If-z-1-z-2-and-z-3-are-the-vertices-of-an-equilateral-triangle-described-in-counterclock-sense-and-w-1-is-a-cube-root-of-unity-then-A-z-1-z-3-z-3-z-2-w-B-z-1-z-2-w-z-3-w-2-0-C-1-z

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Question Number 140107 by EnterUsername last updated on 04/May/21

If z_1 , z_2 , and z_3 are the vertices of an equilateral triangle described in counterclock sense and w≠1 is a cube root of unity, then (A) z_1 −z_3 =(z_3 −z_2 )w (B) z_1 +z_2 w+z_3 w^2 =0 (C) (1/(z_1 −z_2 ))+(1/(z_2 −z_3 ))+(1/(z_3 −z_1 ))=0 (D) z_1 ^2 +z_2 ^2 +z_3 ^2 =z_1 z_2 +z_2 z_3 +z_3 z_1

$$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{counterclock}\:\mathrm{sense}\:\mathrm{and}\:{w}\neq\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{unity},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:{z}_{\mathrm{1}} −{z}_{\mathrm{3}} =\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right){w} \\ $$$$\left(\mathrm{B}\right)\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} {w}+{z}_{\mathrm{3}} {w}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }+\frac{\mathrm{1}}{{z}_{\mathrm{2}} −{z}_{\mathrm{3}} }+\frac{\mathrm{1}}{{z}_{\mathrm{3}} −{z}_{\mathrm{1}} }=\mathrm{0} \\ $$$$\left(\mathrm{D}\right)\:{z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{3}} ^{\mathrm{2}} ={z}_{\mathrm{1}} {z}_{\mathrm{2}} +{z}_{\mathrm{2}} {z}_{\mathrm{3}} +{z}_{\mathrm{3}} {z}_{\mathrm{1}} \\ $$

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If-z-1-z-2-and-z-3-are-the-vertices-of-an-equilateral-triangle-described-in-counterclock-sense-and-w-1-is-a-cube-root-of-unity-then-A-z-1-z-3-z-3-z-2-w-B-z-1-z-2-w-z-3-w-2-0-C-1-z

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