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Question Number 77991 by peter frank last updated on 12/Jan/20

If P_1 P_2 P_3 will be taken as point in an Argand diagram representing complex number Z_1 ,Z_2 ,Z_3 and point P_(1 ) ,P_2 ,P_3 is an equalateral triangle.show that (Z_2 −Z_3 )^2 +(Z_3 −Z_1 )^2 +(Z_1 −Z_2 )^2 =0

$${If}\:\:{P}_{\mathrm{1}} \:\:{P}_{\mathrm{2}} \:\:{P}_{\mathrm{3}} \:\:{will}\:{be}\:{taken} \\ $$$${as}\:{point}\:{in}\:{an}\:{Argand} \\ $$$${diagram}\:{representing} \\ $$$${complex}\:{number} \\ $$$${Z}_{\mathrm{1}} ,{Z}_{\mathrm{2}} ,{Z}_{\mathrm{3}} \:\:{and}\:{point} \\ $$$${P}_{\mathrm{1}\:} ,{P}_{\mathrm{2}} ,{P}_{\mathrm{3}} \:{is}\:{an}\:{equalateral} \\ $$$${triangle}.{show}\:{that} \\ $$$$\left({Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by mind is power last updated on 13/Jan/20

the equality is symetrice in sens of P(z_1 ,z_2 ,z_3 )=(z_1 −z_2 )^2 +(z_2 −z_3 )^2 +(z_1 −z_3 )^2 =p(z_1 ,z_3 ,z_2 )..... P_1 P_2 P_3 equalateral ⇒((z_3 −z_2 )/(z_3 −z_1 ))=e^((iπ)/3) ,we can chose this cause of the symetri ⇒z_3 =((z_2 −z_1 e^((iπ)/3) )/(1−e^((iπ)/3) ))=(z_2 −z_1 e^((iπ)/3) )e^((iπ)/3) z_3 =z_2 e^((iπ)/3) −z_1 e^((2iπ)/3) z_3 −z_2 =(z_2 −z_1 )e^((2iπ)/3) z_3 −z_1 =(z_2 −z_1 )e^((iπ)/3) (z_3 −z_2 )^2 +(z_3 −z_1 )^2 +(z_1 −z_2 )^2 =(z_2 −z_1 )^2 e^((4iπ)/3) +(z_2 −z_1 )^2 e^((2iπ)/3) +(z_2 −z_1 )^2 letj=e^((2iπ)/3) =(z_2 −z_1 )^2 (1+j+j^2 ) since j^2 +j+1=0 ⇒(z_3 −z_2 )^2 +(z_3 −z_1 )^2 +(z_2 −z_1 )^2 =0

$$\mathrm{the}\:\mathrm{equality}\:\mathrm{is}\:\mathrm{symetrice}\:\mathrm{in}\:\mathrm{sens}\:\mathrm{of}\: \\ $$$$\mathrm{P}\left(\mathrm{z}_{\mathrm{1}} ,\mathrm{z}_{\mathrm{2}} ,\mathrm{z}_{\mathrm{3}} \right)=\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{3}} \right)^{\mathrm{2}} +\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{p}\left(\mathrm{z}_{\mathrm{1}} ,\mathrm{z}_{\mathrm{3}} ,\mathrm{z}_{\mathrm{2}} \right)..... \\ $$$$\mathrm{P}_{\mathrm{1}} \mathrm{P}_{\mathrm{2}} \mathrm{P}_{\mathrm{3}} \:\:\mathrm{equalateral}\:\Rightarrow\frac{\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{2}} }{\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{1}} }=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} ,\mathrm{we}\:\mathrm{can}\:\mathrm{chose}\:\mathrm{this}\: \\ $$$$\mathrm{cause}\:\mathrm{of}\:\mathrm{the}\:\mathrm{symetri}\:\:\:\: \\ $$$$\Rightarrow\mathrm{z}_{\mathrm{3}} =\frac{\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }{\mathrm{1}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} }=\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{3}} =\mathrm{z}_{\mathrm{2}} \mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} −\mathrm{z}_{\mathrm{1}} \mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{2}} =\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{1}} =\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\left(\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} \mathrm{e}^{\frac{\mathrm{4i}\pi}{\mathrm{3}}} +\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} \mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} +\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\mathrm{letj}=\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$=\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{j}+\mathrm{j}^{\mathrm{2}} \right) \\ $$$$\mathrm{since}\:\mathrm{j}^{\mathrm{2}} +\mathrm{j}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{z}_{\mathrm{3}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$

Answered by MJS last updated on 13/Jan/20

start with P_(j+1) ^� = (((acos (α+((2π)/3)j))),((asin (α+((2π)/3)j))) ) ; j=0, 1, 2 obviously this gives an equilateral triangle for any value of α now shift by s^→ = ((p),(q) ) ⇒ P_(j+1) =P_(j+1) ^� +s^→ = (((p+acos (α+((2π)/3)j))),((q+asin (α+((2π)/3)j))) ) ; j=0, 1, 2 ⇒ Z_1 =p+acos α +i(q+asin α) Z_2 =p+acos (α+((2π)/3)) +i(q+asin (α+((2π)/3))) Z_3 =p+acos (α+((4π)/3)) +i(q+asin (α+((4π)/3))) cos (α+((2π)/3)) =−((cos α +(√3)sin α)/2) sin (α+((2π)/3)) =(((√3)cos α −sin α)/2) cos (α+((4π)/3)) =−((cos α −(√3)sin α)/2) sin (α+((4π)/3)) =−(((√3)cos α +sin α)/2) let u=cos α ∧v=sin α Z_1 =p+au+i(q+av) Z_2 =p−(a/2)(u+(√3)v)+i(q+(a/2)((√3)u−v)) Z_3 =p−(a/2)(u−(√3)v)+i(q−(a/2)((√3)u+v)) (Z_1 −Z_2 )^2 =((((√3)a)/2)(((√3)u+v)−i(u−(√3)v)))^2 = =((3a^2 )/2)((u^2 +2(√3)uv−v^2 )−i((√3)u^2 −2uv−(√3)v^2 )) (Z_1 −Z_3 )^2 =((((√3)a)/2)(((√3)u−v)+i(u+(√3)v)))^2 = =((3a^2 )/2)((u^2 −2(√3)uv−v^2 )+i((√3)u^2 +2uv−(√3)v^2 )) (Z_2 −Z_3 )^2 =((√3)a(−v+iu))^2 = =3a^2 ((v^2 −u^2 )−2iuv) adding them gives 0

$$\mathrm{start}\:\mathrm{with} \\ $$$$\bar {{P}}_{{j}+\mathrm{1}} =\begin{pmatrix}{{a}\mathrm{cos}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}{j}\right)}\\{{a}\mathrm{sin}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}{j}\right)}\end{pmatrix}\:;\:{j}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\mathrm{obviously}\:\mathrm{this}\:\mathrm{gives}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{value}\:\mathrm{of}\:\alpha \\ $$$$\mathrm{now}\:\mathrm{shift}\:\mathrm{by}\:\overset{\rightarrow} {{s}}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix} \\ $$$$\Rightarrow\:{P}_{{j}+\mathrm{1}} =\bar {{P}}_{{j}+\mathrm{1}} +\overset{\rightarrow} {{s}}=\begin{pmatrix}{{p}+{a}\mathrm{cos}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}{j}\right)}\\{{q}+{a}\mathrm{sin}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}{j}\right)}\end{pmatrix}\:;\:{j}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\Rightarrow \\ $$$${Z}_{\mathrm{1}} ={p}+{a}\mathrm{cos}\:\alpha\:+\mathrm{i}\left({q}+{a}\mathrm{sin}\:\alpha\right) \\ $$$${Z}_{\mathrm{2}} ={p}+{a}\mathrm{cos}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:+\mathrm{i}\left({q}+{a}\mathrm{sin}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right) \\ $$$${Z}_{\mathrm{3}} ={p}+{a}\mathrm{cos}\:\left(\alpha+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)\:+\mathrm{i}\left({q}+{a}\mathrm{sin}\:\left(\alpha+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)\right) \\ $$$$\mathrm{cos}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=−\frac{\mathrm{cos}\:\alpha\:+\sqrt{\mathrm{3}}\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha\:−\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\left(\alpha+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)\:=−\frac{\mathrm{cos}\:\alpha\:−\sqrt{\mathrm{3}}\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\left(\alpha+\frac{\mathrm{4}\pi}{\mathrm{3}}\right)\:=−\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha\:+\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\mathrm{let}\:{u}=\mathrm{cos}\:\alpha\:\wedge{v}=\mathrm{sin}\:\alpha \\ $$$${Z}_{\mathrm{1}} ={p}+{au}+\mathrm{i}\left({q}+{av}\right) \\ $$$${Z}_{\mathrm{2}} ={p}−\frac{{a}}{\mathrm{2}}\left({u}+\sqrt{\mathrm{3}}{v}\right)+\mathrm{i}\left({q}+\frac{{a}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}{u}−{v}\right)\right) \\ $$$${Z}_{\mathrm{3}} ={p}−\frac{{a}}{\mathrm{2}}\left({u}−\sqrt{\mathrm{3}}{v}\right)+\mathrm{i}\left({q}−\frac{{a}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}{u}+{v}\right)\right) \\ $$$$\left({Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \right)^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\left(\left(\sqrt{\mathrm{3}}{u}+{v}\right)−\mathrm{i}\left({u}−\sqrt{\mathrm{3}}{v}\right)\right)\right)^{\mathrm{2}} = \\ $$$$=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{2}}\left(\left({u}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}{uv}−{v}^{\mathrm{2}} \right)−\mathrm{i}\left(\sqrt{\mathrm{3}}{u}^{\mathrm{2}} −\mathrm{2}{uv}−\sqrt{\mathrm{3}}{v}^{\mathrm{2}} \right)\right) \\ $$$$\left({Z}_{\mathrm{1}} −{Z}_{\mathrm{3}} \right)^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\left(\left(\sqrt{\mathrm{3}}{u}−{v}\right)+\mathrm{i}\left({u}+\sqrt{\mathrm{3}}{v}\right)\right)\right)^{\mathrm{2}} = \\ $$$$=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{2}}\left(\left({u}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{uv}−{v}^{\mathrm{2}} \right)+\mathrm{i}\left(\sqrt{\mathrm{3}}{u}^{\mathrm{2}} +\mathrm{2}{uv}−\sqrt{\mathrm{3}}{v}^{\mathrm{2}} \right)\right) \\ $$$$\left({Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}{a}\left(−{v}+\mathrm{i}{u}\right)\right)^{\mathrm{2}} = \\ $$$$=\mathrm{3}{a}^{\mathrm{2}} \left(\left({v}^{\mathrm{2}} −{u}^{\mathrm{2}} \right)−\mathrm{2i}{uv}\right) \\ $$$$\mathrm{adding}\:\mathrm{them}\:\mathrm{gives}\:\mathrm{0} \\ $$

Commented by mind is power last updated on 13/Jan/20

hello sir Mjs its been long times hope all going good for you

$$\mathrm{hello}\:\mathrm{sir}\:\mathrm{Mjs}\:\mathrm{its}\:\mathrm{been}\:\mathrm{long}\:\mathrm{times} \\ $$$$\mathrm{hope}\:\mathrm{all}\:\mathrm{going}\:\mathrm{good}\:\mathrm{for}\:\mathrm{you} \\ $$$$ \\ $$

Commented by MJS last updated on 13/Jan/20

thank you, I hope 2020 is going to be a good year for you too

$$\mathrm{thank}\:\mathrm{you},\:\mathrm{I}\:\mathrm{hope}\:\mathrm{2020}\:\mathrm{is}\:\mathrm{going}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{good} \\ $$$$\mathrm{year}\:\mathrm{for}\:\mathrm{you}\:\mathrm{too} \\ $$

Commented by mind is power last updated on 13/Jan/20

thanx sir

$$\mathrm{thanx}\:\mathrm{sir}\:\: \\ $$

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