Simplifying ${\left| {1 - \bar zw} \right|^2} - {\left| {z - w} \right|^2}$ I need to show that ${\left| {1 - \bar zw} \right|^2} - {\left| {z - w} \right|^2} = \left( {1 - {{\left| z \right|}^2}} \right)\left( {1 - {{\left| w \right|}^2}} \right)$ using the identity ${\left| {z - w} \right|^2} = {\left| z \right|^2} + {\left| w \right|^2} - 2\left\langle {z,w} \right\rangle$.
 Note: Here $\left\langle {\, \cdot \,,\, \cdot \,} \right\rangle$ denotes the scalar product of two complex numbers, that is, for two complex numbers $z=x+iy$ and $w=a+ib$ the scalar product is defined as $\left\langle {z,w} \right\rangle  = xa + yb$.
So far I have the following:
$\begin{gathered}
  RHS = \left( {1 - {{\left| z \right|}^2}} \right)\left( {1 - {{\left| w \right|}^2}} \right) \hfill \\
   = 1 - {\left| w \right|^2} - {\left| z \right|^2} + {\left| z \right|^2}{\left| w \right|^2} \hfill \\
   = 1 - \left( {{{\left| z \right|}^2} + {{\left| w \right|}^2}} \right) + {\left| z \right|^2}{\left| w \right|^2} \hfill \\
   = 1 - \left[ {{{\left| {z - w} \right|}^2} + 2\left\langle {z,w} \right\rangle } \right] + {\left| z \right|^2}{\left| w \right|^2} \hfill \\
   = 1 - 2\left\langle {z,w} \right\rangle  + {\left| z \right|^2}{\left| w \right|^2} - {\left| {z - w} \right|^2} \hfill \\ 
\end{gathered}$ 
 How should I continue? Should I show that ${\left| {1 - \bar zw} \right|^2}\mathop  = \limits^? 1 - 2\left\langle {z,w} \right\rangle  + {\left| z \right|^2}{\left| w \right|^2}$?
 A: I think it is easier to use the (equivalent) 
identity $|a-b|^2 = |a|^2 - 2\mathrm{Re} \bar a b + |b|^2.$
Then $$|1 - \bar z w|^2 = 1 - 2\mathrm{Re} \bar z w + |\bar z w|^2$$
and
$$|z- w|^2 = |z|^2 - 2\mathrm{Re} \bar zw + |w|^2$$ so that
$$|1 - \bar z w|^2 - |z- w|^2 = 1 - |z|^2 - |w|^2 + |\bar z w|^2 = 1 - |z|^2 - |w|^2 + |z|^2 |w|^2.$$
Now factor.
A: Use systematically that the modulus of a complex number, squared, is the product of this complex number by its conjugate. Namely,
\begin{align}
&{}\phantom{={}\;}|1 - \bar zw|^2 - |z - w |^2=(1-\bar zw)(1-z\bar w)-(z-w)(\bar z-\bar w)\\
&=(1 - \bar zw -z\bar w+ z\bar z w\bar w)-(z\bar z-z\bar w-w\bar z+w\bar w) \\
&=1 -z\bar z - w\bar w +  z\bar z w\bar w= (1 -z\bar z )(1- w\bar w )
\end{align}
by a well-known highschool identity.
A: Your way works as well. You need to show:
$${\left| {1 - \bar zw} \right|^2} =  1 - 2\left\langle {z,w} \right\rangle  + {\left| z \right|^2}{\left| w \right|^2}$$
Using the same identity (from the hypothesis):
$$\left| {1 - \bar zw} \right|^2 = 1+|\bar zw|^2-2\left\langle {1,\bar z w} \right\rangle$$
It's easy to see that $|\bar zw|^2 = {\left| z \right|^2}{\left| w \right|^2}$, so it remains only to prove:
$$\left\langle {z,w} \right\rangle = \left\langle {1,\bar z w}\right\rangle $$
This is easy to prove using the following formula for the scalar product:
$$\left\langle {z_1,z_2} \right\rangle=\frac{1}{2}(\bar z_1z_2+z_1\bar z_2)$$
I leave the rest to you.
A: I think you can just use this fact: $|x|^2 = \bar{x}x$. Then
\begin{equation}
|1-\bar{z}w|^2 = (1-\bar{z}w)(1-z\bar{w}) =  1-\bar{z}w -z\bar{w} + |z|^2|w|^2
\end{equation}
and
\begin{equation}
|z-w|^2 = (z-w)(\bar{z}-\bar{w}) = |z|^2 + |w|^2 - \bar{z}w - z \bar{w},
\end{equation}
hence
\begin{equation}
1 - |z|^2 - |w|^2 + |z|^2|w|^2 = (1-|z|^2)(1-|w|^2).
\end{equation}
Hope this will help you!
