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...
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Maybe from $f(w^) = \left(\frac{1}{f(w)}\right)^$? – Feb 24 '12 at 23:56
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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 $\text{Im}\ z = - \text{Im}\ a$.
Robert Israel
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+1: Nice. This caters to both interpretations of the question: $a$, $b$ fixed or allowed to vary. – Aryabhata Feb 25 '12 at 01:15
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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 $z = \overline{z}$
Aryabhata
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