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

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Maybe from $f(w^*) = \left(\frac{1}{f(w)}\right)^*$? –  user13838 Feb 24 '12 at 23:56

2 Answers 2

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 $\text{Im}\ z = - \text{Im}\ a$.

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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 1:15
    
Thank you, Robert! –  Aida Feb 25 '12 at 1:34

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 $z = \overline{z}$

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Thank you, Aryabhata! –  Aida Feb 25 '12 at 1:35

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