Simplifying ${\left| {1 - \bar zw} \right|^2} - {\left| {z - w} \right|^2}$

Question

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 ${\left| {1 - \bar zw} \right|^2}\mathop = \limits^? 1 - 2\left\langle {z,w} \right\rangle + {\left| z \right|^2}{\left| w \right|^2}$?

Answer 1

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.

Answer 2

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.

Answer 3

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!

Answer 4

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.