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If $z_{1}$ and $z_{2}$ be 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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Verified Answer
The correct answer is:
form an equilateral triangle
We know that, if $z_{1}, z_{2}$ and $z_{3}$ are the vertices of an equilateral triangle. Then,
$$
z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1} z_{2}-z_{2} z_{3}-z_{3} z_{1}=0
$$
Now, but we have
$$
\begin{aligned}
\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}} &=1 \\
\Rightarrow \quad z_{1}^{2}+z_{2}^{2} &=z_{1} z_{2} \\
\Rightarrow \quad z_{1}^{2}+z_{2}^{2}-z_{1} z_{2} &=0 \\
z_{3} &=0
\end{aligned}
$$
Hence, given points form an equilateral triangle.
$$
z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1} z_{2}-z_{2} z_{3}-z_{3} z_{1}=0
$$
Now, but we have
$$
\begin{aligned}
\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}} &=1 \\
\Rightarrow \quad z_{1}^{2}+z_{2}^{2} &=z_{1} z_{2} \\
\Rightarrow \quad z_{1}^{2}+z_{2}^{2}-z_{1} z_{2} &=0 \\
z_{3} &=0
\end{aligned}
$$
Hence, given points form an equilateral triangle.
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