Question

Prove that
(−1 + i√3)^n + (−1−i√3)^n has either the value 2^(n+1) or the value −2^n if n is any integer (positive, negative or zero).

Answer 1

Given question is

$(-1+i\sqrt3)^n + (-1-i\sqrt3)^n\\or,\bigg(\frac{2(-1+i\sqrt3)}{2}\bigg)^n + \bigg(\frac{2(-1-i\sqrt3)}{2}\bigg)^n\\or, 2^n\bigg(\frac{-1}{2} + i\frac{\sqrt3}{2}\bigg)^n + 2^n\bigg(\frac{-1}{2} - i\frac{\sqrt3}{2}\bigg)^n\\or, 2^n\bigg(cos\frac{2\pi}{3} + i \space sin\frac{2\pi}{3}\bigg)^n + 2^n\bigg(cos\frac{2\pi}{3} - i \space sin\frac{2\pi}{3}\bigg)^n\\ Now \space using \space DeMoivre's \space Theorem \space which \space says \space that \\(cos\theta +i \space sin\theta)^n = (cosn\theta + i \space sin n \theta) \space , n\in I$

$or, 2^n\bigg(cos\frac{2n\pi}{3}+i\space sin\frac{2n\pi}{3}\bigg)+ 2^n\bigg(cos\frac{2n\pi}{3}-i\space sin\frac{2n\pi}{3}\bigg)\\or, 2^n\bigg(cos\frac{2n\pi}{3}+i\space sin\frac{2n\pi}{3}+ cos\frac{2n\pi}{3}-i\space sin\frac{2n\pi}{3}\bigg)\\or, 2^n \bigg(2cos \frac{2n\pi}{3}\bigg)\\or, 2^{n+1}cos\frac{2n\pi}{3}$

Noe there are two cases:

Case 1: when n is a multiple of 3 (whether odd integer (like 3,9,15...) or even integer (like 6,12,18....)

Case 2: when n is not a multiple of 3 (like 1,2,4,5,7,.....)

Case 1: When n is a multiple of 3 as discussed above ..in this case the value of $cos\frac{2n\pi}{3}$ is always 1.

$so \space 2^{n+1}cos\frac{2n\pi}{3}= 2^{n+1}*1 = 2^{n+1}. (Proved)$

Case 2: When n is not a multiple of 3..in this case for all those values of n whether positive or negative $cos\frac{2n\pi}{3}$ will always lies either in 2^nd quadrant or in 3^rd quadrant and in these two quadrants cosine is negative and value of $cos\frac{2n\pi}{3}$ is always $-\frac{1}{2}$ .

Moreover in case of negative integers there will be no effect as cosine absorbs negative sign.

$so \space 2^{n+1}*cos\frac{2n\pi}{3}=2^{n+1}*(-\frac{1}{2})$

$= -2^{(n+1-1)}$

$= -2^n (Proved).$