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Let $z_{1}$ be a fixed point on the circle of radius 1 centred at the origin in the argand plane and $z_{1} \neq \pm 1 .$ Consider an equilateral triangle inscribed in the circle with $z_{1}, z_{2}, z_{3}$ as the
direction. Then, $z_{1} z_{2} z_{3}$ is equal to
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direction. Then, $z_{1} z_{2} z_{3}$ is equal to
Solution:
2173 Upvotes
Verified Answer
The correct answer is:
$z_{1}^{3}$
Given, $z_{1} \neq\pm 1$
Since, $z_{2}$ and $z_{3}$ can be obtained by rotating
vector representing through $\frac{2 \pi}{3}$ and $\frac{4 \pi}{3}$
respectively
$\therefore \quad z_{2}=z_{1} \omega$
and
$$
z_{3}=z_{1} \omega^{2}
$$
$$
\begin{aligned}
\therefore \quad z, z_{2} z_{3} &=z_{1} \times z_{1} \omega \times z_{1} \omega^{2} \\
&=z_{1}^{3} \omega^{3} \\
&=z_{1}^{3}
\end{aligned}
$$
Since, $z_{2}$ and $z_{3}$ can be obtained by rotating
vector representing through $\frac{2 \pi}{3}$ and $\frac{4 \pi}{3}$
respectively
$\therefore \quad z_{2}=z_{1} \omega$
and
$$
z_{3}=z_{1} \omega^{2}
$$
$$
\begin{aligned}
\therefore \quad z, z_{2} z_{3} &=z_{1} \times z_{1} \omega \times z_{1} \omega^{2} \\
&=z_{1}^{3} \omega^{3} \\
&=z_{1}^{3}
\end{aligned}
$$
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