Let z,w belong to all C, complex numbers. Prove that |z + w|^2 - |z- \(\displaystyle \overline{w}\) |^2 = 4Re(z)Re(w). ........ edited
\(~I~|z+w|^2=(z+w)(\overline{\;z\;}+\overline{\;w\;})=|z|^2+z\overline{\;w\;}+w\overline{\,z\,}+|w|^2\)
\(II~|z-\overline{\,w\,}|^2=(z-\overline{\,w\,})(\overline{\,z\,}-w)=|z^2|-z\,w-\overline{\,w\,}~\overline{\,z\,}+|w|^2\)
Now subtract I-II to get:
\(z\,\overline{\,w\,}+z\,w+w\overline{\,z\,}+\overline{\,z\,}\,\overline{\,w\,}\)
\(z(w+\overline{\,w\,})+\overline{\,z\,}(w+\overline{\,w\,})\)
\((z+\overline{\,z\,})(w+\overline{\,w\,})\)
\((2\Re(z))(2\Re(w))\)
\(4\Re(z)\Re(w)\)