If z_{1} and z_{2} are two non-zero complex n | vedclass

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(C) We know that if $z_{1}, z_{2},$ and $z_{3}$ are the vertices of an equilateral triangle, then $z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1}z_{2}-z_{2}z_{3}

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form an equilateral triangle

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(C) We know that if $z_{1}, z_{2},$ and $z_{3}$ are the vertices of an equilateral triangle, then $z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}=0$. Given that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$, we can multiply by $z_{1}z_{2}$ to get $z_{1}^{2}+z_{2}^{2}=z_{1}z_{2}$, which implies $z_{1}^{2}+z_{2}^{2}-z_{1}z_{2}=0$. If we consider the origin as the third point $z_{3}=0$, the condition becomes $z_{1}^{2}+z_{2}^{2}+0^{2}-z_{1}z_{2}-z_{2}(0)-0(z_{1})=0$, which simplifies to $z_{1}^{2}+z_{2}^{2}-z_{1}z_{2}=0$. This matches the condition for an equilateral triangle. Thus, the origin and the points $z_{1}$ and $z_{2}$ form an equilateral triangle.

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$, then the origin and the points represented by $z_{1}$ and $z_{2}$:

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