Answer to Question #124663 in Algebra for desmond

Question #124663
(a) Prove that
(−1 + i

3)n + (−1 − i

3)n
has either the value 2n+1 or the value −2
n
if n is any integer (positive, negative or zero).
1
Expert's answer
2020-07-05T17:35:15-0400

(-1+i√3)"^n" + (-1-i√3)"^n"

= 2"^n" ("\\frac{-1+i\u221a3}{2})^n" + 2"^n" ("\\frac{-1-i\u221a3}{2})^n"

= "2^n" "(\\frac{-1}{2}+\\frac{i\u221a3}{2})^n" + "2^n" "(\\frac{-1}{2}-\\frac{i\u221a3}{2})^n"

= "2^n" ( cos "\\frac{2\u03c0}{3}+i" sin"\\frac{2\u03c0}{3}" )"^n" + 2"^n" ( cos"\\frac{2\u03c0}{3}-i"sin"\\frac{2\u03c0}{3}" )"^n"

= "2^n" ( cos "\\frac{2n\u03c0}{3}+" isin"\\frac{2n\u03c0}{3}" ) + "2^n" ( cos"\\frac{2n\u03c0}{3}-" isin"\\frac{2n\u03c0}{3}" ) by D'Moivres theorem

= "2^n" . 2 cos"\\frac{2n\u03c0}{3}"

= "2^{n+1}" cos"\\frac{2n\u03c0}{3}"

Now as n is an integer, n can be any one of the forms 3k, 3k+1, 3k+2 where k is an integer.

When n = 3k

"cos \\frac{2n\u03c0}{3}" = "cos \\frac{2(3k)\u03c0}{3}" = cos 2kπ = 1

So "(-1+i\u221a3)^n+(-1-i\u221a3)^n"

= 2"^{n+1}"

When n = 3k+1

"cos \\frac{2n\u03c0}{3}" = "cos \\frac{2(3k+1)\u03c0}{3}" = cos(2kπ+"\\frac{2\u03c0}{3})" = cos"\\frac{2\u03c0}{3} = cos(\u03c0-\\frac{\u03c0}{3})= -cos\\frac{\u03c0}{3}" = "-\\frac{1}{2}"

So "(-1+i\u221a3)^n+(-1-i\u221a3)^n"

= 2"^{n+1}(-\\frac{1}{2})"

= "-2^n"

When n = 3k+2

"cos \\frac{2n\u03c0}{3}" = "cos \\frac{2(3k+2)\u03c0}{3}" = cos(2kπ +"\\frac{4\u03c0}{3})" = cos"\\frac{4\u03c0}{3}" = cos(π+"\\frac{\u03c0}{3}" )= -cos"\\frac{\u03c0}{3}" = "-\\frac{1}{2}"

So "(-1+i\u221a3)^n+(-1-i\u221a3)^n"

= "2^{n+1}(-\\frac{1}{2})"

= "-2^n"

Finally we can conclude that

"(-1+i\u221a3)^n+(-1-i\u221a3)^n"

= 2"^{n+1}" or "-" 2"^{n}" if n is an integer (positive, negative or zero)



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