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If z1, z2, z3...

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If z1, z2, z3 are three non-zero complex number such that z3=(1-x)z1 + x z2, where x belong to R-{0}, then determine the curve on which the points z1, z2.z3 lies.

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z_{1}, z_{2}, z_{3}

non-zero, non-collinear complex numbers such that

\frac{2}{z_{1}} = \frac{1}{z_{2}} + \frac{1}{z_{3}},

then the points

z_{1}, z_{2}, z_{3}

z_{1}, z_{2}, z_{3}

are three distinct complex numbers and

p, q, r

are three positive real numbers such that

\frac{p}{| z_{2} - z_{3} |} = \frac{q}{| z_{3} - z_{1} |} = \frac{r}{| z_{1} - z_{2} |}

\frac{p^{2}}{z_{2} - z_{3}} + \frac{q^{2}}{z_{3} - z_{1}} + \frac{r^{2}}{z_{1} - z_{2}} = 0

z_{1}, z_{2}, z_{3}

are non - zero complex such that

z_{1} + z_{2} + z_{3} = 0

z_{1}^{- 1} + z_{2}^{- 1} + z_{3}^{- 1} = 0

then prove that the given points are the verices of an equilateral triangle. Also show that

| z_{1} | = | z_{2} | = | z_{3} |

z_{1}, z_{2}, z_{3}

are complex numbers such that

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ ∣ = 1

| z_{1} + z_{2} + z_{3} |

z_{1,} z_{_{2},} z_{3}

are complex number such that

| z_{1} | = | z_{2} | =

| z_{3} |

∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ ∣ = 1,

| z_{1 +} z_{2} + z_{3} |

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