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Can someone show me how $$\frac{1-\bar{z}}{(1-z)(1-\bar{z})}=\frac{1-\bar{z}}{2(1-Re(z))}.$$

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If you did any work at all (like, perhaps, multiplying out the bottom) it would be good to include it to convince readers you made some progress :) – rschwieb Nov 6 '12 at 17:38
Do you wish to include that $|z|=1$? If you let $z=3$, then $(1-3)(1-3)=4$, but $2(1-3)=-4$. – Joe Johnson 126 Nov 6 '12 at 17:42
up vote 0 down vote accepted

For every $z \in\mathbb{C}\setminus \{1\}$: $$\frac{1}{1-z}=\frac{1-\bar{z}}{(1-z)(1-\bar{z})}=\frac{1-\bar{z}}{1-z-\bar{z}+z\bar{z}}=\frac{1-\bar{z}}{1-2\Re{z}+|z|^2},$$ therefore $$\frac{1-\bar{z}}{(1-z)(1-\bar{z})}=\frac{1-\bar{z}}{2(1-\Re(z))}$$ if $|z|=1.$

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You should add that $z \ne 1$. – Mercy King Nov 6 '12 at 18:04
@Mercy Thanks for remark! – M. Strochyk Nov 6 '12 at 18:19

Hint: Use the fact that $$Re(z) = \frac{z + \bar{z}}{2}$$

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Let $z=x+iy$

Cancelling the numerators we get, $\{1-(x+iy)\}\{1-(x-iy)\}=2(1-x)$

or, $(1-x)^2-(iy)^2=2-2x$

or, $x^2+y^2=1\implies \mid z\mid =1$

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