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I'm trying to show that given three distinct points $z_1,z_2,z_3\in\mathbb C$, the rational function $$ f(z) = \frac{(z-z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \frac{(z_2 - z_3)z + (z_1z_3 - z_1z_2)}{(z_2 - z_1)z + (z_1z_3 - z_2z_3)} = \frac{az + b}{cz + d} $$ is a Möbius transformation. That is, I must show that $ad - bc \neq 0$. I worked out that \begin{align*} ad - bc & = (z_1z_2z_3 - z_2^2z_3 - z_1z_3^2 + z_2z_3^2) - (z_1z_2z_3 - z_1^2z_3 - z_1z_2^2 + z_1^2z_2) \\ & = - z_2^2z_3 - z_1z_3^2 + z_2z_3^2 + z_1^2z_3 + z_1z_2^2 - z_1^2z_2 \\ & = (z_2z_3^2 - z_2^2z_3) + (z_1^2z_3 - z_1z_3^2) + (z_1z_2^2 - z_1^2z_2) \\ & = z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1), \end{align*} but I'm unsure of where to go from here to show that this expression is nonzero.

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  • $\begingroup$ The determinant of a two by two matriz is zero iff its two columns are linearly dependent. $\endgroup$ – Mariano Suárez-Álvarez Jun 28 '15 at 4:09
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To show that your final expression is non-zero, note that

\begin{align*} (z_3 - z_2)(z_3 - z_1)(z_2 - z_1) &= (z_3^2 - z_1z_3 - z_2z_3 + z_1z_2)(z_2 - z_1)\\ &= z_2z_3^2 - z_1z_2z_3 - z_2^2z_3 + z_1z_2^2 - z_1z_3^2 + z_1^2z_3 + z_1z_2z_3 - z_1^2z_2\\ &= z_2z_3^2 - z_2^2z_3 + z_1z_2^2 - z_1z_3^2 + z_1^2z_3 - z_1^2z_2\\ &= z_1^2(z_3 - z_2) + z_2^2(z_1 - z_3) + z_3^2(z_2 - z_1). \end{align*}

As $z_1, z_2, z_3$ are all distinct, the expression is non-zero.


Here is an alternative approach to the initial problem.

If $g(z) = \frac{az + b}{cz + d}$ is a Möbius transformation ($ad - bc \neq 0$), then $f(z) = kg(z) = \frac{kaz + kb}{cz + d}$ is also a Möbius transformation for $k \neq 0$; note that $(ka)d - (kb)c = k(ad - bc) \neq 0$.

The cross-ratio can be written as

$$f(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \left(\frac{z_2 - z_3}{z_2 - z_1}\right)\frac{z - z_1}{z - z_3} = kg(z)$$

where $k = \frac{z_2 - z_3}{z_2 - z_1}$ and $g(z) = \frac{z - z_1}{z - z_3}$. As $g$ is a Möbius transformation ($z_1 - z_3 \neq 0$), and $k \neq 0$, $f$ is a Möbius transformation by the above.

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We can factor $z_3$ from the expression for $d$ and $z_1$ from the expression for $c$ and then group: \begin{align} ad-bc &= (z_2 - z_3) (z_1 z_3 - z_2 z_3) - (z_2 - z_1) (z_1 z_3 - z_1 z_2) \\ &= (z_1 - z_2) (z_2 - z_3) z_3 - (z_1 - z_2) (z_2 - z_3) z_1 \\ &= (z_1 - z_2) (z_2 - z_3) (z_3 - z_1) . \end{align} Since $z_1, z_2, z_3$ are distinct, $ad - bc$ is a product of three nonzero expressions and hence nonzero.

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