Question Number 164473 by mathocean1 last updated on 17/Jan/22
$${Show}\:{for}\:{z}_{\mathrm{1}} ;\:{z}_{\mathrm{2}\:} \:\in\:\mathbb{C}\:{that}: \\ $$$$\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid^{\mathrm{2}} +\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\mathrm{2}\left(\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \right). \\ $$
Answered by puissant last updated on 19/Jan/22
$$\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid^{\mathrm{2}} +\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mathrm{2}\mid{z}_{\mathrm{1}} \mid\mid{z}_{\mathrm{2}} \mid+\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} +\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} −\mathrm{2}\mid{z}_{\mathrm{1}} \mid\mid{z}_{\mathrm{2}} \mid+\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left(\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \right) \\ $$
Commented by mathocean1 last updated on 19/Jan/22
$${thanks} \\ $$