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Problem

Let $C$ be a circle or a line belonging to $\overline{\mathbb C}$ and let $z_2,z_3,z_4$. Two points $z$ and $z^*$ are said to be symmetric with respecto to $C$ if $\overline{(z,z_2,z_3,z_4)}=(z^*,z_2,z_3,z_4)$.

$(i)$ Prove that the previous definition doesn't depend on the chosen points $z_2,z_3,z_4 \in C$ but of $C$.

$(ii)$ Prove that for each $z \in \overline{\mathbb C}$ there is a unique symmetric point $z^*$ with respect to $C$. The function that assigns to each $z$ its correspondent $z^*$ with respect to $C$ is called symmetry with respect to $C$. Show that for each Möbius transformation $T$ which maps $\overline{\mathbb R}$ to $C$, the function $$T \circ \overline{T^{-1}}:\overline{\mathbb C} \to \overline{\mathbb C}$$ is the symmetry with respect to $C$.

My attempt

$(i)$ Let $C$ be a circle centered at $c$ of radius $R$ Using invariance of cross ratio under Möbius transformations, and using that $z_i-c=R$ for $i=2,3,4$ and $z\overline{z}=|z|^2$ we get $$\overline{(z,z_2,z_3,z_4)}=\overline{(z-c,z_2-c,z_3-c,z_4-c)}=(\overline{z}-\overline{c},\overline{z_2-c},\overline{z_3-c},\overline{z_4-c})=(\overline{z}-\overline{c},\dfrac{R^2}{z_2-c},\dfrac{R^2}{z_3-c},\dfrac{R^2}{z_4-c})=(\dfrac{R^2}{\overline{z}-\overline{c}},z_2-c,z_3-c,z_4-c)=(\dfrac{R^2}{\overline{z}-\overline{c}}+c,z_2,z_3,z_4)$$

So if $C$ is a circle, from this equation one deduces the dependence only on $C$.

$(ii)$ If $C$ is a circle, from the formula $z^*=\dfrac{R^2}{\overline{z}-\overline{c}}+c$ it follows the uniqueness of $z^*$.

I need help to show $(i)$ and uniqueness of $C$ if $C$ is a line. I also don't know what to do to show that $T \circ \overline{T^{-1}}:\overline{\mathbb C} \to \overline{\mathbb C}$ is the symmetry with respect to $C$, I would appreciate any suggestions.

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    $\begingroup$ Presumably $T\circ T^{-1}$ should be $T\circ B\circ T^{-1}$ where $B$ is complex conjugation. ($T\circ T^{-1}$ is just the identity map.) $\endgroup$ – Andreas Blass Jun 24 '14 at 2:30
  • $\begingroup$ It is indeed, thanks for the correction $\endgroup$ – user156441 Jun 24 '14 at 2:45
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Let $C$ be any circle and $R$ be the real line

[I'll use $z^*_C$ to denote the symmetric point of $z$ with respect to the $C$, and $z^* = z^*_R$ to denote the complex conjugate of $z$.]

Let $f(z) = (z,a;b,c)$ where $a,b,c$ are distinct points in $C$

Let $g(z) = (z,d;e,f)$ where $d,e,f$ are distinct points in $C$

Then $f$ is an invertible Mobius transformation that sends $(a,b,c)$ to $(1,0,\infty)$ and hence $C$ to $R$

Similarly for $g$

Let $h = g f^{-1}$

Then $h$ is an invertible Mobius transformation from $R$ to itself

Thus $h(z) = \frac{pz+q}{rz+s}$ for some $p,q,r,s \in \mathbb{R}$ because:

  $\frac{p}{r} = h(\infty) \in R$ and hence WLOG $p,r \in \mathbb{R}$

  $h(z) = 0$ for some $z \in R$ and hence $s \in \mathbb{R}$

  $\frac{q}{s} = h(0) \in R$ and hence $q \in \mathbb{R}$

Thus $h(z^*) = \frac{pz^*+q}{rz^*+s} = \frac{p^*z^*+q^*}{r^*z^*+s^*} = (\frac{pz+q}{rz+s})^* = h(z)^*$

For any $z,z^*_C$ such that $f(z^*_C) = f(z)^*$:

  $g(z^*_C) = g(f^{-1}(f(z^*_C))) = h(f(z)^*) = h(f(z))^* = g(z)^*$

Therefore the definition is independent of the three distinct points in $C$

For any Mobius transformation $T$ that maps $R$ to $C$:

  Let $U = T^{-1} f^{-1}$ [You could use $U = f T$ and it would work as well.]

  Then as before $U(z^*) = U(z)^*$

  $T(T^{-1}(z)^*) = T(T^{-1}(f^{-1}(f(z)))^*) = T(U(f(z))^*) = T(U(f(z)^*)) = f^{-1}(f(z)^*) = z^*_C$

The uniqueness of the symmetric points was already evident from the invertibility of $f$

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    $\begingroup$ By the way, this method is independent of the type of circle that $C$ is. It could be a finite circle or a line and it still works. In my opinion, the underlying structure of many such problems is the property of a Mobius transformation that maps the real line to itself. $\endgroup$ – user21820 Jun 24 '14 at 3:20
  • $\begingroup$ I like your proposed solution a lot, there is just one little thing that I don't understand, you say $\dfrac{p}{r}=h(\infty) \in R$, so $p,r \in \mathbb R$. Is this the right argument to be able to say $p,r \in \mathbb R$? For example, consider a complex number $z$, then $z\overline{z}=|z|^2 \in \mathbb R$, but this doesn't mean $z \in \mathbb R$. Maybe I didn't understand your argument. In any case, I've recently asked here how to prove that a Möbius transformation that sends $\overline{\mathbb R}$ to itself can be reduced to real coefficients, so I can use that result in this exercise. $\endgroup$ – user156441 Jun 24 '14 at 20:16
  • $\begingroup$ @user156441: I said "WLOG", and note that I said "$R$" and not "$\mathbb{R}$". If $r \ne 0$, then we can scale $(p,q,r,s)$ so that $r$ becomes $1$, and then $\frac{p}{r} \in R$ implies that now $p \in R$. If $r = 0$, we can scale $(p,q,r,s)$ so that $p$ becomes $1$, and so now $p \in R$. I did not say that $p,r \in R$ but that we could assume WLOG that it was so. If you want to, you could include this little argument to make the proof more complete. $\endgroup$ – user21820 Jun 26 '14 at 16:24

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