Q113 / 200 · WBJEE 2014 · Mathematics

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

  • A
    $z_{1}^{2}$
  • Correct answer
    B
    $z_{1}^{3}$
  • C
    $z_{1}^{4}$
  • D
    $z_{1}$

Solution

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 $$ z_{3}=z_{1} \omega^{2} $$ $$ \begin{aligned} \therefore \quad z, z_{2} z_{3} &=z_{1} \times z_{1} \omega \times z_{1} \omega^{2} \\ &=z_{1}^{3} \omega^{3} \\ &=z_{1}^{3} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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