Answer to Question #282430 in Complex Analysis for iqa

Question #282430

Show that | z + w |^2 − | z − w |^2 = 4Re (zŵ).


1
Expert's answer
2021-12-27T18:19:56-0500

Let z=a+bi,w=c+di,a,b,c,dR.z=a+bi, w=c+di, a,b,c,d\in\R.

Then


z+w2=(a+c)2+(b+d)2|z+w|^2=(a+c)^2+(b+d)^2

zw2=(ac)2+(bd)2|z-w|^2=(a-c)^2+(b-d)^2

z+w2zw2=(a+c)2+(b+d)2|z+w|^2-|z-w|^2=(a+c)^2+(b+d)^2

((ac)2+(bd)2)=4ac+4bd-((a-c)^2+(b-d)^2)=4ac+4bd

=4(ac+bd)=4(ac+bd)

zw^=(a+bi)(cdi)=ac+bd+(ad+bc)izŵ=(a+bi)(c-di)=ac+bd+(-ad+bc)i

Re(zw^)=ac+bdRe(zŵ)=ac+bd

Hence


z+w2zw2=4(ac+bd)=4Re(zw^),|z+w|^2-|z-w|^2=4(ac+bd)=4Re(zŵ),

a,b,c,dRa,b,c,d\in \R

Therefore


z+w2zw2=4Re(zw^)|z+w|^2-|z-w|^2=4Re(zŵ)


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