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For $z\in { C }, \Re (z)\neq 2$ we have $F(z)=\frac { 4-z\overline { z } }{ 4-z-\overline { z } }$. I'm trying to prove the equality between the modulus of these numbers without using the exponential form, so how can I prove that?

$$|F(z)-z|=|F(z)-2|$$

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Hint: $|w|^2=w\bar w$ for a complex number $w$. –  Berci Feb 12 '13 at 22:34
    
I don't think would help. –  pourjour Feb 12 '13 at 22:35
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Nothing else matters. –  Berci Feb 12 '13 at 22:41

2 Answers 2

up vote 1 down vote accepted

By a simple calculation, we have: $$ |F(z) - z| = \left|\frac{(z-2)^2}{\overline{z}+z-4}\right| $$

And: $$ |F(z) - 2| = \left|\frac{(z-2) \left(\overline{z}-2\right)}{\overline{z}+z-4}\right| $$

But: $$ |\overline{z} - 2| = |\overline{z-2}| = |z-2| $$

Plug in and the desired result will immediately follow.

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$$|z|^2 = z\bar{z}$$

$$\begin{align*} |F(z)-z|^2 &= (F(z)-z)(\bar{F(z)}-\bar{z})\\ &= |F(z)|^2-z\bar{F(z)}-\bar{z}F(z)+|z|^2 \end{align*}$$

$$\begin{align*} |F(z)-2|^2 &= (F(z)-2)(\bar{F(z)}-2)\\ &= |F(z)|^2-2\bar{F(z)}-2F(z)+|2|^2 \end{align*}$$

Set them equal, the $|F(z)|^2$ terms cancel, from here it should be simple algebra.

Also, note that $\Re(z) = \frac{z+\bar{z}}{2}$.

This means $2(F(z)+\bar{F(z)}) = 4\Re(F(z))$.

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