Question

A value of n such that ${\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^n=1$ is

• A. $12$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A

$12$

Explanation for correct answer:

Finding the value of n.

Step-1: Simplify the given data.

Given, ${\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)}^n=1$

$\Rightarrow$ $\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)}^n$

Step-2: Apply formula De-Moivre's Theorem.

${\left(\cos \left(\theta \right)+i\sin \left(\theta \right)\right)}^n=\cos \left(n\theta \right)+i\sin \left(n\theta \right)$

$\Rightarrow$ $\mathrm{L}.\mathrm{H}.\mathrm{S}=\left(\cos \left(n\frac{\pi }{6}\right)+i\sin \left(n\frac{\pi }{6}\right)\right)$

Put, $n=12$

$\Rightarrow$ $\mathrm{L}.\mathrm{H}.\mathrm{S}=\left(\cos \left(2\pi \right)+i\sin \left(2\pi \right)\right)$

$\Rightarrow$ $\mathrm{L}.\mathrm{H}.\mathrm{S}=1=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Therefore, the value of $n$ is $12$.

Hence, option $\left(A\right)$ is correct.