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-

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}) \dots . .

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