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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 $Rep(z)$?2011-10-19
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    I'd hazard a guess it's the real part of $p(z)$.2011-10-19
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    If $p(z)=1$ I'm getting 0 on the LHS and not on the RHS, so surely something's wrong.2011-10-19
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    @GerryMyerson, noted that $p(z)= \frac{1+w(z)}{1-w(z)}$ and $Re p(z)>\delta$ where $\delta < \cos{\alpha}, |\alpha|\leq \Pi$2011-10-19
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    @J.M.: Real part of $p(z)$2011-10-19
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    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.2011-10-19

1 Answers 1

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

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|}$2011-10-19