# Using some linear algebra in a complex analysis problem

I'm trying to complete a proof in complex analysis but have realized that I have forgotten some linear algebra which I am trying to use to solve my proof.

Suppose $x_1,x_2,x_3,y_1,y_2,y_3 \in \mathbb{R}$ s.t. $x_1<x_2<x_3$ and $y_1<y_2<y_3$. I am trying to use show there is exactly one function $\phi$ of the form $\frac{az+b}{cz+d}$ with $\phi(x_j)=y_j$ and $ad-bc=1$. So here is where my linear algebra forgefulness comes in. I have four equations and four unknowns so I think I should be able to find a unique solution. However, while I can make the three involving $\phi$ linear, the condition $ad-bc=1$ is not. I think this effects things somehow but can't remember exactly how and why. Also, I am posting this as well with the complex analysis tag since there may be an approach I am missing to this problem. Note, the book which has this statement gives a hint that if for each $j$, $x_j=y_j$ then $\phi$ is the identity.

Thanks for the help

• $\frac{az+b}{cz+d}$ with $ad-bc \neq 0$ gives a Möbius transform, multiplying the coefficients with a constant gives the same map, so you can without loss of generality assume $ad-bc = 1$. You can map any three points to $0,1,\infty$ via a (unique) Möbius transform. Looking at the cross-ratio may be helpful. – Daniel Fischer Nov 17 '13 at 22:35

Let $z_1,z_2,z_3$ be three distinct points in the sphere. There is a unique (see below) Möbius transformation that maps these three points to $0,\,1,\,\infty$ (in that order), this Möbius transformation is the cross ratio $\operatorname{CR}(z,z_1,z_2,z_3)$.

First assume none of the points is $\infty$. Then

$$Tz = \frac{z-z_1}{z-z_3} : \frac{z_2-z_1}{z_2-z_3}\tag{1}$$

is a Möbius transformation that maps $z_1 \mapsto 0$, $z_2 \mapsto 1$, and $z_3 \mapsto \infty$, as can be seen by inserting $z_i$, so $Tz = \operatorname{CR}(z,z_1,z_2,z_3)$. If one of the $z_i$ is $\infty$, then the corresponding parts of the fractions in $(1)$ are removed, for $z_i = \infty$, $i = 1,\,2,\,3$ that yields in order

$$\frac{z_2-z_3}{z-z_3},\quad \frac{z-z_1}{z-z_3},\quad \frac{z-z_1}{z_2-z_1}.$$

Given two triples, $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$, of distinct points, there is a unique Möbius transformation that maps $x_i \mapsto y_i$, $i = 1,\,2,\,3$, namely

$$\operatorname{CR}(\,\cdot\,,y_1,y_2,y_3)^{-1} \circ \operatorname{CR}(\,\cdot\,,x_1,x_2,x_3).$$

The uniqueness remains to be shown: let $S\colon z \mapsto \frac{az+b}{cz+d}$ be a Möbius transformation. $\infty$ is a fixed point of $S$ if and only $c = 0$. A finite $w \in \mathbb{C}$ is a fixed point of $S$ if and only if

\begin{align} &&\frac{aw+b}{cw+d} &= w\\ \iff&& aw+b &= cw^2 + dw\\ \iff&& 0 &= cw^2 - (a-d)w - b.\tag{2} \end{align}

If $\infty$ is not a fixed point of $S$, this is a quadratic equation, and hence $S$ has at most two fixed points. If $\infty$ is a fixed point, $(2)$ is either a linear equation with exactly one solution ($a \neq d$), an impossible equation with no solution ($a = d$ and $b\neq 0$), or the tautological $0 = 0$, when $a = d$ and $b = 0$, then $S$ is the identity.

So, a Möbius transformation other than the identity has at most two fixed points in the sphere, whence the uniqueness follows (if $S$ and $T$ are Möbius transformations mapping the three points as desired, $S^{-1}\circ T$ has (at least) three fixed points).

Multiplying the coefficients of a Möbius transform $\frac{az+b}{cz+d}$ by a common (nonzero) factor $\lambda$ does not change the map, but the determinant by a factor of $\lambda^2$, so every Möbius transform has a unique representation

$$Tz = \frac{az+b}{cz+d}$$

with $ad-bc = 1$.

We want to map $(x_1,x_2,x_3)\to (y_1,y_2,y_3)$ uniquely.

A Moebius Transformation that is not the identity has at most two fixed points(if it has then is the identity,this is another proposition-ask if you need the proof).

Now let $f(z)=\frac {az+b}{cz+d}$ be such a transformation.

If $c\neq 0$ then the equation $f(z)=z$ is equivalent with this:$az+b=cz^2+dc$, which has at most two solutions.Also we have that $f(\infty)=\frac {a}{c}\neq \infty$.

If $c=0$ then $f(z)=\frac {a}{d}z+\frac {b}{d}$.

If $\frac {a}{d}\neq 1$ then $f$ has exactly one fixed point and because $f(\infty)=\infty$ then $\infty$ is a second fixed point.

If $\frac {a}{d}=1$ then $\infty$ is the only fixed point of $f$, unless $b=0$ that means $f$ is the identity.

• Maybe I am missing something obvious but why does what you have shown guarantee the existence of the map or the uniqueness. Your argument focuses on fixed points but how does that relate to where our three points are sent. – Leo Spencer Nov 17 '13 at 23:41
• You all the possible occasions of the relations of $a,b,c,d$ – Haha Nov 18 '13 at 0:08