In the complex plane ,the vertices of an equilateral triangle are represented by the complex numbers $z_{1}, z_{2}$ $z_1,z_2$ and $z_{3}$ $z_3$ ,prove that $\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$ ?

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In the complex plane ,the vertices of an equilateral triangle are represented by the complex numbers $z_{1}, z_{2}$ $z_1,z_2$ and $z_{3}$ $z_3$ ,prove that $\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$ ?

$\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$

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Answer 1 · 2018-01-03T16:59:09

See below.

Explanation:

Calling $u = z_{1} - z_{2}$ $u = z_1-z_2$ with $z_{1}, z_{2}$ $z_1,z_2$ vertices of an equilateral triangle, the other two sides can be represented as

$v = u e^{i ϕ}$ $v = u e^(i phi)$
$w = u e^{2 i ϕ}$ $w = u e^(2i phi)$ with $ϕ = \pm \frac{2}{3} π$ $phi =+-2/3pi$ (triangle can be reflected)
![https://www.geogebra.org/geometry]( useruploads.socratic.org )
Now,

$\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = \frac{1}{z_{1} - z_{2}} (1 + e^{- i ϕ} + e^{- 2 i ϕ}) = \frac{1}{z_{1} - z_{2}} (\frac{e^{- 3 i ϕ} - 1}{e^{- i ϕ} - 1})$ $1/u+1/v+1/w = 1/(z_1-z_2)(1+e^(-iphi) + e^(-2i phi)) = 1/(z_1-z_2)((e^(-3i phi)-1)/(e^(-i phi) -1))$

but the numerator

$e^{- 3 i ϕ} - 1 = e^{\pm 3 i \frac{2}{3} π} - 1 = 1 - 1 = 0$ $e^(-3iphi)-1 = e^(+-3i 2/3pi)-1 = 1 -1 = 0$ then finally

$\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = 0$ $1/u+1/v+1/w =0$

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In the complex plane ,the vertices of an equilateral triangle are represented by the complex numbers $z_{1}, z_{2}$ $z_1,z_2$ and $z_{3}$ $z_3$ ,prove that $\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$ ?

$\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$

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Explanation:

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In the complex plane ,the vertices of an equilateral triangle are represented by the complex numbers z1,z2z_1,z_2 and z3z_3 ,prove that 1z1−z2+1z2−z3+1z3−z1=01/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0?

1z1−z2+1z2−z3+1z3−z1=01/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0

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Explanation:

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In the complex plane ,the vertices of an equilateral triangle are represented by the complex numbers $z_{1}, z_{2}$ $z_1,z_2$ and $z_{3}$ $z_3$ ,prove that $\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$ ?

$\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$ $1/(z_1-z_2)+1/(z_2-z_3)+1/(z_3-z_1)=0$