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 $\delta < \cos{\alpha}, |\alpha|\leq \Pi$