, if you have something more to say. Thanks so much for your help, @Travis. $\endgroup$
    – User001
    Jul 1 '15 at 3:35
  • $\begingroup$ Like you say, it fixes the upper and lower half-planes (as sets, not pointwise). Also, we can think of Mobius transformations as continuous maps from the Riemann sphere, $\Bbb C \cup \{\infty\}$ to itself, so there's no issue with continuity here. $\endgroup$ Jul 1 '15 at 3:37
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    $\begingroup$ Sure, for any region $A$ we can ask what the region $f(A)$ is, and it happens that our particular map $f$ maps deveral familiar regions to other familiar regions. Note that this lets us conclude, e.g., that the first quadrant is mapped the the upper half of the unit disk. $\endgroup$ Jul 1 '15 at 3:55
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    $\begingroup$ You're welcome, I'm glad you found it useful, and you too. $\endgroup$ Jul 1 '15 at 4:57
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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.

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  • $\begingroup$ Hi @AlexS, nice argument - and based on Travis's comment above, then this mapping also maps the imaginary axis back to the unit circle. So, it is its own inverse, sort of? And, how do we know the imaginary axis gets mapped to a unit circle? Thanks, $\endgroup$
    – User001
    Jul 1 '15 at 3:09
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    $\begingroup$ It is not quite its own inverse: $$f(f(z))=\frac{\frac{z-1}{z+1}-1}{\frac{z-1}{z+1}+1}=\frac{-2}{2z}=-1/z.$$ But, if $z$ is on the unit circle, so is $-1/z$. To show directly that $f$ takes the imaginary axis to the unit circle, multiply $f(iy)$ by its complex conjugate, and check that the result is 1. $\endgroup$
    – Alex S
    Jul 1 '15 at 3:17
  • $\begingroup$ ah, nice suggestion about multiplication by complex conjugates to check the norm -- thanks so much @AlexS. $\endgroup$
    – User001
    Jul 1 '15 at 4:17
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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!!

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  • $\begingroup$ ok, got it. Thanks @Gary. I will try to, but with Kyrie and Kevin injured, and me playing 48minutes per game, it's been hard to keep up with complex analysis :-( $\endgroup$
    – User001
    Jul 1 '15 at 4:21

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