TITLE: Suppose $z,w\in\Bbb C$. Prove that if $\Re(z\bar w)=1$, then $|z+w|^4=|z|^4+|w|^4+2|z|^2|w|^2+4|z|^2+4|w|^2+4$
QUESTION [2 upvotes]: Suppose $z,w\in\Bbb C$. Prove that if $\Re(z\bar w)=1$, then $$|z+w|^4=|z|^4+|w|^4+2|z|^2|w|^2+4|z|^2+4|w|^2+4.$$
Let $z=a+bi$,   $w=x+yi$   and  I find $ax+by=1$.
But I don't know know to factorize $|z+w|^4$. Should I combine $z$ and $w$ or change them into $(a+bi)$ and $(x+yi)$?
Picture of the problem statement

REPLY [2 votes]: Hint. Use the fact that 
$$|z+w|^2=(z+w)\overline{(z +w)}=(z+w){(\bar z +\bar w)}=|z|^2+|w|^2+ z\bar w +\bar z w \\=|z|^2+|w|^2+ 2\mbox{Re}(z\bar w)=
|z|^2+|w|^2+ 2.$$
P.S. Note that $\overline{z\bar w}=\bar z w$, so if $z\bar w=x+iy$ then
$$z\bar w +\bar z w=(x+iy)+(x-iy)=2x=2\mbox{Re}(z\bar w).$$

REPLY [1 votes]: HINT: use the fact that $|z+w|^2=(z+w)\overline{(z+w)}$.
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