Test Index

Complex Numbers

Section: Mathematics

© examsnet.com

Question : 2 of 50

Marks: +1, -0

If

(\sqrt{3}-i)^{n}=2^{n}, n \in \mathbb{N}

, then the least possible value of n is

[AP EAPCET 24 May 2025 S1]

3
4
6
12

Solution:

Given,

(\sqrt{3}-i)^n, n \in \mathbb{N}

\;\;\text{Modulus}\;|\sqrt{3}-1|\;=\sqrt{(\sqrt{3})^2+(-1)^2}

\;=\sqrt{3+1}=\sqrt{4}=2

\;\theta=\arg(\sqrt{3}-i)

\;=\tan^{-1}\left(\;\frac{-1}{\sqrt{3}}\right)=\;-\frac{\pi}{6}

Argument :

\theta=\arg(\sqrt{3}-i)=\tan^{-1}\left(\;\frac{-1}{\sqrt{3}}\right)=\;-\frac{\pi}{6}

So, in polar form,

\;\;\sqrt{3}-i=2\cdot(\cos\theta+i\sin\;\theta)

\;\;=2\left[\cos\left(\;-\frac{\pi}{6}\right)+\sin\left(\;-\frac{\pi}{6}\right)\right]

\;\;=2\left[\cos\;\frac{\pi}{6}-i\sin\;\;\frac{\pi}{6}\right]

\;\Rightarrow(\sqrt{3}-i)^n=\left\{2\cdot\left[\cos\left(\;\frac{\pi}{6}\right)-i\sin\left(\;\frac{\pi}{6}\right)\right]\right\}^n

\;\;=2^n\cdot\left[\cos\left(\;\frac{n\pi}{6}\right)-i\sin\left(\;\frac{n\pi}{6}\right)\right].

But given that

(\sqrt{3}-1)^n=2^n

From Eqs. (i) and (ii), we get

\;2^n\left[\cos\left(\;\frac{n\pi}{6}\right)-i\sin\left(\;\frac{n\pi}{6}\right)\right]=2^n

\Rightarrow\cos\left(\;\frac{n\pi}{6}\right)-i\sin\left(\;\frac{n\pi}{6}\right)=1

\Rightarrow\cos\left(\;\frac{n\pi}{6}\right)=1\;\text{ and }\;\sin\left(-\;\frac{n\pi}{6}\right)=1

\Rightarrow-\;\frac{n\pi}{6}=2k\pi

\Rightarrow n=\;\frac{-12k\pi}{\pi}=-12k

Since,

n \in \mathbb{N}

, for the smallest positive

n

, let

k=-1

\Rightarrow\;\;n=-12(-1)=12

∴ The least positive value of

n

is 12 .

© examsnet.com

Go to Question:

Prev Question

Next Question