Answer to Question #282430 in Complex Analysis for iqa

Question

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

Answer 1

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

Then

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

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

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

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

$$=4(ac+bd)$$

$$zŵ=(a+bi)(c-di)=ac+bd+(-ad+bc)i$$

$$Re(zŵ)=ac+bd$$

Hence

$$|z+w|^2-|z-w|^2=4(ac+bd)=4Re(zŵ),$$

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

Therefore

$$|z+w|^2-|z-w|^2=4Re(zŵ)$$