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


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

$I f P_{1} P_{2} P_{3} w i l l b e t a k e n$ $a s p o i n t i n a n A r g a n d$ $d i a g r a m r e p r e s e n t i n g$ $c o m p l e x n u m b e r$ $Z_{1}, Z_{2}, Z_{3} a n d p o i n t$ $P_{1}, P_{2}, P_{3} i s a n e q u a l a t e r a l$ $t r i a n g l e . s h o w t h a t$ ${(Z_{2} - Z_{3})}^{2} + {(Z_{3} - Z_{1})}^{2} + {(Z_{1} - Z_{2})}^{2} = 0$
	Answered by mind is power last updated on 13/Jan/20

	$the equality is symetrice in sens of$ $P (z_{1}, z_{2}, z_{3}) = {(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{2} + {(z_{1} - z_{3})}^{2} = p (z_{1}, z_{3}, z_{2}) . . . . .$ $P_{1} P_{2} P_{3} equalateral \Rightarrow \frac{z_{3} - z_{2}}{z_{3} - z_{1}} = e^{\frac{i π}{3}}, we can chose this$ $cause of the symetri$ $\Rightarrow z_{3} = \frac{z_{2} - z_{1} e^{\frac{i π}{3}}}{1 - e^{\frac{i π}{3}}} = (z_{2} - z_{1} e^{\frac{i π}{3}}) e^{\frac{i π}{3}}$ $z_{3} = z_{2} e^{\frac{i π}{3}} - z_{1} e^{\frac{2 i π}{3}}$ $z_{3} - z_{2} = (z_{2} - z_{1}) e^{\frac{2 i π}{3}}$ $z_{3} - z_{1} = (z_{2} - z_{1}) e^{\frac{i π}{3}}$ ${(z_{3} - z_{2})}^{2} + {(z_{3} - z_{1})}^{2} + {(z_{1} - z_{2})}^{2} = {(z_{2} - z_{1})}^{2} e^{\frac{4 i π}{3}} + {(z_{2} - z_{1})}^{2} e^{\frac{2 i π}{3}} + {(z_{2} - z_{1})}^{2}$ $letj = e^{\frac{2 i π}{3}}$ $= {(z_{2} - z_{1})}^{2} (1 + j + j^{2})$ $since j^{2} + j + 1 = 0$ $\Rightarrow {(z_{3} - z_{2})}^{2} + {(z_{3} - z_{1})}^{2} + {(z_{2} - z_{1})}^{2} = 0$
	Answered by MJS last updated on 13/Jan/20

	$start with$ ${\bar{P}}_{j + 1} = (\begin{matrix} a \cos (α + \frac{2 π}{3} j) \\ a \sin (α + \frac{2 π}{3} j) \end{matrix}); j = 0, 1, 2$ $obviously this gives an equilateral triangle$ $for any value of α$ $now shift by \vec{s} = (\begin{matrix} p \\ q \end{matrix})$ $\Rightarrow P_{j + 1} = {\bar{P}}_{j + 1} + \vec{s} = (\begin{matrix} p + a \cos (α + \frac{2 π}{3} j) \\ q + a \sin (α + \frac{2 π}{3} j) \end{matrix}); j = 0, 1, 2$ $\Rightarrow$ $Z_{1} = p + a \cos α + i (q + a \sin α)$ $Z_{2} = p + a \cos (α + \frac{2 π}{3}) + i (q + a \sin (α + \frac{2 π}{3}))$ $Z_{3} = p + a \cos (α + \frac{4 π}{3}) + i (q + a \sin (α + \frac{4 π}{3}))$ $\cos (α + \frac{2 π}{3}) = - \frac{\cos α + \sqrt{3} \sin α}{2}$ $\sin (α + \frac{2 π}{3}) = \frac{\sqrt{3} \cos α - \sin α}{2}$ $\cos (α + \frac{4 π}{3}) = - \frac{\cos α - \sqrt{3} \sin α}{2}$ $\sin (α + \frac{4 π}{3}) = - \frac{\sqrt{3} \cos α + \sin α}{2}$ $let u = \cos α \land v = \sin α$ $Z_{1} = p + a u + i (q + a v)$ $Z_{2} = p - \frac{a}{2} (u + \sqrt{3} v) + i (q + \frac{a}{2} (\sqrt{3} u - v))$ $Z_{3} = p - \frac{a}{2} (u - \sqrt{3} v) + i (q - \frac{a}{2} (\sqrt{3} u + v))$ ${(Z_{1} - Z_{2})}^{2} = {(\frac{\sqrt{3} a}{2} ((\sqrt{3} u + v) - i (u - \sqrt{3} v)))}^{2} =$ $= \frac{3 a^{2}}{2} ((u^{2} + 2 \sqrt{3} u v - v^{2}) - i (\sqrt{3} u^{2} - 2 u v - \sqrt{3} v^{2}))$ ${(Z_{1} - Z_{3})}^{2} = {(\frac{\sqrt{3} a}{2} ((\sqrt{3} u - v) + i (u + \sqrt{3} v)))}^{2} =$ $= \frac{3 a^{2}}{2} ((u^{2} - 2 \sqrt{3} u v - v^{2}) + i (\sqrt{3} u^{2} + 2 u v - \sqrt{3} v^{2}))$ ${(Z_{2} - Z_{3})}^{2} = {(\sqrt{3} a (- v + i u))}^{2} =$ $= 3 a^{2} ((v^{2} - u^{2}) - 2 i u v)$ $adding them gives 0$
	Commented by mind is power last updated on 13/Jan/20

	$hello sir Mjs its been long times$ $hope all going good for you$
	Commented by MJS last updated on 13/Jan/20

	$thank you, I hope 2020 is going to be a good$ $year for you too$
	Commented by mind is power last updated on 13/Jan/20

	$thanx sir$
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