If n is a positive integer,then (1+i sqrt{3}) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $z = 1 \pm i \sqrt{3}$. We can write $z$ in polar form as $2(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3})$.Then,$(1+i \sqrt{3})^n + (1-i \sqrt

Vedclass · Accepted Answer

$2^{n+1} \cos \frac{n \pi}{3}$

Vedclass · Answer

(C) Let $z = 1 \pm i \sqrt{3}$. We can write $z$ in polar form as $2(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3})$.Then,$(1+i \sqrt{3})^n + (1-i \sqrt{3})^n = [2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})]^n + [2(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})]^n$.Using De Moivre's Theorem,this becomes $2^n(\cos \frac{n \pi}{3} + i \sin \frac{n \pi}{3}) + 2^n(\cos \frac{n \pi}{3} - i \sin \frac{n \pi}{3})$.Simplifying the expression,we get $2^n(2 \cos \frac{n \pi}{3}) = 2^{n+1} \cos \frac{n \pi}{3}$.

If $n$ is a positive integer,then $(1+i \sqrt{3})^n+(1-i \sqrt{3})^n$ is equal to

Similar Questions

If $2 \alpha = -1 - i \sqrt{3}$ and $2 \beta = -1 + i \sqrt{3}$,then $5 \alpha^4 + 5 \beta^4 + 7 \alpha^{-1} \beta^{-1}$ is equal to

The least positive integer $n$ for which $\left( \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} \right)^n = 1$ is?

$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020} =$

Let $\alpha$ be a root of the equation $1+x^{2}+x^{4}=0$. Then the value of $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ is equal to

Consider a regular $10$-gon with its vertices on the unit circle. With one vertex fixed,draw straight lines to the other $9$ vertices. Call them $L_1, L_2, \ldots, L_9$ and denote their lengths by $l_1, l_2, \ldots, l_9$ respectively. Then,the product $l_1 \times l_2 \times \ldots \times l_9$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module