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Show that, for $z,w \in \mathbb{C}$

$$\left|1 - \bar{z}w\right|^2 - \left|z-w\right|^2 = \left(1-\vert z\vert^2\right)\left(1-\vert w\vert^2\right)$$

Please demonstrate this step by step so I can pin point where I have gone wrong.

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    It would be more fruitful if you post your work and we point out where things went south. Use $|z|^2=z\bar z$ and factorize. – 2017-02-15

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Recall the following identities of complex numbers: $$|z|^2=z\overline{z},\quad \overline{zw}=\overline{z}\cdot\overline{w},\quad |zw|=|z||w|,\quad |\overline{z}|=|z|.$$ Thus $$\begin{aligned}|1-\overline{z}w|^2&=(1-\overline{z}w)(\overline{1-\overline{z}w})=(1-\overline{z}w)(1-\overline{\overline{z}w})\\ &=1-\overline{z}w-\overline{\overline{z}w}+|\overline{z}w|^2=1-\overline{z}w-\overline{\overline{z}w}+|z|^2|w|^2. \end{aligned}$$ $$|z-w|^2=(z-w)(\overline{z}-\overline{w})=z\overline|z|+w\overline|w|-w\overline{z}-z\overline{w}=|z|^2+|w|^2-\overline{z\overline{w}}-z\overline{w}.$$ Therefore $$|1-\overline{z}w|^2-|z-w|^2=1-|z|^2-|w|^2+|z|^2|w|^2=(1-|z|^2)(1-|w|^2).$$

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    I prefer to arrive at expression $(1-z\bar z)(1-w\bar w)$ and then transform to norms than interposing it in the middle. – 2017-02-15
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    @zwim Ohh thank you very much! It's been baffling me for hours! Listing the identities was very helpful too as I was struggling on which ones to incorporate. – 2017-02-15