If $x_r = \cos(\pi/3^r) - i\sin(\pi/3^r)$ (where $i = \sqrt{-1}$),then the value of $x_1 \cdot x_2 \cdot x_3 \cdots \infty$ is:

  • A
    $1$
  • B
    $-1$
  • C
    $-i$
  • D
    $i$

Explore More

Similar Questions

The equation whose solutions are the non-zero solutions of the equation $\bar{z}=i z^2$ is:

Find the complex number $z$ satisfying the equations $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.

Difficult
View Solution

Let $f(x) = x^4 + ax^3 + bx^2 + c$ be a polynomial with real coefficients such that $f(1) = -9$. Suppose that $i\sqrt{3}$ is a root of the equation $4x^3 + 3ax^2 + 2bx = 0$,where $i = \sqrt{-1}$. If $\alpha_1, \alpha_2, \alpha_3$,and $\alpha_4$ are all the roots of the equation $f(x) = 0$,then $|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2$ is equal to:

If the point $(x, y)$ satisfies the equation $\frac{x+i(x-2)}{3+i}-i=\frac{2y+i(1-3y)}{i-3}$,then $x+y=$

If $\omega$ is a complex cube root of unity,then $\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} +\sin \left(\sum_{j=1}^6(j+\omega)(j+\omega^2) \frac{3 \pi}{152}\right)=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo