A value of $n$ such that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$ is
A value of $n$ such that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$ is
$12$
$3$
$2$
$1$
Solution
Given,
$\begin{aligned} & \left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1 \\ & \left(\operatorname{cis} \frac{\pi}{6}\right)^n=1\end{aligned}$
$\therefore$ Only $n=12$ satisfies this equation.
Alternative Solution:
Sure, let's break it down.
The given complex number can be written in the polar form. The magnitude of the complex number $\frac{\sqrt{3}}{2} + \frac{i}{2}$ is $1$ and the argument is $\frac{\pi}{6}$, because $\frac{\sqrt{3}}{2}$ is the cosine and $\frac{i}{2}$ is the sine of $\frac{\pi}{6}$.
So, the complex number is $1(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}))$ which can be written as $cis(\frac{\pi}{6})$.
Now, $\left(cis(\frac{\pi}{6})\right)^n = cis(n\frac{\pi}{6})$.
We want this to be equal to $1$. A complex number is $1$ if its magnitude is $1$ and argument is $2\pi$ times an integer.
As the magnitude is already $1$, we set $n\frac{\pi}{6}$ equal to $2\pi k$ where $k$ is an integer. Simplifying this gives $n = 12k$.
So $n$ has to be a multiple of $12$. Out of the given options, only $12$ satisfies this condition. Hence, the answer is (A) $12$.