Answer to Question #124560 in Algebra for nana

Question #124560
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).
1
Expert's answer
2020-07-01T18:59:01-0400

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 2nd quadrant or in 3rd 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)."












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