Complex numbers proof question

PipkinShamrock

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The question is prove that |z+w|2-|z-wbar|2=4Re{z}Re{w}

(wbar means w with the bar over it, sorry I didn't know how to write it...)

I tried multiplying the left side out, and got |z2+2zw+w2|-|z2-2z(wbar)-(wbar)2|, but am confused about how to simplify sums of mods.
On the right side, I converted it to 2(z+zbar)(w+wbar), but that didn't get me anywhere either.

Can anyone guide me on where to go with this?
Thank you!
 
The question is prove that |z+w|2-|z-wbar|2=4Re{z}Re{w}
Let \(\displaystyle z=a+bi~\&~w=c+di\), then \(\displaystyle \mathcal(Re)(z)=a~\&~\mathcal(Re)(w)=c.\)

\(\displaystyle z+w=(a+c)+i(b+d)~\&~(z-\overline{w})=(a-c)+i(b+d)\).

Thus \(\displaystyle |z+w|^2-|(z-\overline{w})|^2=?\).
 
Let \(\displaystyle z=a+bi~\&~w=c+di\), then \(\displaystyle \mathcal(Re)(z)=a~\&~\mathcal(Re)(w)=c.\)

\(\displaystyle z+w=(a+c)+i(b+d)~\&~(z-\overline{w})=(a-c)+i(b+d)\).

Thus \(\displaystyle |z+w|^2-|(z-\overline{w})|^2=?\).

I got that to equal |(a+c)2+2i(a+c)(b+d)-(b+d)2|-|(a-c)2+2i(a-c)(b+d)-(b+d)2|
Can this be simplified further? I'm not sure what the rules are for adding/subtracting mods of things, and can't find anything useful on the internet/in my textbook.
Sorry for being a bit useless!
 
I got that to equal |(a+c)2+2i(a+c)(b+d)-(b+d)2|-|(a-c)2+2i(a-c)(b+d)-(b+d)2|
Can this be simplified further? I'm not sure what the rules are for adding/subtracting mods of things, and can't find anything useful on the internet/in my textbook.
Your trouble is that you do not know the basics of complex numbers.
\(\displaystyle |x+\mathcal{i}y|^2=x^2+y^2\) note there is no consideration of \(\displaystyle \mathcal{i}\) in the absolute value.

So \(\displaystyle |(a+c)+\mathcal{i}(b-d)|^2=(a+c)^2+(b-d)^2\)
 
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