How to prove this?

ray66

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Let z,w belong to all C, complex numbers. Prove that |z + w|^2 - |z- \(\displaystyle \overline{w}\) |^2 = 4Re(z)Re(w). ........ edited

The "w" is supposed to be the conjugate of w.

Thank you!
 
Let z,w belong to all C, complex numbers. Prove that |z + w|^2 - |z- ¯¯¯¯w |^2 = 4Re(z)Re(w).

The "w" is supposed to be the conjugate of w. Thank you!
If I were to do this problem, I would work with CiS [=r {cos(x) + i sin(x)}].

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.
 
Let z,w belong to all C, complex numbers. Prove that |z + w|^2 - |z- ¯¯¯¯w |^2 = 4Re(z)Re(w).
The "w" is supposed to be the conjugate of w.
It is unfortunate that this post is gabbled. I wish Ray66 would try to repost.
It is true that \(|z|^2=z\cdot\overline{\; z\;}\); \(2\Re(z)=z +\overline{\; z\;} \;~\&\;~2i\Im(z)=z -\overline{\; z\;}\)

In the PO it is not clear which it is: \(|z+w|,~|z+\overline{\; w\;}|,~\text{ or }|\overline{\; z\;}-\overline{\; w\;}|~?\)
 
Let z,w belong to all C, complex numbers. Prove that |z + w|^2 - |z- ¯¯¯¯w |^2 = 4Re(z)Re(w).

The "w" is supposed to be the conjugate of w.

Thank you!
I believe the claim is that [MATH]|z + w|^2 - |z - \overline{w}|^2 = 4Re(z)Re(w)[/MATH]. I think that because this interpretation is true, as well as because that is what you appear to be saying about the conjugate of w.

Have you tried expanding the LHS if [MATH]z = x+iy[/MATH] and [MATH]w = u+iv[/MATH]? It's fairly straightforward that way ...
 
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:
\(z\,\overline{\,w\,}+z\,w+w\overline{\,z\,}+\overline{\,z\,}\,\overline{\,w\,}\)
\(z(w+\overline{\,w\,})+\overline{\,z\,}(w+\overline{\,w\,})\)
\((z+\overline{\,z\,})(w+\overline{\,w\,})\)
\((2\Re(z))(2\Re(w))\)
\(4\Re(z)\Re(w)\)
 
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