Complex Numbers Binomial Expansion Proof

Question

$z, w \in \mathbb{C}$. Prove that if $\Re(z\bar{w}) = 1$, then $|z+w|^4=|z|^4+|w|^4+4(|z|^2+|w|^2+1)+2|zw|^2$ Started by using Binomial Expansion and can't get anywhere :(
I got to:
LS $= |z|^4+|w|^4+2|z|^2|w|^2+4|zw|(|z|^2+|w|^2+|zw|)$
Upto this point I haven't used the fact that $\Re(z\bar{w}) = 1$ because I'm not sure how to manipulate that fact.

Answer 1

Hint: $$|z+w|^2 = (z+w)(\bar{z}+\bar{w}) = |z|^2 +|w|^2+z\bar{w}+w\bar{z} = |z|^2+|w|^2+2\Re{(z\bar{w})}.$$