WBJEE · Maths · Complex Number
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 \(z_{1}^{2}\)
- B \(z_{1}^{3}\)
- C \(z_{1}^{4}\)
- D \(z_{1}\)
Answer & Solution
Correct Answer
(B) \(z_{1}^{3}\)
Step-by-step Solution
Detailed explanation
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 \[ z_{3}=z_{1} \omega^{2} \]…
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