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I get the additional $+1$ in the RHS of this equality. Did you can prove this?

$\frac{-(|p(z)-1|^{2}-r^{2}|p(z)+1|^{2})}{4(1-r^{2})|p(z)+h|}=\frac{-|p(z)|^{2}+2(1+r^{2})Re (p(z))}{4|p(z)+h|}$

noted that $p(z)= \frac{1+w(z)}{1-w(z)}$ and $Re (p(z))>\delta$ where $\delta < \cos{\alpha}, |\alpha|\leq \Pi$

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What is $Rep(z)$? – J. M. Oct 19 '11 at 6:08
I'd hazard a guess it's the real part of $p(z)$. – Gerry Myerson Oct 19 '11 at 6:11
If $p(z)=1$ I'm getting 0 on the LHS and not on the RHS, so surely something's wrong. – Gerry Myerson Oct 19 '11 at 6:13
@GerryMyerson, noted that $p(z)= \frac{1+w(z)}{1-w(z)}$ and $Re p(z)>\delta$ where $\delta < \cos{\alpha}, |\alpha|\leq \Pi$ – DRN Oct 19 '11 at 7:25
I think I'll wait to see how many more corrections are made to the statement before I invest too much time in this one. – Gerry Myerson Oct 19 '11 at 9:46

$LHS=\frac{-((1-r^2)|p(z)-1|^2)}{4(1-r^2)|p(z)+h|}=\frac{-|p(z)-1|^2}{4|p(z)+h|} \Rightarrow |p(z)-1|^2=|p(z)|^2-2(1+r^2)Rep(z)$

Let's denote $p(z)=a+bi$ , so we may write:

$|(a-1)+bi|^2=|a+bi|^2-2(1+r^2)a \Rightarrow$

$\Rightarrow (a-1)^2+b^2=a^2+b^2-2a-2r^2a \Rightarrow$

$\Rightarrow 1=-2r^2a\Rightarrow ar^2=\frac{-1}{2}\Rightarrow Rep(z)r^2=\frac{-1}{2}$

So we may conclude that equality is true only under this last condition.

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sorry,typing error.. this is the real question $\frac{-(|p(z)-1|^{2}-r^{2}|p(z)+1|^{2})}{4(1-r^{2})|p(z)+h|}=\frac{-|p(z)|^{2}+‌​2(1+r^{2})Re (p(z))}{4|p(z)+h|}$ – DRN Oct 19 '11 at 7:34

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