Let $z, z^*, z_1, z_2, z_3 \in \mathbb{C}$ be complex numbers with $z \neq 0$ and $|z_1| = |z_2| = |z_3| = K$. Now I want to show that:

$$ \frac{(z^*-z_2)(z_1 - z_3)}{(z^*-z_3)(z_1 - z_2)} = \overline{\frac{(z-z_2)(z_1 - z_3)}{(z-z_3)(z_1 - z_2)}} $$

(the conjugation on the right side spans is meant for the numerator and denominator) already implies $z^* = \frac{K^2}{\bar z}$. I tried expanding all the factors, but this doesn't seem to help. Maybe I'm missing something obvious, but I can't find the right approach.


$$\overline{\frac{(z-z_2)(z_1 - z_3)}{(z-z_3)(z_1 - z_2)}} =\frac{(\overline z-\overline z_2)(\overline z_1 - \overline z_3)}{(\overline z-\overline z_3)(\overline z_1 - \overline z_2)} =\frac{(\overline z-K^2/z_2)(K^2/z_1 - K^2/z_3)}{(\overline z-K^2/z_3)(K^2/ z_1 - K^2/z_2)} =\frac{(K^2/\overline z-z_2)(z_1 -z_3)}{(K^2/\overline z-z_3)(z_1 -z_2)}$$ which implies $z^*=K^2/\overline z$.

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  • $\begingroup$ Thank you really much! $\endgroup$ – user7802048 May 10 '18 at 18:46
  • $\begingroup$ Can you explain the last equality (not the implication) a little further? $\endgroup$ – user7802048 May 10 '18 at 20:04
  • $\begingroup$ I multiplied top and bottom by $z_1z_2z_3/(K^2\overline z)$. $\endgroup$ – Angina Seng May 10 '18 at 20:20

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