2
$\begingroup$

If $z_1$ and $z_2$ are distinct fixed points of a bilinear transformation $w=T(z)$ show that the transformation may be expressed as

$$\frac{w-z_1}{w-z_2}=K\frac{z-z_1}{z-z_2},$$

where $K$ is a complex constant.

I'm trying to use this fact to solve the problem.

enter image description here

enter image description here

So we have $w_2=z_2, w_1=z_1$, in this case, however, plugging these values in, I cannot get the desired expression, so I'm stuck here. How can I get this expression? I would greatly appreciate any help.

$\endgroup$
0
$\begingroup$

If the Möbiustransformation $T$ maps $z_1, z_2, z_3 \in \Bbb C$ to $w_1, w_2, w_3 \in \Bbb C$ then $$ \frac{(T(z)-w_1)(w_3 - w_2)}{(T(z)-w_2)(w_3-w_1)} = \frac{(z-z_1)(z_3 - z_2)}{(z-z_2)(z_3-z_1)} $$ (Note that you can arrange the three points – together with their image points – in any convenient order.)

If $z_1$ and $z_2$ are (finite) fixed points of $T$ then choose any $z_3 \in \Bbb C$ such that $w_3 = T(z_3) \in \Bbb C$, and you obtain $$ \frac{(T(z)-z_1)(w_3 - z_2)}{(T(z)-z_2)(w_3-z_1)} = \frac{(z-z_1)(z_3 - z_2)}{(z-z_2)(z_3-z_1)} $$ which is the desired representation.

Another approach: Define $$ S(z) = \frac{z-z_1}{z-z_2} \, . $$ Then $S \circ T \circ S^{-1}$ has the fixed points $0$ and $\infty$. Setting $K := S \circ T \circ S^{-1}(1)$ it follows that $$ S \circ T \circ S^{-1}(w) = Kw $$ because the left-hand side and the right-hand side are Möbiustransformations mapping $0, 1, \infty$ to $0, K, \infty$. Therefore $$ S(T(z)) = K S(z) $$ which is again the desired representation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.