Question 1207460
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a,b,c,d are real numbers
i = sqrt(-1) represents an imaginary number


z = a+bi
zBar = horizontal bar over "a+bi" = complex conjugate of z
zBar = a-bi
w = c+di
wBar = c-di


z*w = (a+bi)*(c+di)
z*w = a*(c+di)+bi*(c+di)
z*w = ac+adi+bci+bdi^2
z*w = ac+adi+bci+bd*(-1)
z*w = (ac-bd) + (ad+bc)i
(z*w)bar = (ac-bd) - (ad+bc)i
We'll return to this later.


zBar*wBar = (a-bi)*(c-di)
zBar*wBar = a*(c-di)-bi*(c-di)
zBar*wBar = ac-adi-bci+bdi^2
zBar*wBar = ac-adi-bci+bd*(-1)
zBar*wBar = ac-adi-bci-bd
zBar*wBar = (ac-bd)+(-adi-bci)
zBar*wBar = (ac-bd)-(ad+bc)i
This is an identical match with the conclusion of the previous paragraph.


Therefore we have shown that (z*w)bar = zBar*wBar is indeed the case.
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