3
$\begingroup$

Can anyone spot a complex number $z\in\mathbb C$ such that for $a,b\in \mathbb C$ and $f(w)=\dfrac{w-z}{w-\overline{z}}$, we have $\dfrac{f(a)-f(b)}{1-\overline{f(a)}f(b)}=\dfrac{b-a}{b-\overline{a}}$. I have been staring at this for some time now...

  • 0
    Maybe from $f(w^*) = \left(\frac{1}{f(w)}\right)^*$?2012-02-24

2 Answers 2

3

According to Maple, your equation simplifies to

$-{\frac { \left( a-b \right) \left( -a+\overline{a}-z+\overline{z} \right) }{ \left( -b+\overline{a} \right) \left( -a+\overline{z} \right) }}=0 $

So either $a=b$ or $\text{Im}\ z = - \text{Im}\ a$.

  • 0
    +1: Nice. This caters to both interpretations of the question: $a$, $b$ fixed or allowed to vary.2012-02-25
3

By setting $b=1$, and $a=0$, and doing some algebra we see that $z$ must be real and so $f(w) = 1$.

$\frac{z}{\overline{z}} - f(1) = 1 - \frac{\overline{z}}{z}f(1) $

$z^2 - z\overline{z}f(1) = z\overline{z} - \overline{z}^2 f(1)$

$z(z - \overline{z}) = \overline{z}f(1)(z - \overline{z}) $

If $z \neq \overline{z}$, then $f(1) = \frac{z}{\overline{z}}$

Thus $\frac{1-z}{1-\overline{z}} = \frac{z}{\overline{z}} = \frac{1 - z + z}{1-\overline{z} + \overline{z}} = 1$

Thus $z = \overline{z}$