Show-for-z-1-z-2-C-that-z-1-z-2-2-z-1-z-2-2-2-z-1-2-z-2-2-

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

$${thanks} \\ $$

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