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Given $z,w \in \mathbb C$, show that $$ |1-z \bar w|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$$

I think I need to use the equation $|z|^2 = z \bar z$

Thanks for any help.

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$$ |1-z \bar w|^2 - |z-w|^2 = (1-z\bar w)(1-\bar z w) -(z-w)(\bar z-\bar w)= \\ = 1-z\bar w -\bar z w +|z|^2|w|^2 -|z|^2+w\bar z + z \bar w -|w|^2 =\\ = 1 -|z|^2 -|w|^2 +|z|^2|w|^2 = (1-|z|^2)(1-|w|^2) $$

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